<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>> pref;
        for (int i = 0; i < matrix.size(); i++) {
            vector<int> newv(matrix[0].size(), 0);
            pref.push_back(newv);
        }

        int ans = 0;
        for (int i = 0; i < matrix[0].size(); i++) {
            pref[0][i] = (matrix[0][i] == '1') ? 1 : 0;
            if (matrix[0][i] == '1') ans = 1;
        }

        for (int i = 0; i < matrix.size(); i++) {
            pref[i][0] = (matrix[i][0] == '1') ? 1 : 0;
            if (matrix[i][0] == '1') ans = 1;
        }

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

        return ans;

    }
};
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