1222. Queens That Can Attack the King

Description

On a 0-indexed 8 x 8 chessboard, there can be multiple black queens and one white king.

You are given a 2D integer array queens where queens[i] = [xQueeni, yQueeni] represents the position of the ith black queen on the chessboard. You are also given an integer array king of length 2 where king = [xKing, yKing] represents the position of the white king.

Return the coordinates of the black queens that can directly attack the king. You may return the answer in any order.

 

Example 1:

Input: queens = [[0,1],[1,0],[4,0],[0,4],[3,3],[2,4]], king = [0,0]
Output: [[0,1],[1,0],[3,3]]
Explanation: The diagram above shows the three queens that can directly attack the king and the three queens that cannot attack the king (i.e., marked with red dashes).

Example 2:

Input: queens = [[0,0],[1,1],[2,2],[3,4],[3,5],[4,4],[4,5]], king = [3,3]
Output: [[2,2],[3,4],[4,4]]
Explanation: The diagram above shows the three queens that can directly attack the king and the three queens that cannot attack the king (i.e., marked with red dashes).

 

Constraints:

  • 1 <= queens.length < 64
  • queens[i].length == king.length == 2
  • 0 <= xQueeni, yQueeni, xKing, yKing < 8
  • All the given positions are unique.

Solutions

First, we store all the positions of the queens in a hash table or a two-dimensional array $s$.

Next, starting from the position of the king, we search in the eight directions: up, down, left, right, upper left, upper right, lower left, and lower right. If there is a queen in a certain direction, we add its position to the answer and stop continuing to search in that direction.

After the search is over, we return the answer.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. In this problem, $n = 8$.

Python Code
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class Solution:
    def queensAttacktheKing(
        self, queens: List[List[int]], king: List[int]
    ) -> List[List[int]]:
        n = 8
        s = {(i, j) for i, j in queens}
        ans = []
        for a in range(-1, 2):
            for b in range(-1, 2):
                if a or b:
                    x, y = king
                    while 0 <= x + a < n and 0 <= y + b < n:
                        x, y = x + a, y + b
                        if (x, y) in s:
                            ans.append([x, y])
                            break
        return ans

Java Code
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class Solution {
    public List<List<Integer>> queensAttacktheKing(int[][] queens, int[] king) {
        final int n = 8;
        var s = new boolean[n][n];
        for (var q : queens) {
            s[q[0]][q[1]] = true;
        }
        List<List<Integer>> ans = new ArrayList<>();
        for (int a = -1; a <= 1; ++a) {
            for (int b = -1; b <= 1; ++b) {
                if (a != 0 || b != 0) {
                    int x = king[0] + a, y = king[1] + b;
                    while (x >= 0 && x < n && y >= 0 && y < n) {
                        if (s[x][y]) {
                            ans.add(List.of(x, y));
                            break;
                        }
                        x += a;
                        y += b;
                    }
                }
            }
        }
        return ans;
    }
}

C++ Code
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class Solution {
public:
    vector<vector<int>> queensAttacktheKing(vector<vector<int>>& queens, vector<int>& king) {
        int n = 8;
        bool s[8][8]{};
        for (auto& q : queens) {
            s[q[0]][q[1]] = true;
        }
        vector<vector<int>> ans;
        for (int a = -1; a <= 1; ++a) {
            for (int b = -1; b <= 1; ++b) {
                if (a || b) {
                    int x = king[0] + a, y = king[1] + b;
                    while (x >= 0 && x < n && y >= 0 && y < n) {
                        if (s[x][y]) {
                            ans.push_back({x, y});
                            break;
                        }
                        x += a;
                        y += b;
                    }
                }
            }
        }
        return ans;
    }
};

Go Code
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func queensAttacktheKing(queens [][]int, king []int) (ans [][]int) {
	n := 8
	s := [8][8]bool{}
	for _, q := range queens {
		s[q[0]][q[1]] = true
	}
	for a := -1; a <= 1; a++ {
		for b := -1; b <= 1; b++ {
			if a != 0 || b != 0 {
				x, y := king[0]+a, king[1]+b
				for 0 <= x && x < n && 0 <= y && y < n {
					if s[x][y] {
						ans = append(ans, []int{x, y})
						break
					}
					x += a
					y += b
				}
			}
		}
	}
	return
}

TypeScript Code
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function queensAttacktheKing(queens: number[][], king: number[]): number[][] {
    const n = 8;
    const s: boolean[][] = Array.from({ length: n }, () => Array.from({ length: n }, () => false));
    queens.forEach(([x, y]) => (s[x][y] = true));
    const ans: number[][] = [];
    for (let a = -1; a <= 1; ++a) {
        for (let b = -1; b <= 1; ++b) {
            if (a || b) {
                let [x, y] = [king[0] + a, king[1] + b];
                while (x >= 0 && x < n && y >= 0 && y < n) {
                    if (s[x][y]) {
                        ans.push([x, y]);
                        break;
                    }
                    x += a;
                    y += b;
                }
            }
        }
    }
    return ans;
}