<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 m = matrix.size(), n = matrix[0].size(), ans = 0;
        vector<vector<int>> max_sq(m, vector<int> (n,0));
        if(matrix.size() == 1){
            for(int e : matrix[0]){
                if(e - '0'){
                    return e - '0';
                }
            }
            return 0;
        }
        for(int i = 0; i < n; i++){
            max_sq[0][i] = matrix[0][i] - '0';
            if(max_sq[0][i]){
                ans = 1;
            }
        }
        for(int i = 1; i < m; i++){
            for(int j = 0; j < n; j++){
                if(matrix[i][j] - '0'){
                    ans = max(ans, 1);
                }
                if(j == 0){
                    max_sq[i][j] = matrix[i][j] - '0';
                } else {
                    if(matrix[i][j] - '0' && max_sq[i-1][j] && max_sq[i-1][j-1] && max_sq[i][j-1]){
                        max_sq[i][j] = min(min(max_sq[i-1][j], max_sq[i-1][j-1]), max_sq[i][j-1]) +1 ;
                        ans = max(max_sq[i][j], ans);
                    } else {
                        max_sq[i][j] = matrix[i][j] - '0';
                    }
                }
            }
        }
        return ans*ans;
    }
};
24500