<p>Given an <code>m x n</code> binary matrix <code>mat</code>, return <em>the distance of the nearest </em><code>0</code><em> for each cell</em>.</p>

<p>The distance between two adjacent cells is <code>1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-1-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> mat = [[0,0,0],[0,1,0],[0,0,0]]
<strong>Output:</strong> [[0,0,0],[0,1,0],[0,0,0]]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-2-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> mat = [[0,0,0],[0,1,0],[1,1,1]]
<strong>Output:</strong> [[0,0,0],[0,1,0],[1,2,1]]
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>4</sup></code></li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There is at least one <code>0</code> in <code>mat</code>.</li>
</ul>

class Solution:
    def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:

        if len(mat) == 0:
            return mat

        que = deque()
        visited = set()

        result = [row for row in mat]

        for row in range(len(mat)):
            for cell in range(len(mat[0])):
                if mat[row][cell] == 0:
                    visited.add((row,cell))
                    que.append([row, cell, 0])

        # print(visited)
        adj_cells = [[-1,0], [0,1], [1,0], [0,-1]]

        while(que):
            r, c, step = que.popleft()
            # print("RC", r,c, step)
            # print(visited)

            for cell in adj_cells:
                row = r + cell[0]
                col = c + cell[1]

                if 0 <= row < len(mat) and 0<= col<len(mat[0]) and (row,col) not in visited:            
                    mat[row][col] = step + 1
                    que.append([row,col, step +1])
                    visited.add((row,col))
        
        return mat
    






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