<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) {
        vector<vector<int>> dp(matrix.size(), vector<int>(matrix[0].size(), 0));
        int ans = 0;
        for (int i = 0; i < matrix.size(); ++i) {
            dp[i][0] = int(matrix[i][0] - '0');
            ans = max(ans, dp[i][0]);
        }
        for (int i = 0; i < matrix[0].size(); ++i) {
            dp[0][i] = int(matrix[0][i] - '0');
            ans = max(ans, dp[0][i]);
        }

        cout << endl;
        for (int i = 1; i < matrix.size(); ++i) {
            for (int j = 1; j < matrix[0].size(); ++j) {
                if (matrix[i][j] == '1') {
                    dp[i][j] = 1;
                    if (int(matrix[i - 1][j] - '0') &&
                        int(matrix[i - 1][j - 1] - '0') &&
                        int(matrix[i][j - 1] - '0')) {
                        dp[i][j] = 1 + min(min((dp[i - 1][j - 1] ),
                                               (dp[i][j - 1] )),
                                           (dp[i - 1][j] ));
                    }
                }
                ans = max(ans, dp[i][j]);
            }
        }

        for (auto f : dp) {
            for (auto g : f) {
                cout << g << " ";
            }
            cout << endl;
        }
        return ans * ans;
    }
};
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