<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();
        int m = matrix[0].size();
        vector<vector<int>> dp(n, vector<int>(m, 0));

        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if(i == 0) dp[i][j] = matrix[i][j]-'0';
                else {
                    int prevCol = j-1 >= 0 ? dp[i][j-1] : 0;
                    int prevRow = i-1 >= 0 ? dp[i-1][j] : 0;
                    int prevDiag = (i-1 >= 0 && j-1 >= 0) ? dp[i-1][j-1] : 0;
                    if(matrix[i][j]-'0' != 0) {
                        int mini = min(prevCol, min(prevRow, prevDiag));
                        dp[i][j] = mini + 1;
                    }
                }
            }
        }
        int maxi = 0;
        for(auto i : dp) {
            for(auto j : i) {
                cout << j << " ";
                maxi = max(maxi, j);
            }
            cout << endl;
        }
        return maxi * maxi;
    }
};
27000