Algorithm to Solve Sudoku | Sudoku Solver

Last Updated : 15 May, 2023

Given a partially filled 9×9 2D array ‘grid[9][9]’, the goal is to assign digits (from 1 to 9) to the empty cells so that every row, column, and subgrid of size 3×3 contains exactly one instance of the digits from 1 to 9.

Examples:

Input: grid
{ {3, 0, 6, 5, 0, 8, 4, 0, 0},
{5, 2, 0, 0, 0, 0, 0, 0, 0},
{0, 8, 7, 0, 0, 0, 0, 3, 1},
{0, 0, 3, 0, 1, 0, 0, 8, 0},
{9, 0, 0, 8, 6, 3, 0, 0, 5},
{0, 5, 0, 0, 9, 0, 6, 0, 0},
{1, 3, 0, 0, 0, 0, 2, 5, 0},
{0, 0, 0, 0, 0, 0, 0, 7, 4},
{0, 0, 5, 2, 0, 6, 3, 0, 0} }
Output:
3 1 6 5 7 8 4 9 2
5 2 9 1 3 4 7 6 8
4 8 7 6 2 9 5 3 1
2 6 3 4 1 5 9 8 7
9 7 4 8 6 3 1 2 5
8 5 1 7 9 2 6 4 3
1 3 8 9 4 7 2 5 6
6 9 2 3 5 1 8 7 4
7 4 5 2 8 6 3 1 9
Explanation: Each row, column and 3*3 box of the output matrix contains unique numbers.

Input: grid
{ { 3, 1, 6, 5, 7, 8, 4, 9, 2 },
{ 5, 2, 9, 1, 3, 4, 7, 6, 8 },
{ 4, 8, 7, 6, 2, 9, 5, 3, 1 },
{ 2, 6, 3, 0, 1, 5, 9, 8, 7 },
{ 9, 7, 4, 8, 6, 0, 1, 2, 5 },
{ 8, 5, 1, 7, 9, 2, 6, 4, 3 },
{ 1, 3, 8, 0, 4, 7, 2, 0, 6 },
{ 6, 9, 2, 3, 5, 1, 8, 7, 4 },
{ 7, 4, 5, 0, 8, 6, 3, 1, 0 } };
Output:
3 1 6 5 7 8 4 9 2
5 2 9 1 3 4 7 6 8
4 8 7 6 2 9 5 3 1
2 6 3 4 1 5 9 8 7
9 7 4 8 6 3 1 2 5
8 5 1 7 9 2 6 4 3
1 3 8 9 4 7 2 5 6
6 9 2 3 5 1 8 7 4
7 4 5 2 8 6 3 1 9
Explanation: Each row, column and 3*3 box of the output matrix contains unique numbers.

Recommended Practice

Naive Approach:

The naive approach is to generate all possible configurations of numbers from 1 to 9 to fill the empty cells. Try every configuration one by one until the correct configuration is found, i.e. for every unassigned position fill the position with a number from 1 to 9. After filling all the unassigned positions check if the matrix is safe or not. If safe print else recurs for other cases.

Follow the steps below to solve the problem:

• Create a function that checks if the given matrix is valid sudoku or not. Keep Hashmap for the row, column and boxes. If any number has a frequency greater than 1 in the hashMap return false else return true;
• Create a recursive function that takes a grid and the current row and column index.
• Check some base cases.
• If the index is at the end of the matrix, i.e. i=N-1 and j=N then check if the grid is safe or not, if safe print the grid and return true else return false.
• The other base case is when the value of column is N, i.e j = N, then move to next row, i.e. i++ and j = 0.
• If the current index is not assigned then fill the element from 1 to 9 and recur for all 9 cases with the index of next element, i.e. i, j+1. if the recursive call returns true then break the loop and return true.
• If the current index is assigned then call the recursive function with the index of the next element, i.e. i, j+1

Below is the implementation of the above approach:

C++

 #include   using namespace std;   // N is the size of the 2D matrix   N*N #define N 9   /* A utility function to print grid */ void print(int arr[N][N]) {     for (int i = 0; i < N; i++)     {         for (int j = 0; j < N; j++)             cout << arr[i][j] << " ";         cout << endl;     } }   // Checks whether it will be // legal to assign num to the // given row, col bool isSafe(int grid[N][N], int row,                        int col, int num) {           // Check if we find the same num     // in the similar row , we     // return false     for (int x = 0; x <= 8; x++)         if (grid[row][x] == num)             return false;       // Check if we find the same num in     // the similar column , we     // return false     for (int x = 0; x <= 8; x++)         if (grid[x][col] == num)             return false;       // Check if we find the same num in     // the particular 3*3 matrix,     // we return false     int startRow = row - row % 3,             startCol = col - col % 3;         for (int i = 0; i < 3; i++)         for (int j = 0; j < 3; j++)             if (grid[i + startRow][j +                             startCol] == num)                 return false;       return true; }   /* Takes a partially filled-in grid and attempts to assign values to all unassigned locations in such a way to meet the requirements for Sudoku solution (non-duplication across rows, columns, and boxes) */ bool solveSudoku(int grid[N][N], int row, int col) {     // Check if we have reached the 8th     // row and 9th column (0     // indexed matrix) , we are     // returning true to avoid     // further backtracking     if (row == N - 1 && col == N)         return true;       // Check if column value  becomes 9 ,     // we move to next row and     //  column start from 0     if (col == N) {         row++;         col = 0;     }         // Check if the current position of     // the grid already contains     // value >0, we iterate for next column     if (grid[row][col] > 0)         return solveSudoku(grid, row, col + 1);       for (int num = 1; num <= N; num++)     {                   // Check if it is safe to place         // the num (1-9)  in the         // given row ,col  ->we         // move to next column         if (isSafe(grid, row, col, num))         {                          /* Assigning the num in               the current (row,col)               position of the grid               and assuming our assigned               num in the position               is correct     */             grid[row][col] = num;                         //  Checking for next possibility with next             //  column             if (solveSudoku(grid, row, col + 1))                 return true;         }                 // Removing the assigned num ,         // since our assumption         // was wrong , and we go for         // next assumption with         // diff num value         grid[row][col] = 0;     }     return false; }   // Driver Code int main() {     // 0 means unassigned cells     int grid[N][N] = { { 3, 0, 6, 5, 0, 8, 4, 0, 0 },                        { 5, 2, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 8, 7, 0, 0, 0, 0, 3, 1 },                        { 0, 0, 3, 0, 1, 0, 0, 8, 0 },                        { 9, 0, 0, 8, 6, 3, 0, 0, 5 },                        { 0, 5, 0, 0, 9, 0, 6, 0, 0 },                        { 1, 3, 0, 0, 0, 0, 2, 5, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 7, 4 },                        { 0, 0, 5, 2, 0, 6, 3, 0, 0 } };       if (solveSudoku(grid, 0, 0))         print(grid);     else         cout << "no solution  exists " << endl;       return 0;     // This is code is contributed by Pradeep Mondal P }

C

 #include #include   // N is the size of the 2D matrix   N*N #define N 9   /* A utility function to print grid */ void print(int arr[N][N]) {      for (int i = 0; i < N; i++)       {          for (int j = 0; j < N; j++)             printf("%d ",arr[i][j]);          printf("\n");        } }   // Checks whether it will be legal  // to assign num to the // given row, col int isSafe(int grid[N][N], int row,                        int col, int num) {           // Check if we find the same num     // in the similar row , we return 0     for (int x = 0; x <= 8; x++)         if (grid[row][x] == num)             return 0;       // Check if we find the same num in the     // similar column , we return 0     for (int x = 0; x <= 8; x++)         if (grid[x][col] == num)             return 0;       // Check if we find the same num in the     // particular 3*3 matrix, we return 0     int startRow = row - row % 3,                  startCol = col - col % 3;         for (int i = 0; i < 3; i++)         for (int j = 0; j < 3; j++)             if (grid[i + startRow][j +                           startCol] == num)                 return 0;       return 1; }   /* Takes a partially filled-in grid and attempts to assign values to all unassigned locations in such a way to meet the requirements for Sudoku solution (non-duplication across rows, columns, and boxes) */ int solveSudoku(int grid[N][N], int row, int col) {           // Check if we have reached the 8th row     // and 9th column (0     // indexed matrix) , we are     // returning true to avoid     // further backtracking     if (row == N - 1 && col == N)         return 1;       //  Check if column value  becomes 9 ,     //  we move to next row and     //  column start from 0     if (col == N)     {         row++;         col = 0;     }         // Check if the current position     // of the grid already contains     // value >0, we iterate for next column     if (grid[row][col] > 0)         return solveSudoku(grid, row, col + 1);       for (int num = 1; num <= N; num++)     {                   // Check if it is safe to place         // the num (1-9)  in the         // given row ,col  ->we move to next column         if (isSafe(grid, row, col, num)==1)         {             /* assigning the num in the                current (row,col)                position of the grid                and assuming our assigned num                in the position                is correct     */             grid[row][col] = num;                         //  Checking for next possibility with next             //  column             if (solveSudoku(grid, row, col + 1)==1)                 return 1;         }                 // Removing the assigned num ,         // since our assumption         // was wrong , and we go for next         // assumption with         // diff num value         grid[row][col] = 0;     }     return 0; }   int main() {     // 0 means unassigned cells     int grid[N][N] = { { 3, 0, 6, 5, 0, 8, 4, 0, 0 },                        { 5, 2, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 8, 7, 0, 0, 0, 0, 3, 1 },                        { 0, 0, 3, 0, 1, 0, 0, 8, 0 },                        { 9, 0, 0, 8, 6, 3, 0, 0, 5 },                        { 0, 5, 0, 0, 9, 0, 6, 0, 0 },                        { 1, 3, 0, 0, 0, 0, 2, 5, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 7, 4 },                        { 0, 0, 5, 2, 0, 6, 3, 0, 0 } };       if (solveSudoku(grid, 0, 0)==1)         print(grid);     else         printf("No solution exists");       return 0;     // This is code is contributed by Pradeep Mondal P }

