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Formatted question description: https://leetcode.ca/all/1765.html

1765. Map of Highest Peak

Level

Medium

Description

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

  • If isWater[i][j] == 0, cell (i, j) is a land cell.
  • If isWater[i][j] == 1, cell (i, j) is a water cell.

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

  • The height of each cell must be non-negative.
  • If the cell is a water cell, its height must be 0.
  • Any two adjacent cells must have an absolute height difference of at most 1. A cell is adjacent to another cell if the former is directly north, east, south, or west of the latter (i.e., their sides are touching).

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:

Image text

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:

Image text

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:

  • m == isWater.length
  • n == isWater[i].length
  • 1 <= m, n <= 1000
  • isWater[i][j] is 0 or 1.
  • There is at least one water cell.

Solution

Do breadth first search starting from all water cells. All water cells have heights 0. For a cell that has height x, if an adjacent cell has not been visited, then the adjacent cell has height x + 1. Since breadth first search finds the shortest distances of the cells from the water cells, it is guaranteed that any two adjacent cells have an absolute height difference of at most 1.

  • class Solution {
    public int[][] highestPeak(int[][] isWater) {
        int rows = isWater.length, columns = isWater[0].length;
        int[][] heights = new int[rows][columns];
        Queue<int[]> queue = new LinkedList<int[]>();
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                if (isWater[i][j] == 1) {
                    heights[i][j] = 0;
                    queue.offer(new int[]{i, j});
                } else
                    heights[i][j] = -1;
            }
        }
        int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            int row = cell[0], column = cell[1];
            int height = heights[row][column];
            for (int[] direction : directions) {
                int newRow = row + direction[0], newColumn = column + direction[1];
                if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && heights[newRow][newColumn] == -1) {
                    heights[newRow][newColumn] = height + 1;
                    queue.offer(new int[]{newRow, newColumn});
                }
            }
        }
        return heights;
    }
}

############

class Solution {
    public int[][] highestPeak(int[][] isWater) {
        int m = isWater.length, n = isWater[0].length;
        int[][] ans = new int[m][n];
        Deque<int[]> q = new ArrayDeque<>();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans[i][j] = isWater[i][j] - 1;
                if (ans[i][j] == 0) {
                    q.offer(new int[] {i, j});
                }
            }
        }
        int[] dirs = {-1, 0, 1, 0, -1};
        while (!q.isEmpty()) {
            var p = q.poll();
            int i = p[0], j = p[1];
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
                    ans[x][y] = ans[i][j] + 1;
                    q.offer(new int[] {x, y});
                }
            }
        }
        return ans;
    }
}

  • // OJ: https://leetcode.com/problems/map-of-highest-peak/
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    vector<vector<int>> highestPeak(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size();
        vector<vector<int>> H(M, vector<int>(N, -1));
        queue<pair<int, int>> q;
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] == 1) { // BFS from the water cells.
                    q.emplace(i, j);
                    H[i][j] = 0;
                }
            }
        }
        int h = 0, dirs[4][2] = { {0,1},{0,-1},{1,0},{-1,0} };
        while (q.size()) {
            int cnt = q.size();
            while (cnt--) {
                auto [x, y] = q.front();
                q.pop();
                for (auto [dx, dy]: dirs) {
                    int a = x + dx, b = y + dy;
                    if (a < 0 || a >= M || b < 0 || b >= N || H[a][b] != -1) continue;
                    H[a][b] = h + 1;
                    q.emplace(a, b);
                }
            }
            ++h;
        }
        return H;
    }
};

  • class Solution:
    def highestPeak(self, isWater: List[List[int]]) -> List[List[int]]:
        m, n = len(isWater), len(isWater[0])
        ans = [[-1] * n for _ in range(m)]
        q = deque()
        for i, row in enumerate(isWater):
            for j, v in enumerate(row):
                if v:
                    q.append((i, j))
                    ans[i][j] = 0
        while q:
            i, j = q.popleft()
            for a, b in pairwise((-1, 0, 1, 0, -1)):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and ans[x][y] == -1:
                    ans[x][y] = ans[i][j] + 1
                    q.append((x, y))
        return ans

