Leetcode #1765: Map of Highest Peak
In this guide, we solve Leetcode #1765 Map of Highest Peak in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Breadth-First Search, Array, Matrix
Intuition
We need level-by-level exploration or shortest steps, which is ideal for BFS.
A queue naturally models the frontier of the search.
Approach
Push initial nodes into a queue and expand in layers.
Track visited nodes to prevent cycles.
Steps:
- Initialize queue with start nodes.
- Process level by level.
- Track visited nodes.
Example
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.
Python Solution
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
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.