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# N Queen Problem | Backtracking-3

We have discussed Knight’s tour and Rat in a Maze problem earlier as examples of Backtracking problems. Let us discuss N Queen as another example problem that can be solved using backtracking.

### N-Queen Problem:

The N Queen is the problem of placing N chess queens on an N×N chessboard so that no two queens attack each other. For example, the following is a solution for the 4 Queen problem.

Solution for 4-Queen problem

The expected output is in the form of a matrix that has ‘Q‘s for the blocks where queens are placed and the empty spaces are represented by ‘.’ . For example, the following is the output matrix for the above 4-Queen solution.

#### . . Q . Q . . . . . . Q . Q . .

Naive Algorithm:
Generate all possible configurations of queens on board and print a configuration that satisfies the given constraints.

while there are untried configurations
{
generate the next configuration
if queens don’t attack in this configuration then
{
print this configuration;
}
}

Time Complexity: O(N*NCN)
Auxiliary Space: O(N2)

## Algorithm for N queen problem:

Following is the backtracking algorithm for solving the N-Queen problem

1. Initialize an empty chessboard of size NxN.
2. Start with the leftmost column and place a queen in the first row of that column.
3. Move to the next column and place a queen in the first row of that column.
4. Repeat step 3 until either all N queens have been placed or it is impossible to place a queen in the current column without violating the rules of the problem.
5. If all N queens have been placed, print the solution.
6. If it is not possible to place a queen in the current column without violating the rules of the problem, backtrack to the previous column.
7. Remove the queen from the previous column and move it down one row.
8. Repeat steps 4-7 until all possible configurations have been tried.

Pseudo-code implementation:

function solveNQueens(board, col, n):
if col >= n:
print board
return true
for row from 0 to n-1:
if isSafe(board, row, col, n):
board[row][col] = 1
if solveNQueens(board, col+1, n):
return true
board[row][col] = 0
return false

function isSafe(board, row, col, n):
for i from 0 to col-1:
if board[row][i] == 1:
return false
for i,j from row-1, col-1 to 0, 0 by -1:
if board[i][j] == 1:
return false
for i,j from row+1, col-1 to n-1, 0 by 1, -1:
if board[i][j] == 1:
return false
return true

board = empty NxN chessboard
solveNQueens(board, 0, N)

### Backtracking Algorithm by placing queens in columns:

The idea is to place queens one by one in different columns, starting from the leftmost column. When we place a queen in a column, we check for clashes with already placed queens. In the current column, if we find a row for which there is no clash, we mark this row and column as part of the solution. If we do not find such a row due to clashes, then we backtrack and return false.

Follow the steps mentioned below to implement the idea:

• Start in the leftmost column
• If all queens are placed return true
• Try all rows in the current column. Do the following for every tried row.
• If the queen can be placed safely in this row
• Then mark this [row, column] as part of the solution and recursively check if placing queen here leads to a solution.
• If placing the queen in [row, column] leads to a solution then return true.
• If placing queen doesn’t lead to a solution then unmark this [row, column] and track back and try other rows.
• If all rows have been tried and nothing worked return false to trigger backtracking.

Implementation of the above backtracking solution:

## C

 `// C program to solve N Queen Problem using backtracking` `#define N 4``#include ``#include ` `// A utility function to print solution``void` `printSolution(``int` `board[N][N])``{``    ``for` `(``int` `i = 0; i < N; i++) {``        ``for` `(``int` `j = 0; j < N; j++) {``            ``if``(board[i][j])``                ``printf``(``"Q "``);``            ``else``                ``printf``(``". "``);``        ``}``        ``printf``(``"\n"``);``    ``}``}` `// A utility function to check if a queen can``// be placed on board[row][col]. Note that this``// function is called when "col" queens are``// already placed in columns from 0 to col -1.``// So we need to check only left side for``// attacking queens``bool` `isSafe(``int` `board[N][N], ``int` `row, ``int` `col)``{``    ``int` `i, j;` `    ``// Check this row on left side``    ``for` `(i = 0; i < col; i++)``        ``if` `(board[row][i])``            ``return` `false``;` `    ``// Check upper diagonal on left side``    ``for` `(i = row, j = col; i >= 0 && j >= 0; i--, j--)``        ``if` `(board[i][j])``            ``return` `false``;` `    ``// Check lower diagonal on left side``    ``for` `(i = row, j = col; j >= 0 && i < N; i++, j--)``        ``if` `(board[i][j])``            ``return` `false``;` `    ``return` `true``;``}` `// A recursive utility function to solve N``// Queen problem``bool` `solveNQUtil(``int` `board[N][N], ``int` `col)``{``    ``// Base case: If all queens are placed``    ``// then return true``    ``if` `(col >= N)``        ``return` `true``;` `    ``// Consider this column and try placing``    ``// this queen in all rows one by one``    ``for` `(``int` `i = 0; i < N; i++) {``        ` `        ``// Check if the queen can be placed on``        ``// board[i][col]``        ``if` `(isSafe(board, i, col)) {``            ` `            ``// Place this queen in board[i][col]``            ``board[i][col] = 1;` `            ``// Recur to place rest of the queens``            ``if` `(solveNQUtil(board, col + 1))``                ``return` `true``;` `            ``// If placing queen in board[i][col]``            ``// doesn't lead to a solution, then``            ``// remove queen from board[i][col]``            ``board[i][col] = 0; ``// BACKTRACK``        ``}``    ``}` `    ``// If the queen cannot be placed in any row in``    ``// this column col  then return false``    ``return` `false``;``}` `// This function solves the N Queen problem using``// Backtracking. It mainly uses solveNQUtil() to``// solve the problem. It returns false if queens``// cannot be placed, otherwise, return true and``// prints placement of queens in the form of 1s.``// Please note that there may be more than one``// solutions, this function prints one  of the``// feasible solutions.``bool` `solveNQ()``{``    ``int` `board[N][N] = { { 0, 0, 0, 0 },``                        ``{ 0, 0, 0, 0 },``                        ``{ 0, 0, 0, 0 },``                        ``{ 0, 0, 0, 0 } };` `    ``if` `(solveNQUtil(board, 0) == ``false``) {``        ``printf``(``"Solution does not exist"``);``        ``return` `false``;``    ``}` `    ``printSolution(board);``    ``return` `true``;``}` `// Driver program to test above function``int` `main()``{``    ``solveNQ();``    ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## C++

