<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:
    bool valid(int x, int y, int s, int ss) {
        if (x < 0 || y < 0)
            return false;

        if (x >= s || y >= ss)
            return false;

        return true;
    }

    void BFS(queue<pair<pair<int, int>, int>> q, vector<vector<int>>& grid,
             vector<vector<bool>> visit, vector<vector<int>>& ans) {
        int indexX[] = {-1, 1, 0, 0};
        int indexY[] = {0, 0, -1, 1};

        while (!q.empty()) {
            pair<int, int> u = q.front().first;
            int level = q.front().second;

            q.pop();

            visit[u.first][u.second] = true;
            ans[u.first][u.second] = level;

            int x = u.first;
            int y = u.second;

            for (int i = 0; i < 4; i++) {
                int vx = x + indexX[i];
                int vy = y + indexY[i];

                if (valid(vx, vy, grid.size(), grid[0].size())) {
                    if (!visit[vx][vy]) {
                        q.push({{vx, vy}, level + 1});
                        visit[vx][vy] = true;
                    }
                }
            }
        }
    }

    vector<vector<int>> updateMatrix(vector<vector<int>>& grid) {
        queue<pair<pair<int, int>, int>> ones;

        vector<vector<bool>> visit(grid.size(),
                                   vector<bool>(grid[0].size(), false));
        vector<vector<int>> ans(grid.size(), vector<int>(grid[0].size(), -1));

        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[i].size(); j++) {
                if (grid[i][j]==0) {
                    ones.push({{i, j}, 0});
                    visit[i][j] = true;
                }
            }
        }

        BFS(ones, grid, visit, ans);

        return ans;
    }
};
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