<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:
    int maximalSquare(vector<vector<char>>& matrix) {
        int n=matrix.size(),m=matrix[0].size();
        vector<vector<int>> dp(n+1,vector<int>(m+1,0));
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
               dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+(matrix[i-1][j-1]=='1');
            }
        }
        auto isPossible = [&](int ind){
            for(int i=ind;i<=n;i++){
                for(int j=ind;j<=m;j++){
                    int one=dp[i][j]-dp[i-ind][j]-dp[i][j-ind]+dp[i-ind][j-ind];
                    if(one>=ind*ind) return true;
                }
            }
            return false;
        };

        int s=1,e=n,ans=0;
        while(s<=e){
            int mid=s+(e-s)/2;
            if(isPossible(mid)){
                ans=mid;
                s=mid+1;
            }else{
                e=mid-1;
            }
        }
        return ans*ans;
    }
};
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