 `// C++ program to solve N Queen Problem using backtracking` `#include ``#define N 4``using` `namespace` `std;` `// A utility function to print solution``void` `printSolution(``int` `board[N][N])``{``    ``for` `(``int` `i = 0; i < N; i++) {``        ``for` `(``int` `j = 0; j < N; j++)``           ``if``(board[i][j])``            ``cout << ``"Q "``;``           ``else` `cout<<``". "``;``        ``printf``(``"\n"``);``    ``}``}` `// A utility function to check if a queen can``// be placed on board[row][col]. Note that this``// function is called when "col" queens are``// already placed in columns from 0 to col -1.``// So we need to check only left side for``// attacking queens``bool` `isSafe(``int` `board[N][N], ``int` `row, ``int` `col)``{``    ``int` `i, j;` `    ``// Check this row on left side``    ``for` `(i = 0; i < col; i++)``        ``if` `(board[row][i])``            ``return` `false``;` `    ``// Check upper diagonal on left side``    ``for` `(i = row, j = col; i >= 0 && j >= 0; i--, j--)``        ``if` `(board[i][j])``            ``return` `false``;` `    ``// Check lower diagonal on left side``    ``for` `(i = row, j = col; j >= 0 && i < N; i++, j--)``        ``if` `(board[i][j])``            ``return` `false``;` `    ``return` `true``;``}` `// A recursive utility function to solve N``// Queen problem``bool` `solveNQUtil(``int` `board[N][N], ``int` `col)``{``    ``// base case: If all queens are placed``    ``// then return true``    ``if` `(col >= N)``        ``return` `true``;` `    ``// Consider this column and try placing``    ``// this queen in all rows one by one``    ``for` `(``int` `i = 0; i < N; i++) {``        ` `        ``// Check if the queen can be placed on``        ``// board[i][col]``        ``if` `(isSafe(board, i, col)) {``            ` `            ``// Place this queen in board[i][col]``            ``board[i][col] = 1;` `            ``// recur to place rest of the queens``            ``if` `(solveNQUtil(board, col + 1))``                ``return` `true``;` `            ``// If placing queen in board[i][col]``            ``// doesn't lead to a solution, then``            ``// remove queen from board[i][col]``            ``board[i][col] = 0; ``// BACKTRACK``        ``}``    ``}` `    ``// If the queen cannot be placed in any row in``    ``// this column col  then return false``    ``return` `false``;``}` `// This function solves the N Queen problem using``// Backtracking. It mainly uses solveNQUtil() to``// solve the problem. It returns false if queens``// cannot be placed, otherwise, return true and``// prints placement of queens in the form of 1s.``// Please note that there may be more than one``// solutions, this function prints one  of the``// feasible solutions.``bool` `solveNQ()``{``    ``int` `board[N][N] = { { 0, 0, 0, 0 },``                        ``{ 0, 0, 0, 0 },``                        ``{ 0, 0, 0, 0 },``                        ``{ 0, 0, 0, 0 } };` `    ``if` `(solveNQUtil(board, 0) == ``false``) {``        ``cout << ``"Solution does not exist"``;``        ``return` `false``;``    ``}` `    ``printSolution(board);``    ``return` `true``;``}` `// Driver program to test above function``int` `main()``{``    ``solveNQ();``    ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// Java program to solve N Queen Problem using backtracking` `public` `class` `NQueenProblem {``    ``final` `int` `N = ``4``;` `    ``// A utility function to print solution``    ``void` `printSolution(``int` `board[][])``    ``{``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``for` `(``int` `j = ``0``; j < N; j++) {``                ``if` `(board[i][j] == ``1``)``                    ``System.out.print(``"Q "``);``                ``else``                    ``System.out.print(``". "``);``            ``}``            ``System.out.println();``        ``}``    ``}` `    ``// A utility function to check if a queen can``    ``// be placed on board[row][col]. Note that this``    ``// function is called when "col" queens are already``    ``// placeed in columns from 0 to col -1. So we need``    ``// to check only left side for attacking queens``    ``boolean` `isSafe(``int` `board[][], ``int` `row, ``int` `col)``    ``{``        ``int` `i, j;` `        ``// Check this row on left side``        ``for` `(i = ``0``; i < col; i++)``            ``if` `(board[row][i] == ``1``)``                ``return` `false``;` `        ``// Check upper diagonal on left side``        ``for` `(i = row, j = col; i >= ``0` `&& j >= ``0``; i--, j--)``            ``if` `(board[i][j] == ``1``)``                ``return` `false``;` `        ``// Check lower diagonal on left side``        ``for` `(i = row, j = col; j >= ``0` `&& i < N; i++, j--)``            ``if` `(board[i][j] == ``1``)``                ``return` `false``;` `        ``return` `true``;``    ``}` `    ``// A recursive utility function to solve N``    ``// Queen problem``    ``boolean` `solveNQUtil(``int` `board[][], ``int` `col)``    ``{``        ``// Base case: If all queens are placed``        ``// then return true``        ``if` `(col >= N)``            ``return` `true``;` `        ``// Consider this column and try placing``        ``// this queen in all rows one by one``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ` `            ``// Check if the queen can be placed on``            ``// board[i][col]``            ``if` `(isSafe(board, i, col)) {``                ` `                ``// Place this queen in board[i][col]``                ``board[i][col] = ``1``;` `                ``// Recur to place rest of the queens``                ``if` `(solveNQUtil(board, col + ``1``) == ``true``)``                    ``return` `true``;` `                ``// If placing queen in board[i][col]``                ``// doesn't lead to a solution then``                ``// remove queen from board[i][col]``                ``board[i][col] = ``0``; ``// BACKTRACK``            ``}``        ``}` `        ``// If the queen can not be placed in any row in``        ``// this column col, then return false``        ``return` `false``;``    ``}` `    ``// This function solves the N Queen problem using``    ``// Backtracking.  It mainly uses solveNQUtil () to``    ``// solve the problem. It returns false if queens``    ``// cannot be placed, otherwise, return true and``    ``// prints placement of queens in the form of 1s.``    ``// Please note that there may be more than one``    ``// solutions, this function prints one of the``    ``// feasible solutions.``    ``boolean` `solveNQ()``    ``{``        ``int` `board[][] = { { ``0``, ``0``, ``0``, ``0` `},``                          ``{ ``0``, ``0``, ``0``, ``0` `},``                          ``{ ``0``, ``0``, ``0``, ``0` `},``                          ``{ ``0``, ``0``, ``0``, ``0` `} };` `        ``if` `(solveNQUtil(board, ``0``) == ``false``) {``            ``System.out.print(``"Solution does not exist"``);``            ``return` `false``;``        ``}` `        ``printSolution(board);``        ``return` `true``;``    ``}` `    ``// Driver program to test above function``    ``public` `static` `void` `main(String args[])``    ``{``        ``NQueenProblem Queen = ``new` `NQueenProblem();``        ``Queen.solveNQ();``    ``}``}``// This code is contributed by Abhishek Shankhadhar`

