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Sudoku | Backtracking-7
  • Difficulty Level : Hard
  • Last Updated : 24 Feb, 2021

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. 

Example: 

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.
 

Method 1: Simple.
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 position check if the matrix is safe or not. If safe print else recurs for other cases.
Algorithm: 

  1. 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;
  2. Create a recursive function that takes a grid and the current row and column index.
  3. 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.
  4. 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.
  5. if the current index is assigned then call the recursive function with index of next element, i.e. i, j+1

C++




#include <iostream>
 
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 solveSuduko(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 solveSuduko(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 assined
              num in the position
              is correct     */
            grid[row][col] = num;
           
            //  Checking for next possibility with next
            //  column
            if (solveSuduko(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 (solveSuduko(grid, 0, 0))
        print(grid);
    else
        cout << "no solution  exists " << endl;
 
    return 0;
    // This is code is contributed by Pradeep Mondal P
}

C




#include <stdio.h>
#include <stdlib.h>
 
// 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 solveSuduko(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 solveSuduko(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 assined num
               in the position
               is correct     */
            grid[row][col] = num;
           
            //  Checking for next possibility with next
            //  column
            if (solveSuduko(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 (solveSuduko(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 Suduko {
 
    // 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 solveSuduko(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 solveSuduko(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 assined num in the position
                is correct */
                grid[row][col] = num;
 
                // Checking for next
                // possibility with next column
                if (solveSuduko(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 (solveSuduko(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 solveSuduko(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 solveSuduko(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 assined
            # num in the position
            # is correct
            grid[row][col] = num
 
            # Checking for next possibility with next
            # column
            if solveSuduko(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 (solveSuduko(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 solveSuduko(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 solveSuduko(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 assined num in the position
                is correct */
        grid[row,col] = num;
 
        // Checking for next
        // possibility with next column
        if (solveSuduko(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 (solveSuduko(grid, 0, 0))
      print(grid);
    else
      Console.WriteLine("No Solution exists");
  }
}
 
// This code is contributed by divyesh072019.
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

 Complexity Analysis:  

  • 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.

Method 2: Backtracking. 
Approach: 
Like all other Backtracking problems, Sudoku can be solved by one by one assigning numbers 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.
Algorithm: 

  1. 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.
  2. Create a recursive function that takes a grid.
  3. Check for any unassigned location. If present then assign 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.
  4. If there is no unassigned location then return true.

C++




// A Backtracking program in
// C++ to solve Sudoku problem
#include <bits/stdc++.h>
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

C




// A Backtracking program  in C
// to solve Sudoku problem
#include <stdio.h>
 
// UNASSIGNED is used for empty
// cells in sudoku grid
#define UNASSIGNED 0
 
// N is used for the size of
// Sudoku grid. The 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;
 
    // Check If there is no unassigned
    // location, we are done
    if (!FindUnassignedLocation(grid, row, col))
        return true; // success!
 
    //Cconsider 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, yay!
            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++)
            printf("%2d", grid[row][col]);
        printf("\n");
    }
}
 
/* Driver Program to test above functions */
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
        printf("No solution exists");
 
    return 0;
}

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

Python




# 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],
        print ('n')
 
         
# 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
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

Complexity Analysis:  

  • 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.

References: 
http://see.stanford.edu/materials/icspacs106b/H19-RecBacktrackExamples.pdf
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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