############

# 1765. Map of Highest Peak
# https://leetcode.com/problems/map-of-highest-peak/

class Solution:
    def highestPeak(self, water: List[List[int]]) -> List[List[int]]:
        rows, cols = len(water), len(water[0])
        res = [[-1] * cols for _ in range(rows)]
        deq = collections.deque()
        
        for i in range(rows):
            for j in range(cols):
                if water[i][j] == 1:
                    deq.append((i, j))
                    res[i][j] = 0
        
        while deq:
            x, y = deq.popleft()

            for dx,dy in ((x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1)):
                if 0 <= dx < rows and 0 <= dy < cols and res[dx][dy] == -1:
                    res[dx][dy] = res[x][y] + 1
                    deq.append((dx, dy))
        
        return res


  • func highestPeak(isWater [][]int) [][]int {
	m, n := len(isWater), len(isWater[0])
	ans := make([][]int, m)
	type pair struct{ i, j int }
	q := []pair{}
	for i, row := range isWater {
		ans[i] = make([]int, n)
		for j, v := range row {
			ans[i][j] = v - 1
			if v == 1 {
				q = append(q, pair{i, j})
			}
		}
	}
	dirs := []int{-1, 0, 1, 0, -1}
	for len(q) > 0 {
		p := q[0]
		q = q[1:]
		i, j := p.i, p.j
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1 {
				ans[x][y] = ans[i][j] + 1
				q = append(q, pair{x, y})
			}
		}
	}
	return ans
}

  • function highestPeak(isWater: number[][]): number[][] {
    const m = isWater.length;
    const n = isWater[0].length;
    let ans: number[][] = [];
    let q: number[][] = [];
    for (let i = 0; i < m; ++i) {
        ans.push(new Array(n).fill(-1));
        for (let j = 0; j < n; ++j) {
            if (isWater[i][j]) {
                q.push([i, j]);
                ans[i][j] = 0;
            }
        }
    }
    const dirs = [-1, 0, 1, 0, -1];
    while (q.length) {
        let tq: number[][] = [];
        for (const [i, j] of q) {
            for (let k = 0; k < 4; k++) {
                const [x, y] = [i + dirs[k], j + dirs[k + 1]];
                if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
                    tq.push([x, y]);
                    ans[x][y] = ans[i][j] + 1;
                }
            }
        }
        q = tq;
    }
    return ans;
}


  • use std::collections::VecDeque;

impl Solution {
    #[allow(dead_code)]
    pub fn highest_peak(is_water: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let n = is_water.len();
        let m = is_water[0].len();
        let mut ret_vec = vec![vec![-1; m]; n];
        let mut q: VecDeque<(usize, usize)> = VecDeque::new();
        let vis_pair: Vec<(i32, i32)> = vec![(-1, 0), (1, 0), (0, -1), (0 ,1)];

        // Initialize the return vector
        for i in 0..n {
            for j in 0..m {
                if is_water[i][j] == 1 {
                    // This cell is water, the height of which must be 0
                    ret_vec[i][j] = 0;
                    q.push_back((i, j));
                }
            }
        }

        while !q.is_empty() {
            // Get the front X-Y Coordinates
            let (x, y) = q.front().unwrap().clone();
            q.pop_front();
            // Traverse through the vis pair
            for d in &vis_pair {
                let (dx, dy) = *d;
                if Self::check_bounds(x as i32 + dx, y as i32 + dy, n as i32, m as i32) {
                    if ret_vec[(x as i32 + dx) as usize][(y as i32 + dy) as usize] == -1 {
                        // This cell hasn't been visited, update its height
                        ret_vec[(x as i32 + dx) as usize][(y as i32 + dy) as usize] = ret_vec[x][y] + 1;
                        // Enqueue the current cell
                        q.push_back(((x as i32 + dx) as usize, (y as i32 + dy) as usize));
                    }
                }
            }
        }

        ret_vec
    }

    #[allow(dead_code)]
    fn check_bounds(i: i32, j: i32, n: i32, m: i32) -> bool {
        i >= 0 && i < n && j >= 0 && j < m
    }
}

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