## Python3

 `# Python3 program to solve N Queen``# Problem using backtracking` `global` `N``N ``=` `4`  `def` `printSolution(board):``    ``for` `i ``in` `range``(N):``        ``for` `j ``in` `range``(N):``            ``if` `board[i][j] ``=``=` `1``:``                ``print``(``"Q"``,end``=``" "``)``            ``else``:``                ``print``(``"."``,end``=``" "``)``        ``print``()`  `# A utility function to check if a queen can``# be placed on board[row][col]. Note that this``# function is called when "col" queens are``# already placed in columns from 0 to col -1.``# So we need to check only left side for``# attacking queens``def` `isSafe(board, row, col):` `    ``# Check this row on left side``    ``for` `i ``in` `range``(col):``        ``if` `board[row][i] ``=``=` `1``:``            ``return` `False` `    ``# Check upper diagonal on left side``    ``for` `i, j ``in` `zip``(``range``(row, ``-``1``, ``-``1``),``                    ``range``(col, ``-``1``, ``-``1``)):``        ``if` `board[i][j] ``=``=` `1``:``            ``return` `False` `    ``# Check lower diagonal on left side``    ``for` `i, j ``in` `zip``(``range``(row, N, ``1``),``                    ``range``(col, ``-``1``, ``-``1``)):``        ``if` `board[i][j] ``=``=` `1``:``            ``return` `False` `    ``return` `True`  `def` `solveNQUtil(board, col):` `    ``# Base case: If all queens are placed``    ``# then return true``    ``if` `col >``=` `N:``        ``return` `True` `    ``# Consider this column and try placing``    ``# this queen in all rows one by one``    ``for` `i ``in` `range``(N):` `        ``if` `isSafe(board, i, col):` `            ``# Place this queen in board[i][col]``            ``board[i][col] ``=` `1` `            ``# Recur to place rest of the queens``            ``if` `solveNQUtil(board, col ``+` `1``) ``=``=` `True``:``                ``return` `True` `            ``# If placing queen in board[i][col``            ``# doesn't lead to a solution, then``            ``# queen from board[i][col]``            ``board[i][col] ``=` `0` `    ``# If the queen can not be placed in any row in``    ``# this column col then return false``    ``return` `False`  `# This function solves the N Queen problem using``# Backtracking. It mainly uses solveNQUtil() to``# solve the problem. It returns false if queens``# cannot be placed, otherwise return true and``# placement of queens in the form of 1s.``# note that there may be more than one``# solutions, this function prints one of the``# feasible solutions.``def` `solveNQ():``    ``board ``=` `[[``0``, ``0``, ``0``, ``0``],``             ``[``0``, ``0``, ``0``, ``0``],``             ``[``0``, ``0``, ``0``, ``0``],``             ``[``0``, ``0``, ``0``, ``0``]]` `    ``if` `solveNQUtil(board, ``0``) ``=``=` `False``:``        ``print``(``"Solution does not exist"``)``        ``return` `False` `    ``printSolution(board)``    ``return` `True`  `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``solveNQ()` `# This code is contributed by Divyanshu Mehta`

## C#

 `// C# program to solve N Queen Problem``// using backtracking``using` `System;``    ` `class` `GFG``{``    ``readonly` `int` `N = 4;` `    ``// A utility function to print solution``    ``void` `printSolution(``int` `[,]board)``    ``{``        ``for` `(``int` `i = 0; i < N; i++)``        ``{``            ``for` `(``int` `j = 0; j < N; j++)``            ``{``                ``if` `(board[i, j] == 1)``                    ``Console.Write(``"Q "``);``                ``else``                    ``Console.Write(``". "``);``            ``}``            ``Console.WriteLine();``        ``}``    ``}` `    ``// A utility function to check if a queen can``    ``// be placed on board[row,col]. Note that this``    ``// function is called when "col" queens are already``    ``// placeed in columns from 0 to col -1. So we need``    ``// to check only left side for attacking queens``    ``bool` `isSafe(``int` `[,]board, ``int` `row, ``int` `col)``    ``{``        ``int` `i, j;` `        ``// Check this row on left side``        ``for` `(i = 0; i < col; i++)``            ``if` `(board[row,i] == 1)``                ``return` `false``;` `        ``// Check upper diagonal on left side``        ``for` `(i = row, j = col; i >= 0 &&``             ``j >= 0; i--, j--)``            ``if` `(board[i,j] == 1)``                ``return` `false``;` `        ``// Check lower diagonal on left side``        ``for` `(i = row, j = col; j >= 0 &&``                      ``i < N; i++, j--)``            ``if` `(board[i, j] == 1)``                ``return` `false``;` `        ``return` `true``;``    ``}` `    ``// A recursive utility function to solve N``    ``// Queen problem``    ``bool` `solveNQUtil(``int` `[,]board, ``int` `col)``    ``{``        ``// Base case: If all queens are placed``        ``// then return true``        ``if` `(col >= N)``            ``return` `true``;` `        ``// Consider this column and try placing``        ``// this queen in all rows one by one``        ``for` `(``int` `i = 0; i < N; i++)``        ``{``            ``// Check if the queen can be placed on``            ``// board[i,col]``            ``if` `(isSafe(board, i, col))``            ``{``                ``// Place this queen in board[i,col]``                ``board[i, col] = 1;` `                ``// Recur to place rest of the queens``                ``if` `(solveNQUtil(board, col + 1) == ``true``)``                    ``return` `true``;` `                ``// If placing queen in board[i,col]``                ``// doesn't lead to a solution then``                ``// remove queen from board[i,col]``                ``board[i, col] = 0; ``// BACKTRACK``            ``}``        ``}` `        ``// If the queen can not be placed in any row in``        ``// this column col, then return false``        ``return` `false``;``    ``}` `    ``// This function solves the N Queen problem using``    ``// Backtracking. It mainly uses solveNQUtil () to``    ``// solve the problem. It returns false if queens``    ``// cannot be placed, otherwise, return true and``    ``// prints placement of queens in the form of 1s.``    ``// Please note that there may be more than one``    ``// solutions, this function prints one of the``    ``// feasible solutions.``    ``bool` `solveNQ()``    ``{``        ``int` `[,]board = {{ 0, 0, 0, 0 },``                        ``{ 0, 0, 0, 0 },``                        ``{ 0, 0, 0, 0 },``                        ``{ 0, 0, 0, 0 }};` `        ``if` `(solveNQUtil(board, 0) == ``false``)``        ``{``            ``Console.Write(``"Solution does not exist"``);``            ``return` `false``;``        ``}` `        ``printSolution(board);``        ``return` `true``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String []args)``    ``{``        ``GFG Queen = ``new` `GFG();``        ``Queen.solveNQ();``    ``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``

