LeetCode-in-Java

1765. Map of Highest Peak

Medium

You are given an integer matrix isWater of size m x n that represents a map of land and water cells.

You must assign each cell a height in a way that follows these rules:

Find an assignment of heights such that the maximum height in the matrix is maximized.

Return an integer matrix height of size m x n where height[i][j] is cell (i, j)’s height. If there are multiple solutions, return any of them.

Example 1:

Input: isWater = [[0,1],[0,0]]

Output: [[1,0],[2,1]]

Explanation: The image shows the assigned heights of each cell. The blue cell is the water cell, and the green cells are the land cells.

Example 2:

Input: isWater = [[0,0,1],[1,0,0],[0,0,0]]

Output: [[1,1,0],[0,1,1],[1,2,2]]

Explanation: A height of 2 is the maximum possible height of any assignment. Any height assignment that has a maximum height of 2 while still meeting the rules will also be accepted.

Constraints:

Solution

import java.util.LinkedList;
import java.util.Queue;

public class Solution {
    private final int[] dir = {0, 1, 0, -1, 0};

    public int[][] highestPeak(int[][] isWater) {
        int h = 1;
        Queue<int[]> q = new LinkedList<>();
        for (int i = 0; i < isWater.length; i++) {
            for (int j = 0; j < isWater[0].length; j++) {
                isWater[i][j] = isWater[i][j] == 1 ? 0 : -1;
                if (isWater[i][j] == 0) {
                    q.add(new int[] {i, j});
                }
            }
        }
        while (!q.isEmpty()) {
            Queue<int[]> q1 = new LinkedList<>();
            for (int[] cur : q) {
                int x = cur[0];
                int y = cur[1];
                for (int i = 0; i < 4; i++) {
                    int nx = x + dir[i];
                    int ny = y + dir[i + 1];
                    if (nx >= 0
                            && nx < isWater.length
                            && ny >= 0
                            && ny < isWater[0].length
                            && isWater[nx][ny] == -1) {
                        isWater[nx][ny] = h;
                        q1.add(new int[] {nx, ny});
                    }
                }
            }
            h++;
            q = q1;
        }
        return isWater;
    }
}