<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:
   // AC - AKS - solved myself in first attempt -  Grid Prefix sum + Binary Search on answer - O(nlog(m*n))
    bool ispossible(int n, int m, vector<vector<long long>> &prefix, long long mid){
        long long req = mid*mid;
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                int lr = i + mid-1;
                int lc = j + mid-1;
                if(lr <= n and lc <= m){
                    long long sum = prefix[lr][lc] - prefix[lr][j-1] - prefix[i-1][lc] + prefix[i-1][j-1];
                    if(sum == req) return true;
                }
            }
        }
        return false;
    }
    int maximalSquare(vector<vector<char>>& matrix) {
        int n = matrix.size();
        int m = matrix[0].size();

        // prefix[i][j] ==> contain prefix sum untill cell (i,j)
        vector<vector<long long>>prefix(n+1,vector<long long>(m+1,0));

        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                // fill the first row
                if(i == 1) prefix[i][j] = matrix[i-1][j-1] - '0' + prefix[i][j-1];

                // fill the first col
                else if(j == 1) prefix[i][j] = matrix[i-1][j-1] -'0' + prefix[i-1][j];

                // now get for others
                else prefix[i][j] = matrix[i-1][j-1] - '0' + prefix[i][j-1] + prefix[i-1][j] - prefix[i-1][j-1];
            }
        }

        // Now apply Binary Search on the answer
        long long l = 0;
        long long r = min(m,n);
        long long ans = 0;
        while(l <= r){
            long long mid = (l+r) >> 1L;
            if(ispossible(n, m,prefix,mid)){
                ans = mid*mid;
                l = mid+1;
            }
            else r = mid-1;
        }
        return ans;
    }
};
26200