Output

```. . Q .
Q . . .
. . . Q
. Q . . ```

Time Complexity: O(N!)
Auxiliary Space: O(N2)

### Backtracking Algorithm by placing queens in rows:

The idea is to place queens one by one in different rows, starting from the topmost row. When we place a queen in a row, we check for clashes with already placed queens. In the current column, if we find a row for which there is no clash, we mark this row and column as part of the solution. If we do not find such a row due to clashes, then we backtrack and return false.

Follow the steps mentioned below to implement the idea:

• Make a recursive function that takes the state of the board and the current row number as its parameter.
• Start in the topmost row.
• If all queens are placed return true
• Try all columns in the current row. Do the following for every tried column.
• If the queen can be placed safely in this column
• Then mark this [row, column] as part of the solution and recursively check if placing queen here leads to a solution.
• If placing the queen in [row, column] leads to a solution then return true.
• If placing queen doesn’t lead to a solution then unmark this [row, column] and track back and try other columns.
• If all columns have been tried and nothing worked return false to trigger backtracking.

Below is the implementation of the above Backtracking solution:

## C++

 `#include ``using` `namespace` `std;` `// Store all the possible answers``vector > answer;` `// Print the board``void` `print_board()``{``    ``for` `(``auto``& str : answer[1]) {``        ``for` `(``auto``& letter : str)``            ``cout << letter << ``" "``;``        ``cout << endl;``    ``}``    ``return``;``}``// We need to check in three directions``// 1. in the same column above the current position``// 2. in the left top diagonal from the given cell``// 3. in the right top diagonal from the given cell``int` `safe(``int` `row, ``int` `col, vector& board)``{``    ``for` `(``int` `i = 0; i < board.size(); i++) {``        ``if` `(board[i][col] == ``'Q'``)``            ``return` `false``;``    ``}``    ``int` `i = row, j = col;``    ``while` `(i >= 0 && j >= 0)``        ``if` `(board[i--][j--] == ``'Q'``)``            ``return` `false``;``    ``i = row, j = col;``    ``while` `(i >= 0 && j < board.size())``        ``if` `(board[i--][j++] == ``'Q'``)``            ``return` `false``;``    ``return` `true``;``}``// rec function here will fill the queens``// 1. there can be only one queen in one row``// 2. if we filled the final row in the board then row will``// be equal to total number of rows in board``// 3. push that board configuration in answer set because``// there will be more than one answers for filling the board``// with n-queens``void` `rec(vector board, ``int` `row)``{``    ``if` `(row == board.size()) {``        ``answer.push_back(board);``        ``return``;``    ``}``    ``for` `(``int` `i = 0; i < board.size(); i++) {``        ` `        ``// For each position check if it is safe and if it``        ``// safe make a recursive call with``        ``// row+1,board[i][j]='Q' and then revert the change``        ``// in board that is make the board[i][j]='.' again to``        ``// generate more solutions``        ``if` `(safe(row, i, board)) {``            ``board[row][i] = ``'Q'``;``            ``rec(board, row + 1);``            ``board[row][i] = ``'.'``;``        ``}``    ``}``    ``return``;``}``// Function to solve n queens``vector > solveNQueens(``int` `n)``{``    ``string s;``    ``for` `(``int` `i = 0; i < n; i++)``        ``s += ``'.'``;``    ` `    ``// Vector of string will make our board which is``    ``// initially all empty``    ``vector board(n, s);``    ``rec(board, 0);``    ``return` `answer;``}` `// Driver code``int` `main()``{``    ``clock_t` `start, end;``    ``start = ``clock``();``    ``// size 4x4 is taken and we can pass some other``    ``// dimension for chess board as well``    ``cout << solveNQueens(4).size() << endl;``    ``cout << ``"Out of "` `<< answer.size()``         ``<< ``" solutions one is following"` `<< endl;``    ``print_board();``    ` `    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `NQueens {` `    ``// Store all the possible answers``    ``static` `List > answer = ``new` `ArrayList<>();` `    ``// Print the board``    ``static` `void` `print_board()``    ``{``        ``for` `(String str : answer.get(``1``)) {``            ``for` `(Character letter : str.toCharArray())``                ``System.out.print(letter + ``" "``);``            ``System.out.println();``        ``}``        ``return``;``    ``}` `    ``// We need to check in three directions``    ``// 1. in the same column above the current position``    ``// 2. in the left top diagonal from the given cell``    ``// 3. in the right top diagonal from the given cell``    ``static` `boolean` `safe(``int` `row, ``int` `col,``                        ``List board)``    ``{``        ``for` `(``int` `i = ``0``; i < board.size(); i++) {``            ``if` `(board.get(i).charAt(col) == ``'Q'``)``                ``return` `false``;``        ``}``        ``int` `i = row, j = col;``        ``while` `(i >= ``0` `&& j >= ``0``)``            ``if` `(board.get(i--).charAt(j--) == ``'Q'``)``                ``return` `false``;``        ``i = row;``        ``j = col;``        ``while` `(i >= ``0` `&& j < board.size())``            ``if` `(board.get(i--).charAt(j++) == ``'Q'``)``                ``return` `false``;``        ``return` `true``;``    ``}` `    ``// rec function here will fill the queens``    ``// 1. there can be only one queen in one row``    ``// 2. if we filled the final row in the board then row``    ``// will be equal to total number of rows in board``    ``// 3. push that board configuration in answer set``    ``// because