<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]]:
        
        n , m = len(mat) , len(mat[0])

        dp = [[float('inf') for _ in range(m)] for _ in range(n)]

        print(dp)
        q = deque()

        for i in range(n):
            for j in range(m):
                if mat[i][j] == 0:
                    dp[i][j] = 0
                    q.append((i , j))
        
        dirs = [[0 , 1], [1 , 0], [0 , -1], [-1 , 0]]

        while len(q) > 0:
            x , y = q.popleft()

            for dx , dy in dirs:
                tx = x + dx
                ty = y + dy

                if tx >= 0 and ty >= 0 and tx < n and ty < m and mat[tx][ty] == 1:
                    if dp[tx][ty] > dp[x][y] + 1:
                        dp[tx][ty] = dp[x][y] + 1
                        q.append((tx, ty))
        

        return dp

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