LeetCode #1222 — MEDIUM

Queens That Can Attack the King

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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] = [xQueen_i, yQueen_i] represents the position of the i^th 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 <= xQueen_i, yQueen_i, xKing, yKing < 8
All the given positions are unique.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[0,1],[1,0],[4,0],[0,4],[3,3],[2,4]]
[0,0]

Example 2

[[0,0],[1,1],[2,2],[3,4],[3,5],[4,4],[4,5]]
[3,3]

Core Insight

What unlocks the optimal approach

Check 8 directions around the King.
Find the nearest queen in each direction.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1222: Queens That Can Attack the King
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;
    }
}

// Accepted solution for LeetCode #1222: Queens That Can Attack the King
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
}

# Accepted solution for LeetCode #1222: Queens That Can Attack the King
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

// Accepted solution for LeetCode #1222: Queens That Can Attack the King
struct Solution;
use std::collections::HashSet;

type Point = Vec<i32>;

struct Chessboard {
    directions: Vec<Point>,
    queens: HashSet<Point>,
    king: Point,
}

impl Chessboard {
    fn new(queens_vec: Vec<Point>, king: Point) -> Self {
        let directions: Vec<Point> = vec![
            vec![1, 0],
            vec![-1, 0],
            vec![0, 1],
            vec![0, -1],
            vec![1, 1],
            vec![-1, 1],
            vec![1, -1],
            vec![-1, -1],
        ];
        let mut queens: HashSet<Point> = HashSet::new();
        for queen in queens_vec {
            queens.insert(queen);
        }
        Chessboard {
            directions,
            queens,
            king,
        }
    }
    fn contains(&self, point: &[i32]) -> bool {
        point[0] >= 0 && point[1] >= 0 && point[0] < 8 && point[1] < 8
    }
    fn attack(&self, i: usize, step: i32) -> Point {
        let direction = &self.directions[i];
        let king = &self.king;
        vec![king[0] + direction[0] * step, king[1] + direction[1] * step]
    }
}

impl Solution {
    fn queens_attackthe_king(queens: Vec<Vec<i32>>, king: Vec<i32>) -> Vec<Vec<i32>> {
        let mut res = vec![];
        let cb = Chessboard::new(queens, king);
        for i in 0..8 {
            let mut step = 1;
            loop {
                let p = cb.attack(i, step);
                if cb.contains(&p) {
                    if cb.queens.contains(&p) {
                        res.push(p);
                        break;
                    } else {
                        step += 1;
                    }
                } else {
                    break;
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let queens = vec_vec_i32![[0, 1], [1, 0], [4, 0], [0, 4], [3, 3], [2, 4]];
    let king = vec![0, 0];
    let mut res = vec_vec_i32![[0, 1], [1, 0], [3, 3]];
    let mut ans = Solution::queens_attackthe_king(queens, king);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let queens = vec_vec_i32![[0, 0], [1, 1], [2, 2], [3, 4], [3, 5], [4, 4], [4, 5]];
    let king = vec![3, 3];
    let mut res = vec_vec_i32![[2, 2], [3, 4], [4, 4]];
    let mut ans = Solution::queens_attackthe_king(queens, king);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let queens = vec_vec_i32![
        [5, 6],
        [7, 7],
        [2, 1],
        [0, 7],
        [1, 6],
        [5, 1],
        [3, 7],
        [0, 3],
        [4, 0],
        [1, 2],
        [6, 3],
        [5, 0],
        [0, 4],
        [2, 2],
        [1, 1],
        [6, 4],
        [5, 4],
        [0, 0],
        [2, 6],
        [4, 5],
        [5, 2],
        [1, 4],
        [7, 5],
        [2, 3],
        [0, 5],
        [4, 2],
        [1, 0],
        [2, 7],
        [0, 1],
        [4, 6],
        [6, 1],
        [0, 6],
        [4, 3],
        [1, 7]
    ];
    let king = vec![3, 4];
    let mut res = vec_vec_i32![[2, 3], [1, 4], [1, 6], [3, 7], [4, 3], [5, 4], [4, 5]];
    let mut ans = Solution::queens_attackthe_king(queens, king);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #1222: Queens That Can Attack the King
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n^2)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.