there will be more than one answers for``    ``// filling the board with n-queens``    ``static` `void` `rec(List board, ``int` `row)``    ``{``        ``if` `(row == board.size()) {``            ``answer.add(board);``            ``return``;``        ``}``        ``for` `(``int` `i = ``0``; i < board.size(); i++) {` `            ``// For each position check if it is safe and if``            ``// it safe make a recursive call with``            ``// row+1,board[i][j]='Q' and then revert the``            ``// change in board that is make the``            ``// board[i][j]='.' again to generate more``            ``// solutions``            ``if` `(safe(row, i, board)) {``                ``List temp = ``new` `ArrayList<>(board);``                ``temp.set(``                    ``row,``                    ``temp.get(row).substring(``0``, i) + ``"Q"``                        ``+ temp.get(row).substring(i + ``1``));``                ``rec(temp, row + ``1``);``            ``}``        ``}``        ``return``;``    ``}` `    ``// Function to solve n queens``    ``static` `List > solveNQueens(``int` `n)``    ``{``        ``String s``            ``= ``new` `String(``new` `char``[n]).replace(``"\0"``, ``"."``);` `        ``// List of string will make our board which is``        ``// initially all empty``        ``List board = ``new` `ArrayList<>();``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``board.add(s);``        ``rec(board, ``0``);``        ``return` `answer;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Size 4x4 is taken and we can pass some other``        ``// dimension for chess board as well``        ``System.out.println(solveNQueens(``4``).size());``        ``System.out.println(``"Out of "` `+ answer.size()``                           ``+ ``" solutions one is following"``);``        ``print_board();``    ``}``}` `// This code is contributed by surajrasr7277`

## Python3

 `import` `time` `# Print the board``def` `print_board(board, n):``    ``for` `i ``in` `range``(n):``        ``for` `j ``in` `range``(n):``            ``print``(board[i][j], end ``=` `" "``)``        ``print``()` `# Joining '.' and 'Q'``# making combined 2D Array``# For output in desired format``def` `add_sol(board, ans, n):``    ``temp ``=` `[]``    ``for` `i ``in` `range``(n):``        ``string ``=` `""``        ``for` `j ``in` `range``(n):``            ``string ``+``=` `board[i][j]``        ``temp.append(string)``    ``ans.append(temp)``    ` `    ` `# We need to check in three directions``# 1. in the same column above the current position``# 2. in the left top diagonal from the given cell``# 3. in the right top diagonal from the given cell``def`  `is_safe(row, col, board, n):``    ``x ``=` `row``    ``y ``=` `col``    ` `    ``# Check for same upper col``    ``while``(x>``=``0``):``        ``if` `board[x][y] ``=``=` `"Q"``:``            ``return` `False``        ``else``:``            ``x ``-``=` `1``            ` `    ``# Check for Upper Right Diagonal``    ``x ``=` `row``    ``y ``=` `col``    ``while``(y``=``0``):``        ``if` `board[x][y] ``=``=` `"Q"``:``            ``return` `False``        ``else``:``            ``y ``+``=` `1``            ``x ``-``=` `1``            ` `    ``# Check for Upper Left diagonal``    ``x ``=` `row``    ``y ``=` `col``    ``while``(y>``=``0` `and` `x>``=``0``):``        ``if` `board[x][y] ``=``=` `"Q"``:``            ``return` `False``        ``else``:``            ``x ``-``=` `1``            ``y ``-``=` `1``    ``return` `True`     `# Function to solve n queens``# solveNQueens function here will fill the queens``# 1. there can be only one queen in one row``# 2. if we filled the final row in the board then row will``# be equal to total number of rows in board``# 3. push that board configuration in answer set because``# there will be more than one answers for filling the board``# with n-queens``def` `solveNQueens(row, ans, board, n):``    ` `    ``# Base Case``    ``# Queen is depicted by "Q"``    ``# adding solution to final answer array``    ``if` `row ``=``=` `n:``        ``add_sol(board, ans, n)``        ``return``    ` `    ``# Solve 1 case and rest recursion will follow``    ``for` `col ``in` `range``(n):``        ` `        ``# For each position check if it is safe and if it``        ``# is safe make a recursive call with``        ``# row+1, board[i][j]='Q' and then revert the change``        ``# in board that is make the board[i][j]='.' again to``        ``# generate more solutions``        ``if` `is_safe(row, col, board, n):``            ` `            ``# If placing Queen is safe``            ``board[row][col] ``=` `"Q"``            ``solveNQueens(row``+``1``, ans, board, n)``            ` `            ``# Backtrack``            ``board[row][col] ``=` `"."`  `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``# Size 4x4 is taken and we can pass some other``    ``# dimension for chess board as well``    ``n ``=` `4``    ` `    ``# 2D array of string will make our board``    ``# which is initially all empty``    ``board ``=` `[[``"."` `for` `i ``in` `range``(n)] ``for` `j ``in` `range``(n)]``    ` `    ``# Store all the possible answers``    ``ans ``=` `[]``    ``solveNQueens(``0``, ans, board, n)``    ` `    ``if` `ans ``=``=` `[]:``        ``print``(``"Solution does not exist"``)``    ``else``:``        ``print``(``len``(ans))``        ``print``(f``"Out Of {len(ans)} solutions one is following"``)``        ``print_board(ans[``0``], n)``        ` `    ``# This code is contributed by Priyank Namdeo``       `

