How do you solve Map of Highest Peak on LeetCode?

Map of Highest Peak (LeetCode #1765) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: The height of each land cell is determined by its proximity to water cells.

What is the brute force solution for Map of Highest Peak?

The brute force approach for Map of Highest Peak is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each land cell and calculating its height based on the nearest water cell. This is straightforward but inefficient as it requires multiple passes over the matrix. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Map of Highest Peak?

Map of Highest Peak uses the following concepts: Array, Breadth-First Search, Matrix. The recommended patterns to study are: Breadth-First Search, Matrix Traversal.

What are the interview tips for Map of Highest Peak?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as the smallest or largest possible matrices. Explain your thought process clearly while coding; it helps the interviewer understand your approach.

What are common mistakes when solving Map of Highest Peak?

Not initializing the height matrix correctly, leading to incorrect height assignments. Failing to check all adjacent cells during BFS, which can result in missed height updates.

#1765

Map of Highest Peak

Medium

ArrayBreadth-First SearchMatrixBreadth-First SearchMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n)
Space	O(1)	O(m * n)

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Intuition

Time O(m * n)Space O(m * n)

The optimal solution uses a multi-source Breadth-First Search (BFS) starting from all water cells. This efficiently propagates height values to land cells while ensuring the height difference condition is satisfied.

⚙️

Algorithm

3 steps

1Step 1: Initialize a height matrix and a queue for BFS, adding all water cells with height 0.
2Step 2: While the queue is not empty, pop a cell and check its adjacent cells.
3Step 3: For each adjacent land cell, set its height to the current cell's height + 1 if it's not already set, and add it to the queue.

solution.py20 lines

1from collections import deque
2
3def highestPeak(isWater):
4    m, n = len(isWater), len(isWater[0])
5    height = [[-1] * n for _ in range(m)]
6    queue = deque()
7    for i in range(m):
8        for j in range(n):
9            if isWater[i][j] == 1:
10                height[i][j] = 0
11                queue.append((i, j))
12    directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
13    while queue:
14        x, y = queue.popleft()
15        for dx, dy in directions:
16            nx, ny = x + dx, y + dy
17            if 0 <= nx < m and 0 <= ny < n and height[nx][ny] == -1:
18                height[nx][ny] = height[x][y] + 1
19                queue.append((nx, ny))
20    return height

ℹ

Complexity note: The time complexity is O(m * n) because we visit each cell once during the BFS. The space complexity is O(m * n) due to the height matrix and the queue used for BFS.

1The height of each land cell is determined by its proximity to water cells.
2Using BFS allows for efficient propagation of height values while maintaining the height difference constraint.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.