304 North Cardinal St.
Dorchester Center, MA 02124

# N-Queens II LeetCode Solution

## Problem – N-Queens II

The n-queens puzzle is the problem of placing n queens on an n x n chessboard such that no two queens attack each other.

Given an integer n, return the number of distinct solutions to the n-queens puzzle.

Example 1:

Input: n = 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown.

Example 2:

Input: n = 1
Output: 1

Constraints:

• 1 <= n <= 9

### N-Queens II LeetCode Solution in C++

class Solution {
public:
int totalNQueens(int n) {
vector<bool> col(n), diag(2*n-1), anti_diag(2*n-1);
return solve(col, diag, anti_diag, 0);
}

int solve(vector<bool>& col, vector<bool>& diag, vector<bool>& anti_diag, int row) {
int n = size(col), count = 0;
if(row == n) return 1;
for(int column = 0; column < n; column++)
if(!col[column] && !diag[row + column] && !anti_diag[row - column + n - 1]){
col[column] = diag[row + column] = anti_diag[row - column + n - 1] = true;
count += solve(col, diag, anti_diag, row + 1);
col[column] = diag[row + column] = anti_diag[row - column + n - 1] = false;
}
return count;
}
};

### N-Queens II LeetCode Solution in Java

/**
* don't need to actually place the queen,
* instead, for each row, try to place without violation on
* col/ diagonal1/ diagnol2.
* trick: to detect whether 2 positions sit on the same diagnol:
* if delta(col, row) equals, same diagnol1;
* if sum(col, row) equals, same diagnal2.
*/
private final Set<Integer> occupiedCols = new HashSet<Integer>();
private final Set<Integer> occupiedDiag1s = new HashSet<Integer>();
private final Set<Integer> occupiedDiag2s = new HashSet<Integer>();
public int totalNQueens(int n) {
}

private int totalNQueensHelper(int row, int count, int n) {
for (int col = 0; col < n; col++) {
if (occupiedCols.contains(col))
continue;
int diag1 = row - col;
if (occupiedDiag1s.contains(diag1))
continue;
int diag2 = row + col;
if (occupiedDiag2s.contains(diag2))
continue;
// we can now place a queen here
if (row == n-1)
count++;
else {
count = totalNQueensHelper(row+1, count, n);
// recover
occupiedCols.remove(col);
occupiedDiag1s.remove(diag1);
occupiedDiag2s.remove(diag2);
}
}
return count;
}

### N-Queens II LeetCode Solution in Python

class Solution:
def totalNQueens(self, n):
"""
:type n: int
:rtype: int
"""

diag1 = set()
diag2 = set()
usedCols = set()

return self.helper(n, diag1, diag2, usedCols, 0)

def helper(self, n, diag1, diag2, usedCols, row):
if row == n:
return 1

solutions = 0

for col in range(n):
if row + col in diag1 or row - col in diag2 or col in usedCols:
continue

solutions += self.helper(n, diag1, diag2, usedCols, row + 1)

diag1.remove(row + col)
diag2.remove(row - col)
usedCols.remove(col)

return solutions
##### N-Queens II LeetCode Solution Review:

In our experience, we suggest you solve this N-Queens II LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the N-Queens II LeetCode Solution

Find on LeetCode

##### Conclusion:

I hope this N-Queens II LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

More Coding Solutions >>

LeetCode Solutions

Hacker Rank Solutions

CodeChef Solutions