On an 8x8 chessboard, there can be multiple Black Queens and one White King.
Given an array of integer coordinates queens
that represents the positions of the Black Queens, and a pair of coordinates king
that represent the position of the White King, return the coordinates of all the queens (in any order) that can attack the King.
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 queen at [0,1] can attack the king cause they're in the same row. The queen at [1,0] can attack the king cause they're in the same column. The queen at [3,3] can attack the king cause they're in the same diagnal. The queen at [0,4] can't attack the king cause it's blocked by the queen at [0,1]. The queen at [4,0] can't attack the king cause it's blocked by the queen at [1,0]. The queen at [2,4] can't attack the king cause it's not in the same row/column/diagnal as the king.
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]]
Example 3:
Input: queens = [[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]], king = [3,4] Output: [[2,3],[1,4],[1,6],[3,7],[4,3],[5,4],[4,5]]
Constraints:
1 <= queens.length <= 63
queens[0].length == 2
0 <= queens[i][j] < 8
king.length == 2
0 <= king[0], king[1] < 8
- At most one piece is allowed in a cell.
Related Topics:
Array
// OJ: https://leetcode.com/problems/queens-that-can-attack-the-king/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N^2)
class Solution {
public:
vector<vector<int>> queensAttacktheKing(vector<vector<int>>& Q, vector<int>& K) {
vector<vector<int>> ans;
int board[8][8] = {};
for (auto &v : Q) board[v[0]][v[1]] = 1;
for (int dx = -1; dx <= 1; ++dx) {
for (int dy = -1; dy <= 1; ++dy) {
if (dx == 0 && dy == 0) continue;
int x = K[0], y = K[1];
while (x >= 0 && y >= 0 && x < 8 && y < 8) {
if (board[x][y]) {
ans.push_back({ x, y });
break;
}
x += dx;
y += dy;
}
}
}
return ans;
}
};