<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 row = matrix.size(), col = matrix[0].size(); 
        vector<int> prev(col,0);
        int maxi = 0;
        prev[0] = matrix[0][0] == '1' ? 1 : 0;
        maxi = prev[0];
        
        for(int i = 1; i < col; ++i)
        {
            prev[i] = matrix[0][i] == '1' ? 1 : 0;
            maxi = max(maxi, prev[i]);
        } 
        //cout << maxi << endl; 
        /*for(int i = 1; i < row; ++i)
        {
            dp[i][0] = (matrix[i][0] == '1') ? 1 : 0;
            maxi = max(maxi, dp[i][0]);
        }*/
        //cout << maxi << endl; 
        for(int i = 1; i < row; ++i)
        {
            vector<int> curr(col,0);
            for(int j = 0; j < col; ++j)
            {
                if(j == 0)
                {
                    curr[j] = (matrix[i][j] == '1') ? 1 : 0;

                    maxi = max(maxi, curr[j]);
                    continue;
                }

                if(matrix[i][j] == '1')
                {
                    curr[j] = min({prev[j],curr[j-1],prev[j-1]})+1;
                    maxi = max(maxi, curr[j]);
                }
            }
            prev = curr;
        }
        
        return maxi*maxi;
    }
};
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