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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
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;
}
};
/**
* 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) {
return totalNQueensHelper(0, 0, 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 {
occupiedCols.add(col);
occupiedDiag1s.add(diag1);
occupiedDiag2s.add(diag2);
count = totalNQueensHelper(row+1, count, n);
// recover
occupiedCols.remove(col);
occupiedDiag1s.remove(diag1);
occupiedDiag2s.remove(diag2);
}
}
return count;
}
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
diag1.add(row + col)
diag2.add(row - col)
usedCols.add(col)
solutions += self.helper(n, diag1, diag2, usedCols, row + 1)
diag1.remove(row + col)
diag2.remove(row - col)
usedCols.remove(col)
return solutions
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