<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 {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) 
    {
        deque<tuple<int, int, int>> q;
        for (int i = 0; i < mat.size(); ++i)
        {
            for (int j = 0; j < mat[i].size(); ++j)
            {
                if (mat[i][j] == 0)
                {
                    q.push_back({i, j, 0});
                }
                else
                {
                    mat[i][j] = INT_MAX;
                }
            }
        }

        // BFS
        while (!q.empty())
        {
            tuple<int, int, int> cell = q.front();
            q.pop_front();
            int i = get<0>(cell), j = get<1>(cell), d = get<2>(cell);

            mat[i][j] = d < mat[i][j] ? d : mat[i][j];

            array<pair<int, int>, 4> dirs = { make_pair(-1, 0), make_pair(0, -1), make_pair(1, 0), make_pair(0, 1) };
            for (const pair<int,int>& dir : dirs)
            {
                int newI = i + dir.first, newJ = j + dir.second, newD = d + 1;
                if (newI < 0 || newJ < 0 || newI >= mat.size() || newJ >= mat[i].size())
                {
                    continue;
                }

                if (mat[newI][newJ] == INT_MAX)
                {
                    q.push_back({newI, newJ, newD});
                }
            }
        }

        return mat;
    }
};
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