Java

 // Java program for above approach public class Sudoku {       // N is the size of the 2D matrix   N*N     static int N = 9;       /* Takes a partially filled-in grid and attempts     to assign values to all unassigned locations in     such a way to meet the requirements for     Sudoku solution (non-duplication across rows,     columns, and boxes) */     static boolean solveSudoku(int grid[][], int row,                                int col)     {           /*if we have reached the 8th            row and 9th column (0            indexed matrix) ,            we are returning true to avoid further            backtracking       */         if (row == N - 1 && col == N)             return true;           // Check if column value  becomes 9 ,         // we move to next row         // and column start from 0         if (col == N) {             row++;             col = 0;         }           // Check if the current position         // of the grid already         // contains value >0, we iterate         // for next column         if (grid[row][col] != 0)             return solveSudoku(grid, row, col + 1);           for (int num = 1; num < 10; num++) {               // Check if it is safe to place             // the num (1-9)  in the             // given row ,col ->we move to next column             if (isSafe(grid, row, col, num)) {                   /*  assigning the num in the current                 (row,col)  position of the grid and                 assuming our assigned num in the position                 is correct */                 grid[row][col] = num;                   // Checking for next                 // possibility with next column                 if (solveSudoku(grid, row, col + 1))                     return true;             }             /* removing the assigned num , since our                assumption was wrong , and we go for next                assumption with diff num value   */             grid[row][col] = 0;         }         return false;     }       /* A utility function to print grid */     static void print(int[][] grid)     {         for (int i = 0; i < N; i++) {             for (int j = 0; j < N; j++)                 System.out.print(grid[i][j] + " ");             System.out.println();         }     }       // Check whether it will be legal     // to assign num to the     // given row, col     static boolean isSafe(int[][] grid, int row, int col,                           int num)     {           // Check if we find the same num         // in the similar row , we         // return false         for (int x = 0; x <= 8; x++)             if (grid[row][x] == num)                 return false;           // Check if we find the same num         // in the similar column ,         // we return false         for (int x = 0; x <= 8; x++)             if (grid[x][col] == num)                 return false;           // Check if we find the same num         // in the particular 3*3         // matrix, we return false         int startRow = row - row % 3, startCol                                       = col - col % 3;         for (int i = 0; i < 3; i++)             for (int j = 0; j < 3; j++)                 if (grid[i + startRow][j + startCol] == num)                     return false;           return true;     }        // Driver Code     public static void main(String[] args)     {         int grid[][] = { { 3, 0, 6, 5, 0, 8, 4, 0, 0 },                          { 5, 2, 0, 0, 0, 0, 0, 0, 0 },                          { 0, 8, 7, 0, 0, 0, 0, 3, 1 },                          { 0, 0, 3, 0, 1, 0, 0, 8, 0 },                          { 9, 0, 0, 8, 6, 3, 0, 0, 5 },                          { 0, 5, 0, 0, 9, 0, 6, 0, 0 },                          { 1, 3, 0, 0, 0, 0, 2, 5, 0 },                          { 0, 0, 0, 0, 0, 0, 0, 7, 4 },                          { 0, 0, 5, 2, 0, 6, 3, 0, 0 } };           if (solveSudoku(grid, 0, 0))             print(grid);         else             System.out.println("No Solution exists");     }     // This is code is contributed by Pradeep Mondal P }

Python3

 # N is the size of the 2D matrix   N*N N = 9   # A utility function to print grid def printing(arr):     for i in range(N):         for j in range(N):             print(arr[i][j], end = " ")         print()   # Checks whether it will be # legal to assign num to the # given row, col def isSafe(grid, row, col, num):         # Check if we find the same num     # in the similar row , we     # return false     for x in range(9):         if grid[row][x] == num:             return False       # Check if we find the same num in     # the similar column , we     # return false     for x in range(9):         if grid[x][col] == num:             return False       # Check if we find the same num in     # the particular 3*3 matrix,     # we return false     startRow = row - row % 3     startCol = col - col % 3     for i in range(3):         for j in range(3):             if grid[i + startRow][j + startCol] == num:                 return False     return True   # Takes a partially filled-in grid and attempts # to assign values to all unassigned locations in # such a way to meet the requirements for # Sudoku solution (non-duplication across rows, # columns, and boxes) */ def solveSudoku(grid, row, col):         # Check if we have reached the 8th     # row and 9th column (0     # indexed matrix) , we are     # returning true to avoid     # further backtracking     if (row == N - 1 and col == N):         return True             # Check if column value  becomes 9 ,     # we move to next row and     # column start from 0     if col == N:         row += 1         col = 0       # Check if the current position of     # the grid already contains     # value >0, we iterate for next column     if grid[row][col] > 0:         return solveSudoku(grid, row, col + 1)     for num in range(1, N + 1, 1):                 # Check if it is safe to place         # the num (1-9)  in the         # given row ,col  ->we         # move to next column         if isSafe(grid, row, col, num):                         # Assigning the num in             # the current (row,col)             # position of the grid             # and assuming our assigned             # num in the position             # is correct             grid[row][col] = num               # Checking for next possibility with next             # column             if solveSudoku(grid, row, col + 1):                 return True           # Removing the assigned num ,         # since our assumption         # was wrong , and we go for         # next assumption with         # diff num value         grid[row][col] = 0     return False   # Driver Code   # 0 means unassigned cells grid = [[3, 0, 6, 5, 0, 8, 4, 0, 0],         [5, 2, 0, 0, 0, 0, 0, 0, 0],         [0, 8, 7, 0, 0, 0, 0, 3, 1],         [0, 0, 3, 0, 1, 0, 0, 8, 0],         [9, 0, 0, 8, 6, 3, 0, 0, 5],         [0, 5, 0, 0, 9, 0, 6, 0, 0],         [1, 3, 0, 0, 0, 0, 2, 5, 0],         [0, 0, 0, 0, 0, 0, 0, 7, 4],         [0, 0, 5, 2, 0, 6, 3, 0, 0]]   if (solveSudoku(grid, 0, 0)):     printing(grid) else:     print("no solution  exists ")       # This code is contributed by sudhanshgupta2019a

C#

 // C# program for above approach using System; class GFG {     // N is the size of the 2D matrix   N*N   static int N = 9;     /* Takes a partially filled-in grid and attempts     to assign values to all unassigned locations in     such a way to meet the requirements for     Sudoku solution (non-duplication across rows,     columns, and boxes) */   static bool solveSudoku(int[,] grid, int row,                           int col)   {       /*if we have reached the 8th            row and 9th column (0            indexed matrix) ,            we are returning true to avoid further            backtracking       */     if (row == N - 1 && col == N)       return true;       // Check if column value  becomes 9 ,     // we move to next row     // and column start from 0     if (col == N) {       row++;       col = 0;     }       // Check if the current position     // of the grid already     // contains value >0, we iterate     // for next column     if (grid[row,col] != 0)       return solveSudoku(grid, row, col + 1);       for (int num = 1; num < 10; num++) {         // Check if it is safe to place       // the num (1-9)  in the       // given row ,col ->we move to next column       if (isSafe(grid, row, col, num)) {           /*  assigning the num in the current                 (row,col)  position of the grid and                 assuming our assigned num in the position                 is correct */         grid[row,col] = num;           // Checking for next         // possibility with next column         if (solveSudoku(grid, row, col + 1))           return true;       }       /* removing the assigned num , since our                assumption was wrong , and we go for next                assumption with diff num value   */       grid[row,col] = 0;     }     return false;   }     /* A utility function to print grid */   static void print(int[,] grid)   {     for (int i = 0; i < N; i++) {       for (int j = 0; j < N; j++)         Console.Write(grid[i,j] + " ");       Console.WriteLine();     }   }     // Check whether it will be legal   // to assign num to the   // given row, col   static bool isSafe(int[,] grid, int row, int col,                      int num)   {       // Check if we find the same num     // in the similar row , we     // return false     for (int x = 0; x <= 8; x++)       if (grid[row,x] == num)         return false;       // Check if we find the same num     // in the similar column ,     // we return false     for (int x = 0; x <= 8; x++)       if (grid[x,col] == num)         return false;       // Check if we find the same num     // in the particular 3*3     // matrix, we return false     int startRow = row - row % 3, startCol       = col - col % 3;     for (int i = 0; i < 3; i++)       for (int j = 0; j < 3; j++)         if (grid[i + startRow,j + startCol] == num)           return false;       return true;   }     // Driver code   static void Main() {     int[,] grid = { { 3, 0, 6, 5, 0, 8, 4, 0, 0 },                    { 5, 2, 0, 0, 0, 0, 0, 0, 0 },                    { 0, 8, 7, 0, 0, 0, 0, 3, 1 },                    { 0, 0, 3, 0, 1, 0, 0, 8, 0 },                    { 9, 0, 0, 8, 6, 3, 0, 0, 5 },                    { 0, 5, 0, 0, 9, 0, 6, 0, 0 },                    { 1, 3, 0, 0, 0, 0, 2, 5, 0 },                    { 0, 0, 0, 0, 0, 0, 0, 7, 4 },                    { 0, 0, 5, 2, 0, 6, 3, 0, 0 } };       if (solveSudoku(grid, 0, 0))       print(grid);     else       Console.WriteLine("No Solution exists");   } }   // This code is contributed by divyesh072019.

