Map of Highest Peak | Multi-Source BFS | Leetcode 1765 | Graph Concept | CodeWithShaaz

🔍 Problem: 1765. Map of Highest Peak
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 such a way that:

The height of each cell is non-negative.
If a cell is a water cell, its height must be 0.
Any two adjacent cells (north, east, south, or west) must have an absolute height difference of at most 1.
The goal is to maximize the maximum height in the matrix.

Return an integer matrix height of size m x n where height[i][j] represents the height of cell (i, j). If there are multiple solutions, return any of them.

Key Observations:
Multi-Source BFS:

Water cells are initialized to height 0, and they act as the starting points for a BFS.
The BFS expands from these water cells and assigns increasing heights to adjacent land cells.
Height Assignment:

Each level of BFS corresponds to incrementing the height by 1.
This ensures the height difference between adjacent cells is at most 1.
Approach:
Initialize Data Structures:

Create a height matrix of size m x n initialized with -1.
Add all water cells (i, j) with height 0 to a queue.
Perform Multi-Source BFS:

Define directions for moving north, south, east, and west.
For each cell in the queue, visit its neighbors.
If a neighbor is unvisited (i.e., height[i][j] == -1), assign it a height that is 1 greater than the current cell, and add it to the queue.
Return the Result:

After the BFS completes, all cells will have their heights assigned.

Complexity Analysis:
Time Complexity:

BFS processes each cell exactly once.
For m x n grid, the complexity is O(m * n).
Space Complexity:

The queue stores at most m * n cells in the worst case.
The height matrix uses O(m * n) space.
Total: O(m * n).
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