## C#

 `// C# program implementation of above approach``using` `System;``using` `System.Collections.Generic;` `namespace` `NQueensProblem {``class` `Program {``    ` `    ``// Store all the possible answers``    ``static` `List > answer``        ``= ``new` `List >();` `    ``// Print the board``    ``static` `void` `PrintBoard()``    ``{``        ``foreach``(``var` `str ``in` `answer[1])``        ``{``            ``foreach``(``var` `letter ``in` `str)``                ``Console.Write(letter + ``" "``);``            ``Console.WriteLine();``        ``}``    ``}` `    ``// We need to check in three directions``    ``// 1. in the same column above the current position``    ``// 2. in the left top diagonal from the given cell``    ``// 3. in the right top diagonal from the given cell``    ``static` `bool` `Safe(``int` `row, ``int` `col, List<``string``> board)``    ``{``        ``for` `(``int` `i = 0; i < board.Count; i++) {``            ``if` `(board[i][col] == ``'Q'``)``                ``return` `false``;``        ``}``        ``int` `x = row, y = col;``        ``while` `(x >= 0 && y >= 0)``            ``if` `(board[x--][y--] == ``'Q'``)``                ``return` `false``;``        ``x = row;``        ``y = col;``        ``while` `(x >= 0 && y < board.Count)``            ``if` `(board[x--][y++] == ``'Q'``)``                ``return` `false``;``        ``return` `true``;``    ``}` `    ``// rec function here will fill the queens``    ``// 1. there can be only one queen in one row``    ``// 2. if we filled the final row in the board then row``    ``// will be equal to total number of rows in board``    ``// 3. push that board configuration in answer set``    ``// because there will be more than one answers for``    ``// filling the board with n-queens``    ``static` `void` `Rec(List<``string``> board, ``int` `row)``    ``{``        ``if` `(row == board.Count) {``            ``answer.Add(``new` `List<``string``>(board));``            ``return``;``        ``}``        ``for` `(``int` `i = 0; i < board.Count; i++) {` `            ``// For each position check if it is safe and if``            ``// it``            ``// safe make a recursive call with``            ``// row+1,board[i][j]='Q' and then revert the``            ``// change in board that is make the``            ``// board[i][j]='.' again to generate more``            ``// solutions``            ``if` `(Safe(row, i, board)) {``                ``char``[] rowArr = board[row].ToCharArray();``                ``rowArr[i] = ``'Q'``;``                ``board[row] = ``new` `string``(rowArr);``                ``Rec(board, row + 1);``                ``rowArr[i] = ``'.'``;``                ``board[row] = ``new` `string``(rowArr);``            ``}``        ``}``    ``}` `    ``// Function to solve n queens``    ``static` `List > SolveNQueens(``int` `n)``    ``{``        ``string` `s = ``""``;``        ``for` `(``int` `i = 0; i < n; i++)``            ``s += ``'.'``;` `        ``// List of string will make our board which is``        ``// initially all empty``        ``List<``string``> board = ``new` `List<``string``>();``        ``for` `(``int` `i = 0; i < n; i++)``            ``board.Add(s);``        ``Rec(board, 0);``        ``return` `answer;``    ``}` `    ``static` `void` `Main(``string``[] args)``    ``{``        ``// Size 4x4 is taken and we can pass some other``        ``// dimension for chess board as well``        ``Console.WriteLine(SolveNQueens(4).Count);``        ``Console.WriteLine(``"Out of "` `+ answer.Count``                          ``+ ``" solutions one is following"``);``        ``PrintBoard();``    ``}``}``}`

## Javascript

 `// store all the possible answers``let answer = [];` `// Print the board``function` `print_board() {``  ``for` `(let str of answer[1]) {``    ``for` `(let letter of str) console.log(letter + ``' '``);``    ``console.log(``'\n'``);``  ``}``}` `// We need to check in three directions``// 1. in the same column above the current position``// 2. in the left top diagonal from the given cell``// 3. in the right top diagonal from the given cell``function` `safe(row, col, board) {``  ``for` `(let i = 0; i < board.length; i++) {``    ``if` `(board[i][col] == ``'Q'``) ``return` `false``;``  ``}``  ``let i = row,``    ``j = col;``  ``while` `(i >= 0 && j >= 0) ``if` `(board[i--][j--] == ``'Q'``) ``return` `false``;``  ``i = row;``  ``j = col;``  ``while` `(i >= 0 && j < board.length) ``if` `(board[i--][j++] == ``'Q'``) ``return` `false``;``  ``return` `true``;``}` `// rec function here will fill the queens``// 1. there can be only one queen in one row``// 2. if we filled the final row in the board then row will``// be equal to total number of rows in board``// 3. push that board configuration in answer set because``// there will be more than one answers for filling the board``// with n-queens``function` `rec(board, row) {``  ``if` `(row == board.length) {``    ``answer.push([...board]);``    ``return``;``  ``}``  ``for` `(let i = 0; i < board.length; i++) {``    ` `    ``// For each position check if it is safe and if it``    ``// safe make a recursive call with``    ``// row+1,board[i][j]='Q' and then revert the change``    ``// in board that is make the board[i][j]='.' again to``    ``// generate more solutions``    ``if` `(safe(row, i, board)) {``      ``board[row] = board[row].substring(0, i) + ``'Q'` `+ board[row].substring(i + 1);``      ``rec(board, row + 1);``      ``board[row] = board[row].substring(0, i) + ``'.'` `+ board[row].substring(i + 1);``    ``}``  ``}``}` `// Function to solve n queens``function` `solveNQueens(n) {``  ``let board = ``new` `Array(n).fill(``new` `Array(n).fill(``'.'``).join(``''``));``  ``rec(board, 0);``  ``return` `answer;``}`  `// Size 4x4 is taken and we can pass some other``// dimension for chess board as well``console.log(solveNQueens(4).length);``console.log(`Out of \${answer.length} solutions, one is following:`);``print_board();`  `// This code is contributed by Prajwal Kandekar`

Output

