<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>

#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("O3", "unroll-loops", "-ffloat-store")
#pragma GCC target("avx,mmx,sse2,sse3,sse4")

// "_": A lambda function to enable faster I/O operations
auto _=[]()noexcept{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);return 0;}();

#include <vector>
#include <algorithm>

using namespace std;

// 1D DP: [O(m*n), O(n)]
class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        int m = matrix.size();
        int n = matrix[0].size();

        if (m == 0 || n == 0) return 0;

        vector<int> dp(n + 1, 0);
        int max_side = 0;
        int prev = 0; // store dp[j-1]

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

                prev = temp;
            }
        }

        return max_side * max_side;
    }
};
22300