<p>You are given an integer matrix <code>isWater</code> of size <code>m x n</code> that represents a map of <strong>land</strong> and <strong>water</strong> cells.</p>

<ul>
	<li>If <code>isWater[i][j] == 0</code>, cell <code>(i, j)</code> is a <strong>land</strong> cell.</li>
	<li>If <code>isWater[i][j] == 1</code>, cell <code>(i, j)</code> is a <strong>water</strong> cell.</li>
</ul>

<p>You must assign each cell a height in a way that follows these rules:</p>

<ul>
	<li>The height of each cell must be non-negative.</li>
	<li>If the cell is a <strong>water</strong> cell, its height must be <code>0</code>.</li>
	<li>Any two adjacent cells must have an absolute height difference of <strong>at most</strong> <code>1</code>. 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).</li>
</ul>

<p>Find an assignment of heights such that the maximum height in the matrix is <strong>maximized</strong>.</p>

<p>Return <em>an integer matrix </em><code>height</code><em> of size </em><code>m x n</code><em> where </em><code>height[i][j]</code><em> is cell </em><code>(i, j)</code><em>&#39;s height. If there are multiple solutions, return <strong>any</strong> of them</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-82045-am.png" style="width: 220px; height: 219px;" /></strong></p>

<pre>
<strong>Input:</strong> isWater = [[0,1],[0,0]]
<strong>Output:</strong> [[1,0],[2,1]]
<strong>Explanation:</strong> The image shows the assigned heights of each cell.
The blue cell is the water cell, and the green cells are the land cells.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-82050-am.png" style="width: 300px; height: 296px;" /></strong></p>

<pre>
<strong>Input:</strong> isWater = [[0,0,1],[1,0,0],[0,0,0]]
<strong>Output:</strong> [[1,1,0],[0,1,1],[1,2,2]]
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == isWater.length</code></li>
	<li><code>n == isWater[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 1000</code></li>
	<li><code>isWater[i][j]</code> is <code>0</code> or <code>1</code>.</li>
	<li>There is at least <strong>one</strong> water cell.</li>
</ul>

from dataclasses import dataclass
import sys

@dataclass
class Node:
    row: int
    col: int
    
    def neighbours(self, n_rows: int, n_cols: int) -> List[Tuple[int]]:
        neigh = []
        
        if self.row - 1 >= 0:
            neigh.append((self.row - 1, self.col))
            
        if self.col + 1 < n_cols:
            neigh.append((self.row, self.col + 1))
        
        if self.row + 1 < n_rows:
            neigh.append((self.row + 1, self.col))
        
        if self.col -1 >= 0:
            neigh.append((self.row, self.col - 1))
        
        return neigh


class Solution:
    def highestPeak(self, isWater: List[List[int]]) -> List[List[int]]:
        n_rows, n_cols = len(isWater), len(isWater[0])
        
        height: List[List[int]] = [[sys.maxsize for _ in range(n_cols)] for _ in range(n_rows)]
        
        q: List[Node] = []
        
        starting_water = None
        for i in range(n_rows):
            for j in range(n_cols):
                if isWater[i][j]:
                    starting_water = Node(i, j)
                    height[i][j] = 0
                    break
            if starting_water:
                break
        
        q.append(starting_water)
        
        while len(q) > 0:
            nd = q.pop(0)
            
            for n_x, n_y in nd.neighbours(n_rows, n_cols):
                proposed = height[nd.row][nd.col] + 1
                if isWater[n_x][n_y] and height[n_x][n_y] > 0:
                    height[n_x][n_y] = 0
                    q.append(Node(n_x, n_y))
                elif proposed < height[n_x][n_y]:
                    height[n_x][n_y] = proposed
                    q.append(Node(n_x, n_y))

                
                if height[nd.row][nd.col] > height[n_x][n_y] + 1:
                    height[nd.row][nd.col] = height[n_x][n_y] + 1
                    q.append(nd)
                
        
        return height
                
                
                
                
                
                
                
                
                
                
                
                
        
        
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