<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) {
        return getUpdatedMatrix(mat);
    }

    vector<vector<int>> getUpdatedMatrix(vector<vector<int>>& mat)
    {
      //  vector<vector<int>> ansMat(mat.size(), vector<int>(mat[0].size(), 0));
        vector<vector<bool>> seen(mat.size(), vector<bool>(mat[0].size(), false));
        queue<vector<int>> que; 

        for (int i = 0; i < mat.size(); i++)
        {
            for (int j = 0; j < mat[0].size(); j++)
            {
                if (mat[i][j])
                {
                    continue;
                }

                que.push({i, j, 1});
                seen[i][j] = true;
            }
        }

        vector<int> rowDiff = {0, -1, 0, 1};
        vector<int> colDiff = {-1, 0, 1, 0};

        while (!que.empty())
        {
            auto node = que.front();
            int i = node[0];
            int j = node[1];
            int steps = node[2];
            que.pop();

            for (int m = 0; m < 4; m++)
            {
                int newI = i + rowDiff[m];
                int newJ = j + colDiff[m];

                if (newI < 0 || newI >= mat.size() || newJ < 0 || newJ >= mat[0].size() || mat[newI][newJ] == 0
                || seen[newI][newJ])
                {
                    continue;
                }

                seen[newI][newJ] = 1;
                que.push({newI, newJ, steps+1});
                mat[newI][newJ] = steps;
            }
        }

        return mat;
    }

    
   
};
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