 Open in App
Not now

# N Queen Problem | Backtracking-3

• Difficulty Level : Hard
• Last Updated : 03 Feb, 2023

We have discussed Knight’s tour and Rat in a Maze problem in Set 1 and Set 2 respectively. Let us discuss N Queen as another example problem that can be solved using backtracking.
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. The expected output is in form of a matrix that has ‘Q’s for the blocks where queens are placed and the empty spaces  are represented by ‘.’s . 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;
}
}```

Backtracking Algorithm Method 1:
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.

```Method 1:
1) Start in the leftmost column
2) If all queens are placed
return true
3) Try all rows in the current column.
Do following for every tried row.
a) 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.
b) If placing the queen in [row, column] leads to
a solution then return true.
c) If placing queen doesn't lead to a solution then
unmark this [row, column] (Backtrack) and go to
step (a) to try other rows.
4) If all rows have been tried and nothing worked,
return false to trigger backtracking.```

Implementation of Backtracking solution by method 1:

## 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)`

## 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++)` `            ``printf``(``" %d "``, board[i][j]);` `        ``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)`

## 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++)` `                ``System.out.print(``" "` `+ board[i][j]` `                                 ``+ ``" "``);` `            ``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):` `            ``print``(board[i][j], 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` `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++)` `                ``Console.Write(``" "` `+ board[i, j]` `                                  ``+ ``" "``);` `            ``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 Method 2:
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.

Method 2:
0) Make a board, make a space to collect all solution states.
1) Start in the topmost row.
2) Make a recursive function which takes state of board and the current row number
as its parameter.
3) Fill a queen in a safe place and use this state of board to advance to next recursive
call, add 1 to the current row. Revert the state of board after making the call.
a) Safe function checks the current column, left top diagonal and right top diagonal.
b) If no queen is present then fill else return false and stop exploring that state
and track back to the next possible solution state
4) Keep calling the function till the current row is out of bound.
5) If current row reaches the number of rows in the board then the board is filled.
6) Store the state and return.

Implementation of Backtracking solution by Method 2:

## C++

 `#include ` `using` `namespace` `std;` `// store all the possible answers` `vector > answer;` `// print the board` `void` `print_board()` `{` `    ``for` `(``auto``& str : answer) {` `        ``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;` `}` `int` `main()` `{` `    ``clock_t` `start, end; ``// this is to calculate the` `                        ``// execution time for n-queens` `    ``start = ``clock``();` `    ``// size 4x4 is taken and we can pass some other` `    ``// dimension for chess board as well` `    ``cout << solveNQueens(4).size() << endl;` `    ``end = ``clock``();` `    ``double` `time_taken` `        ``= ``double``(end - start) / ``double``(CLOCKS_PER_SEC);` `    ``cout << time_taken << ``" time was taken(in miliseconds)"` `         ``<< endl;` `    ``cout << ``"Out of "` `<< answer.size()` `         ``<< ``" solutions one is following"` `<< endl;` `    ``print_board();` `}`

## 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"``, ``"."``);` `      `  `        ``// vector 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;` `    ``}`   `    ``public` `static` `void` `main(String[] args) {` `        ``long` `start, end; ` `      `  `      ``// this is to calculate the` `        ``// execution time for n-queens` `        ``start = System.currentTimeMillis();` `      `  `        ``// size 4x4 is taken and we can pass some other` `        ``// dimension for chess board as well` `        ``System.out.println(solveNQueens(``4``).size());` `        ``end = System.currentTimeMillis();` `        ``double` `time_taken = (end - start);` `        ``System.out.println(time_taken + ``" time was taken(in miliseconds)"``);` `        ``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 ``=` `[]` `    ``start ``=` `time.time()` `    `  `    ``solveNQueens(``0``, ans, board, n)` `    ``end ``=` `time.time()` `    ``time_taken ``=` `end ``-` `start` `    `  `    ``if` `ans ``=``=` `[]:` `        ``print``(``"Solution does not exist"``)` `    ``else``:` `        ``print``(``len``(ans))` `        ``print``(f``"{time_taken:.06f} time was taken(in miliseconds)"``)` `        ``print``(f``"Out Of {len(ans)} solutions one is following"``)` `        ``print_board(ans[``0``], n)` `        `  `    ``# This code is contributed by Priyank Namdeo` `       `

Output

```2
0.000107 time was taken(in miliseconds)
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 right and left diagonal, instead use the property of diagonals:
1. 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.
2. 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.
Implementation of 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 = { 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 = { 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 = { 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<= 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] */` `        ``/* A 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)`

## 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++)` `            ``printf``(``" %d "``, board[i][j]);` `        ``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)`

## 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] */` `        ``/* A 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] """` `        ``""" A 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` `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``;`   `/* 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``;`   `/*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``;`   `/* 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] */` `        ``/* A 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)

Printing all solutions in N-Queen Problem
Sources:
http://see.stanford.edu/materials/icspacs106b/H19-RecBacktrackExamples.pdf
http://en.literateprograms.org/Eight_queens_puzzle_%28C%29
http://en.wikipedia.org/wiki/Eight_queens_puzzle