Javascript



Output

3 1 6 5 7 8 4 9 2
5 2 9 1 3 4 7 6 8
4 8 7 6 2 9 5 3 1
2 6 3 4 1 5 9 8 7
9 7 4 8 6 3 1 2 5
8 5 1 7 9 2 6 4 3
1 3 8 9 4 7 2 5 6
6 9 2 3 5 1 8 7 4
7 4 5 2 8 6 3 1 9

Time complexity: O(9(N*N)), For every unassigned index, there are 9 possible options so the time complexity is O(9^(n*n)).
Space Complexity: O(N*N), To store the output array a matrix is needed.

Sudoku using Backtracking:

Like all other Backtracking problems, Sudoku can be solved by assigning numbers one by one to empty cells. Before assigning a number, check whether it is safe to assign.

Check that the same number is not present in the current row, current column and current 3X3 subgrid. After checking for safety, assign the number, and recursively check whether this assignment leads to a solution or not. If the assignment doesn’t lead to a solution, then try the next number for the current empty cell. And if none of the number (1 to 9) leads to a solution, return false and print no solution exists.

Follow the steps below to solve the problem:

• Create a function that checks after assigning the current index the grid becomes unsafe or not. Keep Hashmap for a row, column and boxes. If any number has a frequency greater than 1 in the hashMap return false else return true; hashMap can be avoided by using loops.
• Create a recursive function that takes a grid.
• Check for any unassigned location.
• If present then assigns a number from 1 to 9.
• Check if assigning the number to current index makes the grid unsafe or not.
• If safe then recursively call the function for all safe cases from 0 to 9.
• If any recursive call returns true, end the loop and return true. If no recursive call returns true then return false.
• If there is no unassigned location then return true.

Below is the implementation of the above approach:

C++

 // A Backtracking program in // C++ to solve Sudoku problem #include using namespace std;   // UNASSIGNED is used for empty // cells in sudoku grid #define UNASSIGNED 0   // N is used for the size of Sudoku grid. // Size will be NxN #define N 9   // This function finds an entry in grid // that is still unassigned bool FindUnassignedLocation(int grid[N][N],                             int& row, int& col);   // Checks whether it will be legal // to assign num to the given row, col bool isSafe(int grid[N][N], int row,             int col, int num);   /* Takes a partially filled-in grid and attempts to assign values to all unassigned locations in such a way to meet the requirements for Sudoku solution (non-duplication across rows, columns, and boxes) */ bool SolveSudoku(int grid[N][N]) {     int row, col;       // If there is no unassigned location,     // we are done     if (!FindUnassignedLocation(grid, row, col))         // success!         return true;       // Consider digits 1 to 9     for (int num = 1; num <= 9; num++)     {                   // Check if looks promising         if (isSafe(grid, row, col, num))         {                           // Make tentative assignment             grid[row][col] = num;               // Return, if success             if (SolveSudoku(grid))                 return true;               // Failure, unmake & try again             grid[row][col] = UNASSIGNED;         }     }          // This triggers backtracking     return false; }   /* Searches the grid to find an entry that is still unassigned. If found, the reference parameters row, col will be set the location that is unassigned, and true is returned. If no unassigned entries remain, false is returned. */ bool FindUnassignedLocation(int grid[N][N],                             int& row, int& col) {     for (row = 0; row < N; row++)         for (col = 0; col < N; col++)             if (grid[row][col] == UNASSIGNED)                 return true;     return false; }   /* Returns a boolean which indicates whether an assigned entry in the specified row matches the given number. */ bool UsedInRow(int grid[N][N], int row, int num) {     for (int col = 0; col < N; col++)         if (grid[row][col] == num)             return true;     return false; }   /* Returns a boolean which indicates whether an assigned entry in the specified column matches the given number. */ bool UsedInCol(int grid[N][N], int col, int num) {     for (int row = 0; row < N; row++)         if (grid[row][col] == num)             return true;     return false; }   /* Returns a boolean which indicates whether an assigned entry within the specified 3x3 box matches the given number. */ bool UsedInBox(int grid[N][N], int boxStartRow,                int boxStartCol, int num) {     for (int row = 0; row < 3; row++)         for (int col = 0; col < 3; col++)             if (grid[row + boxStartRow]                     [col + boxStartCol] ==                                        num)                 return true;     return false; }   /* Returns a boolean which indicates whether it will be legal to assign num to the given row, col location. */ bool isSafe(int grid[N][N], int row,             int col, int num) {     /* Check if 'num' is not already placed in     current row, current column     and current 3x3 box */     return !UsedInRow(grid, row, num)            && !UsedInCol(grid, col, num)            && !UsedInBox(grid, row - row % 3,                          col - col % 3, num)            && grid[row][col] == UNASSIGNED; }   /* A utility function to print grid */ void printGrid(int grid[N][N]) {     for (int row = 0; row < N; row++)     {         for (int col = 0; col < N; col++)             cout << grid[row][col] << " ";         cout << endl;     } }   // Driver Code int main() {     // 0 means unassigned cells     int grid[N][N] = { { 3, 0, 6, 5, 0, 8, 4, 0, 0 },                        { 5, 2, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 8, 7, 0, 0, 0, 0, 3, 1 },                        { 0, 0, 3, 0, 1, 0, 0, 8, 0 },                        { 9, 0, 0, 8, 6, 3, 0, 0, 5 },                        { 0, 5, 0, 0, 9, 0, 6, 0, 0 },                        { 1, 3, 0, 0, 0, 0, 2, 5, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 7, 4 },                        { 0, 0, 5, 2, 0, 6, 3, 0, 0 } };     if (SolveSudoku(grid) == true)         printGrid(grid);     else         cout << "No solution exists";       return 0; }   // This is code is contributed by rathbhupendra

Java

 /* A Backtracking program in Java to solve Sudoku problem */ class GFG {     public static boolean isSafe(int[][] board,                                  int row, int col,                                  int num)     {         // Row has the unique (row-clash)         for (int d = 0; d < board.length; d++)         {                           // Check if the number we are trying to             // place is already present in             // that row, return false;             if (board[row][d] == num) {                 return false;             }         }           // Column has the unique numbers (column-clash)         for (int r = 0; r < board.length; r++)         {                           // Check if the number             // we are trying to             // place is already present in             // that column, return false;             if (board[r][col] == num)             {                 return false;             }         }           // Corresponding square has         // unique number (box-clash)         int sqrt = (int)Math.sqrt(board.length);         int boxRowStart = row - row % sqrt;         int boxColStart = col - col % sqrt;           for (int r = boxRowStart;              r < boxRowStart + sqrt; r++)         {             for (int d = boxColStart;                  d < boxColStart + sqrt; d++)             {                 if (board[r][d] == num)                 {                     return false;                 }             }         }           // if there is no clash, it's safe         return true;     }       public static boolean solveSudoku(         int[][] board, int n)     {         int row = -1;         int col = -1;         boolean isEmpty = true;         for (int i = 0; i < n; i++)         {             for (int j = 0; j < n; j++)             {                 if (board[i][j] == 0)                 {                     row = i;                     col = j;                       // We still have some remaining                     // missing values in Sudoku                     isEmpty = false;                     break;                 }             }             if (!isEmpty) {                 break;             }         }           // No empty space left         if (isEmpty)         {             return true;         }           // Else for each-row backtrack         for (int num = 1; num <= n; num++)         {             if (isSafe(board, row, col, num))             {                 board[row][col] = num;                 if (solveSudoku(board, n))                 {                     // print(board, n);                     return true;                 }                 else                 {                     // replace it                     board[row][col] = 0;                 }             }         }         return false;     }       public static void print(         int[][] board, int N)     {                   // We got the answer, just print it         for (int r = 0; r < N; r++)         {             for (int d = 0; d < N; d++)             {                 System.out.print(board[r][d]);                 System.out.print(" ");             }             System.out.print("\n");               if ((r + 1) % (int)Math.sqrt(N) == 0)             {                 System.out.print("");             }         }     }       // Driver Code     public static void main(String args[])     {           int[][] board = new int[][] {             { 3, 0, 6, 5, 0, 8, 4, 0, 0 },             { 5, 2, 0, 0, 0, 0, 0, 0, 0 },             { 0, 8, 7, 0, 0, 0, 0, 3, 1 },             { 0, 0, 3, 0, 1, 0, 0, 8, 0 },             { 9, 0, 0, 8, 6, 3, 0, 0, 5 },             { 0, 5, 0, 0, 9, 0, 6, 0, 0 },             { 1, 3, 0, 0, 0, 0, 2, 5, 0 },             { 0, 0, 0, 0, 0, 0, 0, 7, 4 },             { 0, 0, 5, 2, 0, 6, 3, 0, 0 }         };         int N = board.length;           if (solveSudoku(board, N))         {             // print solution             print(board, N);         }         else {             System.out.println("No solution");         }     } }   // This code is contributed // by MohanDas

