LeetCode #1765 — MEDIUM

Map of Highest Peak

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

Solve on LeetCode

The Problem

Problem Statement

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:

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:

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.

Note: This question is the same as 542: https://leetcode.com/problems/01-matrix/

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

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

Pattern signal: Array

Example 1

[[0,1],[0,0]]

Example 2

[[0,0,1],[1,0,0],[0,0,0]]

Step 02

Core Insight

What unlocks the optimal approach

Set each water cell to be 0. The height of each cell is limited by its closest water cell.
Perform a multi-source BFS with all the water cells as sources.

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 #1765: Map of Highest Peak
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;
    }
}

// Accepted solution for LeetCode #1765: Map of Highest Peak
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
}

# Accepted solution for LeetCode #1765: Map of Highest Peak
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

// Accepted solution for LeetCode #1765: Map of Highest Peak
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
    }
}

// Accepted solution for LeetCode #1765: Map of Highest Peak
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;
}

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)

Space

O(1)

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.