N Queens Puzzle - Amazon Top Interview Questions
Problem Statement :
The n queens puzzle asks to place n queens on an n×n chessboard so that no two queens are attacking each other.
Given a partially filled two-dimensional integer matrix where 1 represents a queen and 0 represents an empty cell, return whether this configuration of the board can solve the puzzle.
Constraints
1 ≤ n ≤ 15
Example 1
Input
matrix = [
[1, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 1],
[0, 0, 0, 0, 0],
[0, 0, 0, 1, 0]
]
Output
True
Explanation
One solution is:
[1, 0, 0, 0, 0]
[0, 0, 1, 0, 0]
[0, 0, 0, 0, 1]
[0, 1, 0, 0, 0]
[0, 0, 0, 1, 0]
Solution :
Solution in C++ :
int N;
bool canPlace(vector<vector<int>>& matrix, int row, int col) {
for (int i = 0; i < N; ++i) {
if (matrix[i][col] or matrix[row][i]) return false;
}
int x = row, y = col;
while (x >= 0 and y >= 0) {
if (matrix[x--][y--]) return false;
}
x = row, y = col;
while (x >= 0 and y < N) {
if (matrix[x--][y++]) return false;
}
x = row, y = col;
while (x < N and y >= 0) {
if (matrix[x++][y--]) return false;
}
x = row, y = col;
while (x < N and y < N) {
if (matrix[x++][y++]) return false;
}
return true;
}
bool alreadyFilled(vector<vector<int>>& matrix, int row) {
for (int j = 0; j < N; ++j) {
if (matrix[row][j]) return true;
}
return false;
}
bool nQueen(vector<vector<int>>& matrix, int row) {
if (row == N) {
return true;
}
if (alreadyFilled(matrix, row)) return nQueen(matrix, row + 1);
for (int j = 0; j < N; ++j) {
if (canPlace(matrix, row, j)) {
matrix[row][j] = 1;
if (nQueen(matrix, row + 1)) return true;
matrix[row][j] = 0;
}
}
return false;
}
bool solve(vector<vector<int>>& matrix) {
N = matrix.size();
return nQueen(matrix, 0);
;
}
Solution in Java :
import java.util.*;
class Solution {
// TOP LEFT TO BOTTOM RIGHT
private Set<Integer> ord;
// bottom left to top right
private Set<Integer> wack;
private Set<Integer> rows;
private Set<Integer> columns;
private int[][] matrix;
public boolean solve(int[][] matrix) {
this.matrix = matrix;
ord = new HashSet();
wack = new HashSet();
rows = new HashSet();
columns = new HashSet();
int count = 0;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
if (matrix[i][j] == 1) {
count++;
rows.add(i);
columns.add(j);
wack.add(i + j);
ord.add(i - j);
}
}
}
return recurse(0, 0, count);
}
public boolean recurse(int i, int j, int count) {
if (count == matrix.length) {
return true;
}
if (j == matrix[0].length) {
i++;
j = 0;
}
if (i == matrix.length)
return false;
if (!rows.contains(i) && !columns.contains(j) && !wack.contains(i + j)
&& !ord.contains(i - j)) {
rows.add(i);
columns.add(j);
wack.add(i + j);
ord.add(i - j);
if (recurse(i, j + 1, count + 1))
return true;
rows.remove(i);
columns.remove(j);
wack.remove(i + j);
ord.remove(i - j);
}
return recurse(i, j + 1, count);
}
}
Solution in Python :
class Solution:
def solve(self, board):
n = len(board)
rows, cols = [False] * n, [False] * n
diagll, diagrr = [False] * (2 * n - 1), [False] * (2 * n - 1)
def handleBlocking(row, col, action):
board[row][col] = rows[row] = cols[col] = action
diagll[row + col] = diagrr[row - col + n - 1] = action
def isSafe(row, col):
ans = (
board[row][col]
+ rows[row]
+ cols[col]
+ diagll[row + col]
+ diagrr[row - col + n - 1]
)
return ans == 0
for i in range(n):
for j in range(n):
if board[i][j] == 1:
handleBlocking(i, j, 1)
def nqueen(row=0):
if row == n:
return True
if 1 in board[row]:
return nqueen(row + 1)
for col in range(n):
if isSafe(row, col):
handleBlocking(row, col, 1)
if nqueen(row + 1):
return True
handleBlocking(row, col, 0)
return False
ans = nqueen()
return ans
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