```2
Out of 2 solutions one is following
. . Q .
Q . . .
. . . Q
. Q . .
```

Time Complexity: O(N!)
Auxiliary Space: O(N2)

### Optimization in is_safe() function:

The idea is not to check every element in the right and left diagonal, instead use the property of diagonals:

• The sum of i and j is constant and unique for each right diagonal, where i is the row of elements and j is the
column of elements.
• The difference between i and j is constant and unique for each left diagonal, where i and j are row and column of element respectively.

Below is the implementation of the Backtracking solution(with optimization):

## C++

 `// C++ program to solve N Queen Problem using backtracking` `#include ``using` `namespace` `std;``#define N 4` `// ld is an array where its indices indicate row-col+N-1``// (N-1) is for shifting the difference to store negative``// indices``int` `ld[30] = { 0 };` `// rd is an array where its indices indicate row+col``// and used to check whether a queen can be placed on``// right diagonal or not``int` `rd[30] = { 0 };` `// Column array where its indices indicates column and``// used to check whether a queen can be placed in that``// row or not*/``int` `cl[30] = { 0 };` `// A utility function to print solution``void` `printSolution(``int` `board[N][N])``{``    ``for` `(``int` `i = 0; i < N; i++) {``        ``for` `(``int` `j = 0; j < N; j++)``            ``cout << ``" "` `<< board[i][j] << ``" "``;``        ``cout << endl;``    ``}``}` `// A recursive utility function to solve N``// Queen problem``bool` `solveNQUtil(``int` `board[N][N], ``int` `col)``{``    ``// Base case: If all queens are placed``    ``// then return true``    ``if` `(col >= N)``        ``return` `true``;` `    ``// Consider this column and try placing``    ``// this queen in all rows one by one``    ``for` `(``int` `i = 0; i < N; i++) {``        ` `        ``// Check if the queen can be placed on``        ``// board[i][col]``        ` `        ``// To check if a queen can be placed on``        ``// board[row][col].We just need to check``        ``// ld[row-col+n-1] and rd[row+coln] where``        ``// ld and rd are for left and right``        ``// diagonal respectively``        ``if` `((ld[i - col + N - 1] != 1 && rd[i + col] != 1)``            ``&& cl[i] != 1) {``            ` `            ``// Place this queen in board[i][col]``            ``board[i][col] = 1;``            ``ld[i - col + N - 1] = rd[i + col] = cl[i] = 1;` `            ``// Recur to place rest of the queens``            ``if` `(solveNQUtil(board, col + 1))``                ``return` `true``;` `            ``// If placing queen in board[i][col]``            ``// doesn't lead to a solution, then``            ``// remove queen from board[i][col]``            ``board[i][col] = 0; ``// BACKTRACK``            ``ld[i - col + N - 1] = rd[i + col] = cl[i] = 0;``        ``}``    ``}` `    ``// If the queen cannot be placed in any row in``    ``// this column col  then return false``    ``return` `false``;``}` `// This function solves the N Queen problem using``// Backtracking. It mainly uses solveNQUtil() to``// solve the problem. It returns false if queens``// cannot be placed, otherwise, return true and``// prints placement of queens in the form of 1s.``// Please note that there may be more than one``// solutions, this function prints one  of the``// feasible solutions.``bool` `solveNQ()``{``    ``int` `board[N][N] = { { 0, 0, 0, 0 },``                        ``{ 0, 0, 0, 0 },``                        ``{ 0, 0, 0, 0 },``                        ``{ 0, 0, 0, 0 } };` `    ``if` `(solveNQUtil(board, 0) == ``false``) {``        ``cout << ``"Solution does not exist"``;``        ``return` `false``;``    ``}` `    ``printSolution(board);``    ``return` `true``;``}` `// Driver program to test above function``int` `main()``{``    ``solveNQ();``    ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// Java program to solve N Queen Problem using backtracking` `import` `java.util.*;` `class` `GFG {``    ``static` `int` `N = ``4``;` `    ``// ld is an array where its indices indicate row-col+N-1``    ``// (N-1) is for shifting the difference to store``    ``// negative indices``    ``static` `int``[] ld = ``new` `int``[``30``];` `    ``// rd is an array where its indices indicate row+col``    ``// and used to check whether a queen can be placed on``    ``// right diagonal or not``    ``static` `int``[] rd = ``new` `int``[``30``];` `    ``// Column array where its indices indicates column and``    ``// used to check whether a queen can be placed in that``    ``// row or not``    ``static` `int``[] cl = ``new` `int``[``30``];` `    ``// A utility function to print solution``    ``static` `void` `printSolution(``int` `board[][])``    ``{``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``for` `(``int` `j = ``0``; j < N; j++)``                ``System.out.printf(``" %d "``, board[i][j]);``            ``System.out.printf(``"\n"``);``        ``}``    ``}` `    ``// A recursive utility function to solve N``    ``// Queen problem``    ``static` `boolean` `solveNQUtil(``int` `board[][], ``int` `col)``    ``{``        ``// Base case: If all queens are placed``        ``// then return true``        ``if` `(col >= N)``            ``return` `true``;` `        ``// Consider this column and try placing``        ``// this queen in all rows one by one``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``// Check if the queen can be placed on``            ``// board[i][col]` `            ``// To check if a queen can be placed on``            ``// board[row][col].We just need to check``            ``// ld[row-col+n-1] and rd[row+coln] where``            ``// ld and rd are for left and right``            ``// diagonal respectively``            ``if` `((ld[i - col + N - ``1``] != ``1``                 ``&& rd[i + col] != ``1``)``                ``&& cl[i] != ``1``) {``                ``// Place this queen in board[i][col]``                ``board[i][col] = ``1``;``                ``ld[i - col + N - ``1``] = rd[i + col] = cl[i]``                    ``= ``1``;` `                ``// Recur to place rest of the queens``                ``if` `(solveNQUtil(board, col + ``1``))``                    ``return` `true``;` `                ``// If placing queen in board[i][col]``                ``// doesn't lead to a solution, then``                ``// remove queen from board[i][col]``                ``board[i][col] = ``0``; ``// BACKTRACK``                ``ld[i - col + N - ``1``] = rd[i + col] = cl[i]``                    ``= ``0``;``            ``}``        ``}` `        ``// If the queen cannot be placed in any row in``        ``// this column col then return false``        ``return` `false``;``    ``}` `    ``// This function solves the N Queen problem using``    ``// Backtracking. It mainly uses solveNQUtil() to``    ``// solve the problem. It returns false if queens``    ``// cannot be placed, otherwise, return true and``    ``// prints placement of queens in the form of 1s.``    ``// Please note that there may be more than one``    ``// solutions, this function prints one of the``    ``// feasible solutions.