Python3

 # A Backtracking program # in Python to solve Sudoku problem   # A Utility Function to print the Grid def print_grid(arr):     for i in range(9):         for j in range(9):             print (arr[i][j], end = " "),         print ()             # Function to Find the entry in # the Grid that is still  not used # Searches the grid to find an # entry that is still unassigned. If # found, the reference parameters # row, col will be set the location # that is unassigned, and true is # returned. If no unassigned entries # remains, false is returned. # 'l' is a list  variable that has # been passed from the solve_sudoku function # to keep track of incrementation # of Rows and Columns def find_empty_location(arr, l):     for row in range(9):         for col in range(9):             if(arr[row][col]== 0):                 l[0]= row                 l[1]= col                 return True     return False   # Returns a boolean which indicates # whether any assigned entry # in the specified row matches # the given number. def used_in_row(arr, row, num):     for i in range(9):         if(arr[row][i] == num):             return True     return False   # Returns a boolean which indicates # whether any assigned entry # in the specified column matches # the given number. def used_in_col(arr, col, num):     for i in range(9):         if(arr[i][col] == num):             return True     return False   # Returns a boolean which indicates # whether any assigned entry # within the specified 3x3 box # matches the given number def used_in_box(arr, row, col, num):     for i in range(3):         for j in range(3):             if(arr[i + row][j + col] == num):                 return True     return False   # Checks whether it will be legal # to assign num to the given row, col # Returns a boolean which indicates # whether it will be legal to assign # num to the given row, col location. def check_location_is_safe(arr, row, col, num):           # Check if 'num' is not already     # placed in current row,     # current column and current 3x3 box     return (not used_in_row(arr, row, num) and            (not used_in_col(arr, col, num) and            (not used_in_box(arr, row - row % 3,                            col - col % 3, num))))   # Takes a partially filled-in grid # and attempts to assign values to # all unassigned locations in such a # way to meet the requirements # for Sudoku solution (non-duplication # across rows, columns, and boxes) def solve_sudoku(arr):           # 'l' is a list variable that keeps the     # record of row and col in     # find_empty_location Function        l =[0, 0]           # If there is no unassigned     # location, we are done        if(not find_empty_location(arr, l)):         return True           # Assigning list values to row and col     # that we got from the above Function     row = l[0]     col = l[1]           # consider digits 1 to 9     for num in range(1, 10):                   # if looks promising         if(check_location_is_safe(arr,                           row, col, num)):                           # make tentative assignment             arr[row][col]= num               # return, if success,             # ya !             if(solve_sudoku(arr)):                 return True               # failure, unmake & try again             arr[row][col] = 0                   # this triggers backtracking            return False   # Driver main function to test above functions if __name__=="__main__":           # creating a 2D array for the grid     grid =[[0 for x in range(9)]for y in range(9)]           # assigning values to the grid     grid =[[3, 0, 6, 5, 0, 8, 4, 0, 0],           [5, 2, 0, 0, 0, 0, 0, 0, 0],           [0, 8, 7, 0, 0, 0, 0, 3, 1],           [0, 0, 3, 0, 1, 0, 0, 8, 0],           [9, 0, 0, 8, 6, 3, 0, 0, 5],           [0, 5, 0, 0, 9, 0, 6, 0, 0],           [1, 3, 0, 0, 0, 0, 2, 5, 0],           [0, 0, 0, 0, 0, 0, 0, 7, 4],           [0, 0, 5, 2, 0, 6, 3, 0, 0]]           # if success print the grid     if(solve_sudoku(grid)):         print_grid(grid)     else:         print ("No solution exists")   # The above code has been contributed by Harshit Sidhwa.

C#

 /* A Backtracking program in C# to solve Sudoku problem */ using System;   class GFG {       public static bool isSafe(int[, ] board,                               int row, int col,                               int num)     {                   // Row has the unique (row-clash)         for (int d = 0; d < board.GetLength(0); d++)         {                           // Check if the number             // we are trying to             // place is already present in             // that row, return false;             if (board[row, d] == num)             {                 return false;             }         }           // Column has the unique numbers (column-clash)         for (int r = 0; r < board.GetLength(0); r++)         {                           // Check if the number             // we are trying to             // place is already present in             // that column, return false;             if (board[r, col] == num)             {                 return false;             }         }           // corresponding square has         // unique number (box-clash)         int sqrt = (int)Math.Sqrt(board.GetLength(0));         int boxRowStart = row - row % sqrt;         int boxColStart = col - col % sqrt;           for (int r = boxRowStart;              r < boxRowStart + sqrt; r++)         {             for (int d = boxColStart;                  d < boxColStart + sqrt; d++)             {                 if (board[r, d] == num)                 {                     return false;                 }             }         }           // if there is no clash, it's safe         return true;     }       public static bool solveSudoku(int[, ] board,                                            int n)     {         int row = -1;         int col = -1;         bool isEmpty = true;         for (int i = 0; i < n; i++)         {             for (int j = 0; j < n; j++)             {                 if (board[i, j] == 0)                 {                     row = i;                     col = j;                       // We still have some remaining                     // missing values in Sudoku                     isEmpty = false;                     break;                 }             }             if (!isEmpty)             {                 break;             }         }           // no empty space left         if (isEmpty)         {             return true;         }           // else for each-row backtrack         for (int num = 1; num <= n; num++)         {             if (isSafe(board, row, col, num))             {                 board[row, col] = num;                 if (solveSudoku(board, n))                 {                                           // Print(board, n);                     return true;                 }                 else                 {                                           // Replace it                     board[row, col] = 0;                 }             }         }         return false;     }       public static void print(int[, ] board, int N)     {                   // We got the answer, just print it         for (int r = 0; r < N; r++)         {             for (int d = 0; d < N; d++)             {                 Console.Write(board[r, d]);                 Console.Write(" ");             }             Console.Write("\n");               if ((r + 1) % (int)Math.Sqrt(N) == 0)             {                 Console.Write("");             }         }     }       // Driver Code     public static void Main(String[] args)     {           int[, ] board = new int[, ] {             { 3, 0, 6, 5, 0, 8, 4, 0, 0 },             { 5, 2, 0, 0, 0, 0, 0, 0, 0 },             { 0, 8, 7, 0, 0, 0, 0, 3, 1 },             { 0, 0, 3, 0, 1, 0, 0, 8, 0 },             { 9, 0, 0, 8, 6, 3, 0, 0, 5 },             { 0, 5, 0, 0, 9, 0, 6, 0, 0 },             { 1, 3, 0, 0, 0, 0, 2, 5, 0 },             { 0, 0, 0, 0, 0, 0, 0, 7, 4 },             { 0, 0, 5, 2, 0, 6, 3, 0, 0 }         };         int N = board.GetLength(0);           if (solveSudoku(board, N))         {                           // print solution             print(board, N);         }         else {             Console.Write("No solution");         }     } }   // This code has been contributed by 29AjayKumar

Javascript



Output

3 1 6 5 7 8 4 9 2
5 2 9 1 3 4 7 6 8
4 8 7 6 2 9 5 3 1
2 6 3 4 1 5 9 8 7
9 7 4 8 6 3 1 2 5
8 5 1 7 9 2 6 4 3
1 3 8 9 4 7 2 5 6
6 9 2 3 5 1 8 7 4
7 4 5 2 8 6 3 1 9

Time complexity: O(9(N*N)), For every unassigned index, there are 9 possible options so the time complexity is O(9^(n*n)). The time complexity remains the same but there will be some early pruning so the time taken will be much less than the naive algorithm but the upper bound time complexity remains the same.
Space Complexity: O(N*N), To store the output array a matrix is needed.

This method is a slight optimization to the above 2 methods.  For each row/column/box create a bitmask and for each element in the grid set the bit at position ‘value’ to 1 in the corresponding bitmasks, for O(1) checks.

Follow the steps below to solve the problem:

• Create 3 arrays of size N (one for rows, columns, and boxes).
• The boxes are indexed from 0 to 8. (in order to find the box index of an element we use the following formula: row / 3 * 3 + column / 3).
• Map the initial values of the grid first.
• Each time we add/remove an element to/from the grid set the bit to 1/0 to the corresponding bitmasks.

