<p>Given an <code>m x n</code> binary <code>matrix</code> filled with <code>0</code>&#39;s and <code>1</code>&#39;s, <em>find the largest square containing only</em> <code>1</code>&#39;s <em>and return its area</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/max1grid.jpg" style="width: 400px; height: 319px;" />
<pre>
<strong>Input:</strong> matrix = [[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/max2grid.jpg" style="width: 165px; height: 165px;" />
<pre>
<strong>Input:</strong> matrix = [[&quot;0&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;0&quot;]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[&quot;0&quot;]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>matrix[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution {
public:
    using p=pair<int, int>;
    int maximalSquare(vector<vector<char>>& matrix) {
        int r=matrix.size();
        int c=matrix[0].size();
        vector<vector<int>> vec(r, vector<int>(c));
        for(auto y=0; y<r; y++)
        {
            for(auto x=0; x<c;x++)
            {
                vec[y][x]=matrix[y][x]-'0';
                if(x && vec[y][x])
                    vec[y][x]+=vec[y][x-1];
            }
        }
        int ans=0;
        for(auto x=0; x<c;x++)
        {
            int l=0;
            priority_queue<p, vector<p>, greater<p>> pq;
            for(int y=0; y<r;y++)
            {

                if(vec[y][x])
                {
                    pq.push({vec[y][x], y});
                    while(pq.top().first<y-l+1)
                    {
                        l=max(l, pq.top().second+1);
                        pq.pop();
                    }
                    ans=max(ans, y-l+1);
                }
                else
                {
                    while(pq.size())
                        pq.pop();
                    l=y+1;
                }
            }
        }
        return ans*ans;
    }
};
28800