Map of Highest Peak - Problem
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.
Input & Output
Example 1 — Basic Island
$
Input:
isWater = [[0,1],[0,0]]
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Output:
[[1,0],[2,1]]
💡 Note:
Water cell at (0,1) has height 0. Adjacent cells get height 1. Cell (1,0) is distance 2 from water, so height 2.
Example 2 — Multiple Water Sources
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Input:
isWater = [[0,0,1],[1,0,0],[0,0,0]]
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Output:
[[1,1,0],[0,1,1],[1,2,2]]
💡 Note:
Two water cells at (0,2) and (1,0). Heights are minimum distance to any water cell.
Example 3 — All Water
$
Input:
isWater = [[1,1],[1,1]]
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Output:
[[0,0],[0,0]]
💡 Note:
All cells are water, so all heights are 0.
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
Visualization
Tap to expand
Understanding the Visualization
1
Input Matrix
0=land, 1=water cells marked
2
Distance Calculation
Each cell gets height = min distance to water
3
Height Matrix
Final heights maximize the maximum value
Key Takeaway
🎯 Key Insight: Multi-source BFS from all water cells simultaneously gives optimal heights in O(mn) time
💡
Explanation
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