Below is the implementation of the above approach:

C++

 #include using namespace std;   #define N 9   // Bitmasks for each row/column/box int row[N], col[N], box[N]; bool seted = false;   // Utility function to find the box index // of an element at position [i][j] in the grid int getBox(int i, int j) { return i / 3 * 3 + j / 3; }   // Utility function to check if a number // is present in the corresponding row/column/box bool isSafe(int i, int j, int number) {     return !((row[i] >> number) & 1)            && !((col[j] >> number) & 1)            && !((box[getBox(i, j)] >> number) & 1); }   // Utility function to set the initial values of a Sudoku // board (map the values in the bitmasks) void setInitialValues(int grid[N][N]) {     for (int i = 0; i < N; i++)         for (int j = 0; j < N; j++)             row[i] |= 1 << grid[i][j],                 col[j] |= 1 << grid[i][j],                 box[getBox(i, j)] |= 1 << grid[i][j]; }   /* Takes a partially filled-in grid and attempts to assign values to all unassigned locations in such a way to meet the requirements for Sudoku solution (non-duplication across rows, columns, and boxes) */ bool SolveSudoku(int grid[N][N], int i, int j) {     // Set the initial values     if (!seted)         seted = true, setInitialValues(grid);       if (i == N - 1 && j == N)         return true;     if (j == N)         j = 0, i++;       if (grid[i][j])         return SolveSudoku(grid, i, j + 1);       for (int nr = 1; nr <= N; nr++) {         if (isSafe(i, j, nr)) {             /*  Assign nr in the                 current (i, j)                 position and                 add nr to each bitmask             */             grid[i][j] = nr;             row[i] |= 1 << nr;             col[j] |= 1 << nr;             box[getBox(i, j)] |= 1 << nr;               if (SolveSudoku(grid, i, j + 1))                 return true;               // Remove nr from each bitmask             // and search for another possibility             row[i] &= ~(1 << nr);             col[j] &= ~(1 << nr);             box[getBox(i, j)] &= ~(1 << nr);         }           grid[i][j] = 0;     }       return false; }   // Utility function to print the solved grid void print(int grid[N][N]) {     for (int i = 0; i < N; i++, cout << '\n')         for (int j = 0; j < N; j++)             cout << grid[i][j] << ' '; }   // Driver Code int main() {     // 0 means unassigned cells     int grid[N][N] = { { 3, 0, 6, 5, 0, 8, 4, 0, 0 },                        { 5, 2, 0, 0, 0, 0, 0, 0, 0 },                        { 0, 8, 7, 0, 0, 0, 0, 3, 1 },                        { 0, 0, 3, 0, 1, 0, 0, 8, 0 },                        { 9, 0, 0, 8, 6, 3, 0, 0, 5 },                        { 0, 5, 0, 0, 9, 0, 6, 0, 0 },                        { 1, 3, 0, 0, 0, 0, 2, 5, 0 },                        { 0, 0, 0, 0, 0, 0, 0, 7, 4 },                        { 0, 0, 5, 2, 0, 6, 3, 0, 0 } };       if (SolveSudoku(grid, 0, 0))         print(grid);     else         cout << "No solution exists\n";       return 0; } // This code is contributed // by Gatea David

Java

 /*package whatever //do not write package name here */ import java.io.*;   class GFG {     static int N = 9;       // Bitmasks for each row/column/box     static int row[] = new int[N], col[] = new int[N],                box[] = new int[N];     static Boolean seted = false;       // Utility function to find the box index     // of an element at position [i][j] in the grid     static int getBox(int i, int j)     {         return i / 3 * 3 + j / 3;     }       // Utility function to check if a number     // is present in the corresponding row/column/box     static Boolean isSafe(int i, int j, int number)     {         return ((row[i] >> number) & 1) == 0             && ((col[j] >> number) & 1) == 0             && ((box[getBox(i, j)] >> number) & 1) == 0;     }       // Utility function to set the initial values of a     // Sudoku board (map the values in the bitmasks)     static void setInitialValues(int grid[][])     {         for (int i = 0; i < grid.length; i++) {             for (int j = 0; j < grid.length; j++) {                 row[i] |= 1 << grid[i][j];                 col[j] |= 1 << grid[i][j];                 box[getBox(i, j)] |= 1 << grid[i][j];             }         }     }       /* Takes a partially filled-in grid and attempts       to assign values to all unassigned locations in       such a way to meet the requirements for       Sudoku solution (non-duplication across rows,       columns, and boxes) */     static Boolean SolveSudoku(int grid[][], int i, int j)     {         // Set the initial values         if (!seted) {             seted = true;             setInitialValues(grid);         }           if (i == N - 1 && j == N)             return true;         if (j == N) {             j = 0;             i++;         }           if (grid[i][j] > 0)             return SolveSudoku(grid, i, j + 1);           for (int nr = 1; nr <= N; nr++) {             if (isSafe(i, j, nr)) {                 /* Assign nr in the                             current (i, j)                             position and                             add nr to each bitmask                         */                 grid[i][j] = nr;                 row[i] |= 1 << nr;                 col[j] |= 1 << nr;                 box[getBox(i, j)] |= 1 << nr;                   if (SolveSudoku(grid, i, j + 1))                     return true;                   // Remove nr from each bitmask                 // and search for another possibility                 row[i] &= ~(1 << nr);                 col[j] &= ~(1 << nr);                 box[getBox(i, j)] &= ~(1 << nr);             }               grid[i][j] = 0;         }           return false;     }       // Utility function to print the solved grid     static void print(int grid[][])     {         for (int i = 0; i < grid.length; i++) {             for (int j = 0; j < grid[0].length; j++) {                 System.out.printf("%d ", grid[i][j]);             }             System.out.println();         }     }       // Driver code     public static void main(String args[])     {         // 0 means unassigned cells         int grid[][] = { { 3, 0, 6, 5, 0, 8, 4, 0, 0 },                          { 5, 2, 0, 0, 0, 0, 0, 0, 0 },                          { 0, 8, 7, 0, 0, 0, 0, 3, 1 },                          { 0, 0, 3, 0, 1, 0, 0, 8, 0 },                          { 9, 0, 0, 8, 6, 3, 0, 0, 5 },                          { 0, 5, 0, 0, 9, 0, 6, 0, 0 },                          { 1, 3, 0, 0, 0, 0, 2, 5, 0 },                          { 0, 0, 0, 0, 0, 0, 0, 7, 4 },                          { 0, 0, 5, 2, 0, 6, 3, 0, 0 } };           if (SolveSudoku(grid, 0, 0))             print(grid);         else             System.out.println("No solution exists");     } }   // This code is contributed by shinjanpatra.

Python3

 # N is the size of the 2D matrix N*N N = 9   # A utility function to print grid     def printing(arr):     for i in range(N):         for j in range(N):             print(arr[i][j], end=" ")         print()   # Checks whether it will be # legal to assign num to the # given row, col     def isSafe(grid, row, col, num):       # Check if we find the same num     # in the similar row , we     # return false     for x in range(9):         if grid[row][x] == num:             return False       # Check if we find the same num in     # the similar column , we     # return false     for x in range(9):         if grid[x][col] == num:             return False       # Check if we find the same num in     # the particular 3*3 matrix,     # we return false     startRow = row - row % 3     startCol = col - col % 3     for i in range(3):         for j in range(3):             if grid[i + startRow][j + startCol] == num:                 return False     return True   # Takes a partially filled-in grid and attempts # to assign values to all unassigned locations in # such a way to meet the requirements for # Sudoku solution (non-duplication across rows, # columns, and boxes) */     def solveSudoku(grid, row, col):       # Check if we have reached the 8th     # row and 9th column (0     # indexed matrix) , we are     # returning true to avoid     # further backtracking     if (row == N - 1 and col == N):         return True       # Check if column value becomes 9 ,     # we move to next row and     # column start from 0     if col == N:         row += 1         col = 0       # Check if the current position of     # the grid already contains     # value >0, we iterate for next column     if grid[row][col] > 0:         return solveSudoku(grid, row, col + 1)     for num in range(1, N + 1, 1):           # Check if it is safe to place         # the num (1-9) in the         # given row ,col ->we         # move to next column         if isSafe(grid, row, col, num):               # Assigning the num in             # the current (row,col)             # position of the grid             # and assuming our assigned             # num in the position             # is correct             grid[row][col] = num               # Checking for next possibility with next             # column             if solveSudoku(grid, row, col + 1):                 return True           # Removing the assigned num ,         # since our assumption         # was wrong , and we go for         # next assumption with         # diff num value         grid[row][col] = 0     return False   # Driver Code     # 0 means unassigned cells grid = [[3, 0, 6, 5, 0, 8, 4, 0, 0],         [5, 2, 0, 0, 0, 0, 0, 0, 0],         [0, 8, 7, 0, 0, 0, 0, 3, 1],         [0, 0, 3, 0, 1, 0, 0, 8, 0],         [9, 0, 0, 8, 6, 3, 0, 0, 5],         [0, 5, 0, 0, 9, 0, 6, 0, 0],         [1, 3, 0, 0, 0, 0, 2, 5, 0],         [0, 0, 0, 0, 0, 0, 0, 7, 4],         [0, 0, 5, 2, 0, 6, 3, 0, 0]]   if (solveSudoku(grid, 0, 0)):     printing(grid) else:     print("no solution exists ")   # This code is contributed by sanjoy_62.

