<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>

int dp[1000][1000], vertical[1000][1000];

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
    int n = matrix.size(), m = matrix[0].size();

    for (int i = 0; i < n; i++) {
        dp[i][0] = (matrix[i][0] == '1');
        // cout << "HEYY " << matrix[i][0] << endl;
    }
    for (int j = 0; j < m; j++) {
        dp[0][j] = (matrix[0][j] == '1');
    }

    for (int i = 1; i < n; i++) {
        for (int j = 1; j < m; j++) {
            if (matrix[i][j] == '0') {
                dp[i][j] = 0;
            } else {
                dp[i][j] = min(min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]);
                dp[i][j]++;
            }
        }
    } 

    int ans = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            // cout << dp[i][j] << ' ';
            ans = max(ans, dp[i][j]);
        }
       // cout << endl;
    }


    return ans * ans;

    }
};
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