<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]]:
        m = len(mat)
        n = len(mat[0])
        returnmat = [[-1] * n for i in range(m)]
        checkqueue = deque([])

        for i in range(m):
            for j in range(n):
                if mat[i][j] == 0:
                    returnmat[i][j] = 0
                    if i > 0:
                        checkqueue.append((i-1, j, 1))
                    if i < m - 1:
                        checkqueue.append((i+1, j, 1))
                    if j > 0:
                        checkqueue.append((i, j-1, 1))
                    if j < n - 1:
                        checkqueue.append((i, j+1, 1))
        
        while checkqueue:
            node = checkqueue.popleft()
            i = node[0]
            j = node[1]
            dist = node[2]
            if returnmat[i][j] == -1:
                returnmat[i][j] = dist
                if i > 0 and returnmat[i-1][j] == -1:
                    checkqueue.append((i-1, j, dist + 1))
                if i < m - 1 and returnmat[i+1][j] == -1:
                    checkqueue.append((i+1, j, dist + 1))
                if j > 0 and returnmat[i][j-1] == -1:
                    checkqueue.append((i, j-1, dist + 1))
                if j < n - 1 and returnmat[i][j + 1] == -1:
                    checkqueue.append((i, j+1, dist + 1))
        return returnmat
18500