C#

 // C# program for above approach using System; class GFG {       // N is the size of the 2D matrix   N*N     static int N = 9;       /* Takes a partially filled-in grid and attempts       to assign values to all unassigned locations in       such a way to meet the requirements for       Sudoku solution (non-duplication across rows,       columns, and boxes) */     static bool solveSudoku(int[, ] grid, int row, int col)     {           /*if we have reached the 8th                row and 9th column (0                indexed matrix) ,                we are returning true to avoid further                backtracking       */         if (row == N - 1 && col == N)             return true;           // Check if column value  becomes 9 ,         // we move to next row         // and column start from 0         if (col == N) {             row++;             col = 0;         }           // Check if the current position         // of the grid already         // contains value >0, we iterate         // for next column         if (grid[row, col] != 0)             return solveSudoku(grid, row, col + 1);           for (int num = 1; num < 10; num++) {               // Check if it is safe to place             // the num (1-9)  in the             // given row ,col ->we move to next column             if (isSafe(grid, row, col, num)) {                   /*  assigning the num in the current                         (row,col)  position of the grid and                         assuming our assigned num in the                    position is correct */                 grid[row, col] = num;                   // Checking for next                 // possibility with next column                 if (solveSudoku(grid, row, col + 1))                     return true;             }             /* removing the assigned num , since our                      assumption was wrong , and we go for                next assumption with diff num value   */             grid[row, col] = 0;         }         return false;     }       /* A utility function to print grid */     static void print(int[, ] grid)     {         for (int i = 0; i < N; i++) {             for (int j = 0; j < N; j++)                 Console.Write(grid[i, j] + " ");             Console.WriteLine();         }     }       // Check whether it will be legal     // to assign num to the     // given row, col     static bool isSafe(int[, ] grid, int row, int col,                        int num)     {           // Check if we find the same num         // in the similar row , we         // return false         for (int x = 0; x <= 8; x++)             if (grid[row, x] == num)                 return false;           // Check if we find the same num         // in the similar column ,         // we return false         for (int x = 0; x <= 8; x++)             if (grid[x, col] == num)                 return false;           // Check if we find the same num         // in the particular 3*3         // matrix, we return false         int startRow = row - row % 3, startCol                                       = col - col % 3;         for (int i = 0; i < 3; i++)             for (int j = 0; j < 3; j++)                 if (grid[i + startRow, j + startCol] == num)                     return false;           return true;     }       // Driver code     static void Main()     {         int[, ] grid = { { 3, 0, 6, 5, 0, 8, 4, 0, 0 },                          { 5, 2, 0, 0, 0, 0, 0, 0, 0 },                          { 0, 8, 7, 0, 0, 0, 0, 3, 1 },                          { 0, 0, 3, 0, 1, 0, 0, 8, 0 },                          { 9, 0, 0, 8, 6, 3, 0, 0, 5 },                          { 0, 5, 0, 0, 9, 0, 6, 0, 0 },                          { 1, 3, 0, 0, 0, 0, 2, 5, 0 },                          { 0, 0, 0, 0, 0, 0, 0, 7, 4 },                          { 0, 0, 5, 2, 0, 6, 3, 0, 0 } };           if (solveSudoku(grid, 0, 0))             print(grid);         else             Console.WriteLine("No Solution exists");     } }   // This code is contributed by code_hunt.

Javascript



Output

3 1 6 5 7 8 4 9 2
5 2 9 1 3 4 7 6 8
4 8 7 6 2 9 5 3 1
2 6 3 4 1 5 9 8 7
9 7 4 8 6 3 1 2 5
8 5 1 7 9 2 6 4 3
1 3 8 9 4 7 2 5 6
6 9 2 3 5 1 8 7 4
7 4 5 2 8 6 3 1 9

Time complexity: O(9(N*N)). For every unassigned index, there are 9 possible options so the time complexity is O(9^(n*n)). The time complexity remains the same but checking if a number is safe to use is much faster, O(1).
Space Complexity: O(N*N). To store the output array a matrix is needed, and 3 extra arrays of size n are needed for the bitmasks.

Sudoku using Cross-Hatching with backtracking:

This method is an optimization of the above method 2. It runs 5X times faster than method 2. Like we used to fill sudoku by first identifying the element which is almost filled. It starts with identifying the row and column where the element should be placed. Picking the almost-filled elements first will give better pruning.

Follow the steps below to solve the problem:

1. Build a graph with pending elements mapped to row and column coordinates where they can be fitted in the original matrix.
2. Pick the elements from the graph sorted by fewer remaining elements to be filled.
3. Recursively fill the elements using a graph into the matrix. Backtrack once an unsafe position is discovered.

Below is the implementation of the above approach:

C++

 // C++ Code #include #include #include #include   using namespace std;   // Input matrix vector > arr     = { { 3, 0, 6, 5, 0, 8, 4, 0, 0 },         { 5, 2, 0, 0, 0, 0, 0, 0, 0 },         { 0, 8, 7, 0, 0, 0, 0, 3, 1 },         { 0, 0, 3, 0, 1, 0, 0, 8, 0 },         { 9, 0, 0, 8, 6, 3, 0, 0, 5 },         { 0, 5, 0, 0, 9, 0, 6, 0, 0 },         { 1, 3, 0, 0, 0, 0, 2, 5, 0 },         { 0, 0, 0, 0, 0, 0, 0, 7, 4 },         { 0, 0, 5, 2, 0, 6, 3, 0, 0 } };   // Position of the input elements in the arr // pos = { //     element: [[position 1], [position 2]] // } map > > pos;   // Count of the remaining number of the elements // rem = { //     element: pending count // } map rem;   // Graph defining tentative positions of the elements to be // filled graph = { //     key: { //         row1: [columns], //         row2: [columns] //     } // } map > > graph;   // Print the matrix array void printMatrix() {   for (int i = 0; i < 9; i++) {     for (int j = 0; j < 9; j++) {       cout << arr[i][j] << " ";     }     cout << endl;   } }   // Method to check if the inserted element is safe bool is_safe(int x, int y) {   int key = arr[x][y];   for (int i = 0; i < 9; i++) {     if (i != y && arr[x][i] == key) {       return false;     }     if (i != x && arr[i][y] == key) {       return false;     }   }     int r_start = floor(x / 3) * 3;   int r_end = r_start + 3;     int c_start = floor(y / 3) * 3;   int c_end = c_start + 3;     for (int i = r_start; i < r_end; i++) {     for (int j = c_start; j < c_end; j++) {       if (i != x && j != y && arr[i][j] == key) {         return false;       }     }   }   return true; }   // method to fill the matrix // input keys: list of elements to be filled in the matrix //        k   : index number of the element to be picked up //        from keys rows: list of row index where element is //        to be inserted r   : index number of the row to be //        inserted // bool fill_matrix(int k, vector keys, int r,                  vector rows) {   int c = 0;   arr[rows[r]] = keys[k];   if (is_safe(rows[r], c)) {     if (r < rows.size() - 1) {       if (fill_matrix(k, keys, r + 1, rows)) {         return true;       }       else {         arr[rows[r]] = 0;       }     }     else {       if (k < keys.size() - 1) {         if (fill_matrix(           k + 1, keys, 0,           rows)) {           return true;         }         else {           arr[rows[r]] = 0;           }       }       return true;     }   }   arr[rows[r]] = 0;     return false; }   // Fill the pos and rem dictionary. It will be used to build // graph void build_pos_and_rem() {   for (int i = 0; i < 9; i++) {     for (int j = 0; j < 9; j++) {       if (arr[i][j] > 0) {         if (!pos.count(arr[i][j])) {           pos[arr[i][j]] = {};         }         pos[arr[i][j]].push_back({ i, j });         if (!rem.count(arr[i][j])) {           rem[arr[i][j]] = 9;         }         rem[arr[i][j]] -= 1;       }     }   }     // Fill the elements not present in input matrix.   // Example: 1 is missing in input matrix   for (int i = 1; i < 10; i++) {     if (!pos.count(i)) {       pos[i] = {};     }     if (!rem.count(i)) {       rem[i] = 9;     }   } }   int main() { printMatrix(); }   // This code is contributed by ishankhandelwals.

Python3

 # This program works by identifying the remaining elements and backtrack only on those. # The elements are inserted in the increasing order of the elements left to be inserted. And hence runs much faster. # Comparing with other back tracking algorithms, it runs 5X faster.   # Input matrix arr = [     [3, 0, 6, 5, 0, 8, 4, 0, 0],     [5, 2, 0, 0, 0, 0, 0, 0, 0],     [0, 8, 7, 0, 0, 0, 0, 3, 1],     [0, 0, 3, 0, 1, 0, 0, 8, 0],     [9, 0, 0, 8, 6, 3, 0, 0, 5],     [0, 5, 0, 0, 9, 0, 6, 0, 0],     [1, 3, 0, 0, 0, 0, 2, 5, 0],     [0, 0, 0, 0, 0, 0, 0, 7, 4],     [0, 0, 5, 2, 0, 6, 3, 0, 0] ]   # Position of the input elements in the arr # pos = { #     element: [[position 1], [position 2]] # } pos = {}   # Count of the remaining number of the elements # rem = { #     element: pending count # } rem = {}   # Graph defining tentative positions of the elements to be filled # graph = { #     key: { #         row1: [columns], #         row2: [columns] #     } # } graph = {}     # Print the matrix array def printMatrix():     for i in range(0, 9):         for j in range(0, 9):             print(str(arr[i][j]), end=" ")         print()     # Method to check if the inserted element is safe def is_safe(x, y):     key = arr[x][y]     for i in range(0, 9):         if i != y and arr[x][i] == key:             return False         if i != x and arr[i][y] == key:             return False       r_start = int(x / 3) * 3     r_end = r_start + 3       c_start = int(y / 3) * 3     c_end = c_start + 3       for i in range(r_start, r_end):         for j in range(c_start, c_end):             if i != x and j != y and arr[i][j] == key:                 return False     return True     # method to fill the matrix # input keys: list of elements to be filled in the matrix #        k   : index number of the element to be picked up from keys #        rows: list of row index where element is to be inserted #        r   : index number of the row to be inserted # def fill_matrix(k, keys, r, rows):     for c in graph[keys[k]][rows[r]]:         if arr[rows[r]] > 0:             continue         arr[rows[r]] = keys[k]         if is_safe(rows[r], c):             if r < len(rows) - 1:                 if fill_matrix(k, keys, r + 1, rows):                     return True                 else:                     arr[rows[r]] = 0                     continue             else:                 if k < len(keys) - 1:                     if fill_matrix(k + 1, keys, 0, list(graph[keys[k + 1]].keys())):                         return True                     else:                         arr[rows[r]] = 0                         continue                 return True         arr[rows[r]] = 0     return False     # Fill the pos and rem dictionary. It will be used to build graph def build_pos_and_rem():     for i in range(0, 9):         for j in range(0, 9):             if arr[i][j] > 0:                 if arr[i][j] not in pos:                     pos[arr[i][j]] = []                 pos[arr[i][j]].append([i, j])                 if arr[i][j] not in rem:                     rem[arr[i][j]] = 9                 rem[arr[i][j]] -= 1       # Fill the elements not present in input matrix. Example: 1 is missing in input matrix     for i in range(1, 10):         if i not in pos:             pos[i] = []         if i not in rem:             rem[i] = 9   # Build the graph     def build_graph():     for k, v in pos.items():         if k not in graph:             graph[k] = {}           row = list(range(0, 9))         col = list(range(0, 9))           for cord in v:             row.remove(cord[0])             col.remove(cord[1])           if len(row) == 0 or len(col) == 0:             continue           for r in row:             for c in col:                 if arr[r] == 0:                     if r not in graph[k]:                         graph[k][r] = []                     graph[k][r].append(c)     build_pos_and_rem()   # Sort the rem map in order to start with smaller number of elements to be filled first. Optimization for pruning rem = {k: v for k, v in sorted(rem.items(), key=lambda item: item[1])}   build_graph()   key_s = list(rem.keys()) # Util called to fill the matrix fill_matrix(0, key_s, 0, list(graph[key_s[0]].keys()))   printMatrix()   # This code is contributed by Arun Kumar