``    ``static` `boolean` `solveNQ()``    ``{``        ``int` `board[][] = { { ``0``, ``0``, ``0``, ``0` `},``                          ``{ ``0``, ``0``, ``0``, ``0` `},``                          ``{ ``0``, ``0``, ``0``, ``0` `},``                          ``{ ``0``, ``0``, ``0``, ``0` `} };` `        ``if` `(solveNQUtil(board, ``0``) == ``false``) {``            ``System.out.printf(``"Solution does not exist"``);``            ``return` `false``;``        ``}` `        ``printSolution(board);``        ``return` `true``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``solveNQ();``    ``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 program to solve N Queen Problem using``# backtracking``N ``=` `4` `# ld is an array where its indices indicate row-col+N-1``# (N-1) is for shifting the difference to store negative``# indices``ld ``=` `[``0``] ``*` `30` `# rd is an array where its indices indicate row+col``# and used to check whether a queen can be placed on``# right diagonal or not``rd ``=` `[``0``] ``*` `30` `# Column array where its indices indicates column and``# used to check whether a queen can be placed in that``# row or not``cl ``=` `[``0``] ``*` `30`  `# A utility function to print solution``def` `printSolution(board):``    ``for` `i ``in` `range``(N):``        ``for` `j ``in` `range``(N):``            ``print``(board[i][j], end``=``" "``)``        ``print``()`  `# A recursive utility function to solve N``# Queen problem``def` `solveNQUtil(board, col):` `    ``# Base case: If all queens are placed``    ``# then return True``    ``if` `(col >``=` `N):``        ``return` `True` `    ``# Consider this column and try placing``    ``# this queen in all rows one by one``    ``for` `i ``in` `range``(N):` `        ``# Check if the queen can be placed on board[i][col]` `        ``# To check if a queen can be placed on``        ``# board[row][col] We just need to check``        ``# ld[row-col+n-1] and rd[row+coln]``        ``# where ld and rd are for left and``        ``# right diagonal respectively``        ``if` `((ld[i ``-` `col ``+` `N ``-` `1``] !``=` `1` `and``             ``rd[i ``+` `col] !``=` `1``) ``and` `cl[i] !``=` `1``):` `            ``# Place this queen in board[i][col]``            ``board[i][col] ``=` `1``            ``ld[i ``-` `col ``+` `N ``-` `1``] ``=` `rd[i ``+` `col] ``=` `cl[i] ``=` `1` `            ``# Recur to place rest of the queens``            ``if` `(solveNQUtil(board, col ``+` `1``)):``                ``return` `True` `            ``# If placing queen in board[i][col]``            ``# doesn't lead to a solution,``            ``# then remove queen from board[i][col]``            ``board[i][col] ``=` `0`  `# BACKTRACK``            ``ld[i ``-` `col ``+` `N ``-` `1``] ``=` `rd[i ``+` `col] ``=` `cl[i] ``=` `0` `            ``# If the queen cannot be placed in``            ``# any row in this column col then return False``    ``return` `False`  `# This function solves the N Queen problem using``# Backtracking. It mainly uses solveNQUtil() to``# solve the problem. It returns False if queens``# cannot be placed, otherwise, return True and``# prints placement of queens in the form of 1s.``# Please note that there may be more than one``# solutions, this function prints one of the``# feasible solutions.``def` `solveNQ():``    ``board ``=` `[[``0``, ``0``, ``0``, ``0``],``             ``[``0``, ``0``, ``0``, ``0``],``             ``[``0``, ``0``, ``0``, ``0``],``             ``[``0``, ``0``, ``0``, ``0``]]``    ``if` `(solveNQUtil(board, ``0``) ``=``=` `False``):``        ``printf(``"Solution does not exist"``)``        ``return` `False``    ``printSolution(board)``    ``return` `True`  `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``solveNQ()` `# This code is contributed by SHUBHAMSINGH10`

## C#

 `// C# program to solve N Queen Problem using backtracking` `using` `System;` `class` `GFG {``    ``static` `int` `N = 4;` `    ``// ld is an array where its indices indicate row-col+N-1``    ``// (N-1) is for shifting the difference to store``    ``// negative indices``    ``static` `int``[] ld = ``new` `int``[30];` `    ``// rd is an array where its indices indicate row+col``    ``// and used to check whether a queen can be placed on``    ``// right diagonal or not``    ``static` `int``[] rd = ``new` `int``[30];` `    ``// Column array where its indices indicates column and``    ``// used to check whether a queen can be placed in that``    ``// row or not``    ``static` `int``[] cl = ``new` `int``[30];` `    ``// A utility function to print solution``    ``static` `void` `printSolution(``int``[, ] board)``    ``{``        ``for` `(``int` `i = 0; i < N; i++) {``            ``for` `(``int` `j = 0; j < N; j++)``                ``Console.Write(``" {0} "``, board[i, j]);``            ``Console.Write(``"\n"``);``        ``}``    ``}` `    ``// A recursive utility function to solve N``    ``// Queen problem``    ``static` `bool` `solveNQUtil(``int``[, ] board, ``int` `col)``    ``{``        ``// Base case: If all queens are placed``        ``// then return true``        ``if` `(col >= N)``            ``return` `true``;` `        ``// Consider this column and try placing``        ``// this queen in all rows one by one``        ``for` `(``int` `i = 0; i < N; i++) {``            ``// Check if the queen can be placed on``            ``// board[i,col]` `            ``// To check if a queen can be placed on``            ``// board[row,col].We just need to check``            ``// ld[row-col+n-1] and rd[row+coln] where``            ``// ld and rd are for left and right``            ``// diagonal respectively``            ``if` `((ld[i - col + N - 1] != 1``                 ``&& rd[i + col] != 1)``                ``&& cl[i] != 1) {``                ``// Place this queen in board[i,col]``                ``board[i, col] = 1;``                ``ld[i - col + N - 1] = rd[i + col] = cl[i]``                    ``= 1;` `                ``// Recur to place rest of the queens``                ``if` `(solveNQUtil(board, col + 1))``                    ``return` `true``;` `                ``// If placing queen in board[i,col]``                ``// doesn't lead to a solution, then``                ``// remove queen from board[i,col]``                ``board[i, col] = 0; ``// BACKTRACK``                ``ld[i - col + N - 1] = rd[i + col] = cl[i]``                    ``= 0;``            ``}``        ``}` `        ``// If the queen cannot be placed in any row in``        ``// this column col then return false``        ``return` `false``;``    ``}` `    ``// This function solves the N Queen problem using``    ``// Backtracking. It mainly uses solveNQUtil() to``    ``// solve the problem. It returns false if queens``    ``// cannot be placed, otherwise, return true and``    ``// prints placement of queens in the form of 1s.``    ``// Please note that there may be more than one``    ``// solutions, this function prints one of the``    ``// feasible solutions.``    ``static` `bool` `solveNQ()``    ``{``        ``int``[, ] board = { { 0, 0, 0, 0 },``                          ``{ 0, 0, 0, 0 },``                          ``{ 0, 0, 0, 0 },``                          ``{ 0, 0, 0, 0 } };` `        ``if` `(solveNQUtil(board, 0) == ``false``) {``            ``Console.Write(``"Solution does not exist"``);``            ``return` `false``;``        ``}` `        ``printSolution(board);``        ``return` `true``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``solveNQ();``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

``` 0  0  1  0
1  0  0  0
0  0  0  1
0  1  0  0 ```

Time Complexity: O(N!)
Auxiliary Space: O(N)

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