Javascript

 // convert below python code to js code // # This program works by identifying the remaining elements and backtrack only on those. // # The elements are inserted in the increasing order of the elements left to be inserted. And hence runs much faster. // # Comparing with other back tracking algorithms, it runs 5X faster.   // # Input matrix // arr = [ //     [3, 0, 6, 5, 0, 8, 4, 0, 0], //     [5, 2, 0, 0, 0, 0, 0, 0, 0], //     [0, 8, 7, 0, 0, 0, 0, 3, 1], //     [0, 0, 3, 0, 1, 0, 0, 8, 0], //     [9, 0, 0, 8, 6, 3, 0, 0, 5], //     [0, 5, 0, 0, 9, 0, 6, 0, 0], //     [1, 3, 0, 0, 0, 0, 2, 5, 0], //     [0, 0, 0, 0, 0, 0, 0, 7, 4], //     [0, 0, 5, 2, 0, 6, 3, 0, 0] // ]   // # Position of the input elements in the arr // # pos = { // #     element: [[position 1], [position 2]] // # } // pos = {}   // # Count of the remaining number of the elements // # rem = { // #     element: pending count // # } // rem = {}   // # Graph defining tentative positions of the elements to be filled // # graph = { // #     key: { // #         row1: [columns], // #         row2: [columns] // #     } // # } // graph = {}     // # Print the matrix array // def printMatrix(): //     for i in range(0, 9): //         for j in range(0, 9): //             print(str(arr[i][j]), end=" ") //         print()     // # Method to check if the inserted element is safe // def is_safe(x, y): //     key = arr[x][y] //     for i in range(0, 9): //         if i != y and arr[x][i] == key: //             return False //         if i != x and arr[i][y] == key: //             return False   //     r_start = int(x / 3) * 3 //     r_end = r_start + 3   //     c_start = int(y / 3) * 3 //     c_end = c_start + 3   //     for i in range(r_start, r_end): //         for j in range(c_start, c_end): //             if i != x and j != y and arr[i][j] == key: //                 return False //     return True     // # method to fill the matrix // # input keys: list of elements to be filled in the matrix // #        k   : index number of the element to be picked up from keys // #        rows: list of row index where element is to be inserted // #        r   : index number of the row to be inserted // # // def fill_matrix(k, keys, r, rows): //     for c in graph[keys[k]][rows[r]]: //         if arr[rows[r]] > 0: //             continue //         arr[rows[r]] = keys[k] //         if is_safe(rows[r], c): //             if r < len(rows) - 1: //                 if fill_matrix(k, keys, r + 1, rows): //                     return True //                 else: //                     arr[rows[r]] = 0 //                     continue //             else: //                 if k < len(keys) - 1: //                     if fill_matrix(k + 1, keys, 0, list(graph[keys[k + 1]].keys())): //                         return True //                     else: //                         arr[rows[r]] = 0 //                         continue //                 return True //         arr[rows[r]] = 0 //     return False     // # Fill the pos and rem dictionary. It will be used to build graph // def build_pos_and_rem(): //     for i in range(0, 9): //         for j in range(0, 9): //             if arr[i][j] > 0: //                 if arr[i][j] not in pos: //                     pos[arr[i][j]] = [] //                 pos[arr[i][j]].append([i, j]) //                 if arr[i][j] not in rem: //                     rem[arr[i][j]] = 9 //                 rem[arr[i][j]] -= 1   //     # Fill the elements not present in input matrix. Example: 1 is missing in input matrix //     for i in range(1, 10): //         if i not in pos: //             pos[i] = [] //         if i not in rem: //             rem[i] = 9   // # Build the graph     // def build_graph(): //     for k, v in pos.items(): //         if k not in graph: //             graph[k] = {}   //         row = list(range(0, 9)) //         col = list(range(0, 9))   //         for cord in v: //             row.remove(cord[0]) //             col.remove(cord[1])   //         if len(row) == 0 or len(col) == 0: //             continue   //         for r in row: //             for c in col: //                 if arr[r] == 0: //                     if r not in graph[k]: //                         graph[k][r] = [] //                     graph[k][r].append(c)     // build_pos_and_rem()   // # Sort the rem map in order to start with smaller number of elements to be filled first. Optimization for pruning // rem = {k: v for k, v in sorted(rem.items(), key=lambda item: item[1])}   // build_graph()   // key_s = list(rem.keys()) // # Util called to fill the matrix // fill_matrix(0, key_s, 0, list(graph[key_s[0]].keys()))   // printMatrix()   // # This code is contributed by Arun Kumar   // JavaScript Code   // Input matrix let arr = [     [3, 0, 6, 5, 0, 8, 4, 0, 0],     [5, 2, 0, 0, 0, 0, 0, 0, 0],     [0, 8, 7, 0, 0, 0, 0, 3, 1],     [0, 0, 3, 0, 1, 0, 0, 8, 0],     [9, 0, 0, 8, 6, 3, 0, 0, 5],     [0, 5, 0, 0, 9, 0, 6, 0, 0],     [1, 3, 0, 0, 0, 0, 2, 5, 0],     [0, 0, 0, 0, 0, 0, 0, 7, 4],     [0, 0, 5, 2, 0, 6, 3, 0, 0] ]   // Position of the input elements in the arr // pos = { //     element: [[position 1], [position 2]] // } let pos = {};   // Count of the remaining number of the elements // rem = { //     element: pending count // } let rem = {};   // Graph defining tentative positions of the elements to be filled // graph = { //     key: { //         row1: [columns], //         row2: [columns] //     } // } let graph = {};     // Print the matrix array function printMatrix() {     for (let i = 0; i < 9; i++) {         for (let j = 0; j < 9; j++) {             process.stdout.write(arr[i][j]+" ");         }         console.log();     } }     // Method to check if the inserted element is safe function is_safe(x, y) {     let key = arr[x][y];     for (let i = 0; i < 9; i++) {         if (i !== y && arr[x][i] === key) {             return false;         }         if (i !== x && arr[i][y] === key) {             return false;         }     }       let r_start = Math.floor(x / 3) * 3;     let r_end = r_start + 3;       let c_start = Math.floor(y / 3) * 3;     let c_end = c_start + 3;       for (let i = r_start; i < r_end; i++) {         for (let j = c_start; j < c_end; j++) {             if (i !== x && j !== y && arr[i][j] === key) {                 return false;             }         }     }     return true; }     // method to fill the matrix // input keys: list of elements to be filled in the matrix //        k   : index number of the element to be picked up from keys //        rows: list of row index where element is to be inserted //        r   : index number of the row to be inserted // function fill_matrix(k, keys, r, rows) {     for (let c of graph[keys[k]][rows[r]]) {         if (arr[rows[r]] > 0) {             continue;         }         arr[rows[r]] = keys[k];         if (is_safe(rows[r], c)) {             if (r < rows.length - 1) {                 if (fill_matrix(k, keys, r + 1, rows)) {                     return true;                 } else {                     arr[rows[r]] = 0;                     continue;                 }             } else {                 if (k < keys.length - 1) {                     if (fill_matrix(k + 1, keys, 0, list(graph[keys[k + 1]].keys()))) {                         return true;                     } else {                         arr[rows[r]] = 0;                         continue;                     }                 }                 return true;             }         }         arr[rows[r]] = 0;     }     return false; }     // Fill the pos and rem dictionary. It will be used to build graph function build_pos_and_rem() {     for (let i = 0; i < 9; i++) {         for (let j = 0; j < 9; j++) {             if (arr[i][j] > 0) {                 if (!pos.hasOwnProperty(arr[i][j])) {                     pos[arr[i][j]] = [];                 }                 pos[arr[i][j]].push([i, j]);                 if (!rem.hasOwnProperty(arr[i][j])) {                     rem[arr[i][j]] = 9;                 }                 rem[arr[i][j]] -= 1;             }         }     }       // Fill the elements not present in input matrix. Example: 1 is missing in input matrix     for (let i = 1; i < 10; i++) {         if (!pos.hasOwnProperty(i)) {             pos[i] = [];         }         if (!rem.hasOwnProperty(i)) {             rem[i] = 9;         }     } }   // Build the graph function build_graph() {     for (let [k, v] of Object.entries(pos)) {         if (!graph.hasOwnProperty(k)) {             graph[k] = {};         }           let row = [...Array(9).keys()];         let col = [...Array(9).keys()];           for (let cord of v) {             row.splice(row.indexOf(cord[0]), 1);             col.splice(col.indexOf(cord[1]), 1);         }           if (row.length === 0 || col.length === 0) {             continue;         }           for (let r of row) {             for (let c of col) {                 if (arr[r] === 0) {                     if (!graph[k].hasOwnProperty(r)) {                         graph[k][r] = [];                     }                     graph[k][r].push(c);                 }             }         }     } }   build_pos_and_rem();   // Sort the rem map in order to start with smaller number of elements to be filled first. Optimization for pruning rem = Object.fromEntries(Object.entries(rem).sort((a, b) => a[1] - b[1]));   build_graph();   let key_s = Object.keys(rem); // Util called to fill the matrix fill_matrix(0, key_s, 0, Object.keys(graph[key_s[0]]));   printMatrix();

Java

 // Java code import java.util.*; import java.util.ArrayList; import java.util.Collections; import java.util.List;    class Element {         int row, col;         List candidates;           Element(int row, int col) {             this.row = row;             this.col = col;             this.candidates = new ArrayList<>();         }     }   // Input matrix class Main {     static final int N = 9;     static int[][] grid = {             {3, 0, 6, 5, 0, 8, 4, 0, 0},             {5, 2, 0, 0, 0, 0, 0, 0, 0},             {0, 8, 7, 0, 0, 0, 0, 3, 1},             {0, 0, 3, 0, 1, 0, 0, 8, 0},             {9, 0, 0, 8, 6, 3, 0, 0, 5},             {0, 5, 0, 0, 9, 0, 6, 0, 0},             {1, 3, 0, 0, 0, 0, 2, 5, 0},             {0, 0, 0, 0, 0, 0, 0, 7, 4},             {0, 0, 5, 2, 0, 6, 3, 0, 0}     };       static List graph;       // Method to check if the inserted element is safe     static boolean isSafe(int row, int col, int num) {         // check row         for (int i = 0; i < N; i++) {             if (grid[row][i] == num) {                 return false;             }         }           // check column         for (int i = 0; i < N; i++) {             if (grid[i][col] == num) {                 return false;             }         }           // check subgrid         int subgridRow = row - row % 3;         int subgridCol = col - col % 3;         for (int i = subgridRow; i < subgridRow + 3; i++) {             for (int j = subgridCol; j < subgridCol + 3; j++) {                 if (grid[i][j] == num) {                     return false;                 }             }         }           return true;     }     //initialize the graph     static void initializeGraph() {         graph = new ArrayList<>();         for (int row = 0; row < N; ++row) {             for (int col = 0; col < N; ++col) {                 if (grid[row][col] == 0) {                     Element e = new Element(row, col);                     for (int num = 1; num <= N; ++num) {                         if (isSafe(row, col, num)) {                             e.candidates.add(num);                         }                     }                     graph.add(e);                 }             }         }     }       // Sort the rem map in order to start with smaller number of elements   //to be filled first.     static int sortByFewestCandidates(Element a, Element b) {          return a.candidates.size() - b.candidates.size();     }       static boolean solveSudoku() {         if (graph.isEmpty()) {             return true;         }           Collections.sort(graph, Main::sortByFewestCandidates);         Element e = graph.get(graph.size() - 1);         graph.remove(graph.size() -1);        //  graph.pop_back();     int row = e.row;     int col = e.col;               for (int i = 0; i < e.candidates.size(); ++i) {         int num = e.candidates.get(i);         if (isSafe(e.row, e.col, num)) {             grid[e.row][e.col] = num;             graph.remove(e);             if (solveSudoku()) {                 return true;             }             graph.add(e);             Collections.sort(graph, Main::sortByFewestCandidates);             grid[e.row][e.col] = 0;         }     }     return false; }   public static void main(String[] args) {     initializeGraph();     solveSudoku();       // print solved sudoku     for (int[] row : grid) {         for (int num : row) {             System.out.print(num + " ");         }         System.out.println();     } } }

C#

 // C# Code using System; using System.Collections.Generic;   namespace sudoku {     class Program     {         // Input matrix         static int[,] arr = new int[9, 9] {             { 3, 0, 6, 5, 0, 8, 4, 0, 0 },             { 5, 2, 0, 0, 0, 0, 0, 0, 0 },             { 0, 8, 7, 0, 0, 0, 0, 3, 1 },             { 0, 0, 3, 0, 1, 0, 0, 8, 0 },             { 9, 0, 0, 8, 6, 3, 0, 0, 5 },             { 0, 5, 0, 0, 9, 0, 6, 0, 0 },             { 1, 3, 0, 0, 0, 0, 2, 5, 0 },             { 0, 0, 0, 0, 0, 0, 0, 7, 4 },             { 0, 0, 5, 2, 0, 6, 3, 0, 0 }         };           // Position of the input elements in the arr         // pos = {         //     element: [[position 1], [position 2]]         // }         static Dictionary> pos = new Dictionary>();           // Count of the remaining number of the elements         // rem = {         //     element: pending count         // }         static Dictionary rem = new Dictionary();           // Graph defining tentative positions of the elements to be         // filled graph = {         //     key: {         //         row1: [columns],         //         row2: [columns]         //     }         // }         static Dictionary>> graph = new Dictionary>>();           // Print the matrix array         static void PrintMatrix()         {             for (int i = 0; i < 9; i++)             {                 for (int j = 0; j < 9; j++)                 {                     Console.Write(arr[i, j] + " ");                 }                 Console.WriteLine();             }         }           // Method to check if the inserted element is safe         static bool IsSafe(int x, int y)         {             int key = arr[x, y];             for (int i = 0; i < 9; i++)             {                 if (i != y && arr[x, i] == key)                 {                     return false;                 }                 if (i != x && arr[i, y] == key)                 {                     return false;                 }             }               int r_start = (int)Math.Floor((double)x / 3) * 3;             int r_end = r_start + 3;               int c_start = (int)Math.Floor((double)y / 3) * 3;             int c_end = c_start + 3;               for (int i = r_start; i < r_end; i++)             {                 for (int j = c_start; j < c_end; j++)                 {                     if (i != x && j != y && arr[i, j] == key)                     {                         return false;                     }                 }             }             return true;         }           // method to fill the matrix         // input keys: list of elements to be filled in the matrix         //        k   : index number of the element to be picked up         //        from keys rows: list of row index where element is         //        to be inserted r   : index number of the row to be         //        inserted         //         static bool FillMatrix(int k, List keys, int r, List rows)         {             int c = 0;             arr[rows[r], c] = keys[k];             if (IsSafe(rows[r], c))             {                 if (r < rows.Count - 1)                 {                     if (FillMatrix(k, keys, r + 1, rows))                     {                         return true;                     }                     else                     {                         arr[rows[r], c] = 0;                     }                 }                 else                 {                     if (k < keys.Count - 1)                     {                         if (FillMatrix(                             k + 1, keys, 0,                             rows))                         {                             return true;                         }                         else                         {                             arr[rows[r], c] = 0;                           }                     }                     return true;                 }             }             arr[rows[r], c] = 0;               return false;         }           // Fill the pos and rem dictionary. It will be used to build         // graph         static void BuildPosAndRem()         {             for (int i = 0; i < 9; i++)             {                 for (int j = 0; j < 9; j++)                 {                     if (arr[i, j] > 0)                     {                         if (!pos.ContainsKey(arr[i, j]))                         {                             pos[arr[i, j]] = new List();                         }                         pos[arr[i, j]].Add(new int[] { i, j });                         if (!rem.ContainsKey(arr[i, j]))                         {                             rem[arr[i, j]] = 9;                         }                         rem[arr[i, j]] -= 1;                     }                 }             }               // Fill the elements not present in input matrix.             // Example: 1 is missing in input matrix             for (int i = 1; i < 10; i++)             {                 if (!pos.ContainsKey(i))                 {                     pos[i] = new List();                 }                 if (!rem.ContainsKey(i))                 {                     rem[i] = 9;                 }             }         }           static void Main(string[] args)         {             PrintMatrix();         }     } }

Output

3 1 6 5 7 8 4 9 2
5 2 9 1 3 4 7 6 8
4 8 7 6 2 9 5 3 1
2 6 3 4 1 5 9 8 7
9 7 4 8 6 3 1 2 5
8 5 1 7 9 2 6 4 3
1 3 8 9 4 7 2 5 6
6 9 2 3 5 1 8 7 4
7 4 5 2 8 6 3 1 9

Time complexity: O(9^(n*n)).  Due to the element that needs to fit in a cell will come earlier as we are filling almost filled elements first, it will help in less number of recursive calls. So the time taken will be much less than the naive approaches but the upper bound time complexity remains the same.
Auxiliary Space: O(n*n).  A graph of the remaining positions to be filled for the respected elements is created.

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