<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>>& m) {
                vector<vector<int>> dp(m.size()+1, vector<int>(m[0].size()));
	
	    // preprocess from left side to to signify i,j showing consecutive 1s from horizontal line starting from i,j
	    for(int i=0;i<m.size();i++) {
		    for(int j=m[0].size()-1;j>=0;j--) {
			    if(j == m[0].size()-1) {
				    dp[i][j] = m[i][j]=='1' ? 1 : 0;
                } else if (m[i][j] == '1') {
				    dp[i][j] = dp[i][j+1]+1;
                } else {
	                dp[i][j] = 0;
                }
            }
        }


        // use monotonic stack to calculate area
        int maxi = 0;

        for(int j=0;j<m[0].size();j++) {
            stack<pair<int,int>> s;
            for(int i=0; i<=m.size();i++) {
                // if stack is empty or the new value satisfies increasing stack invariant, push value on stack
                if(s.empty() || s.top().first <= dp[i][j]) {
                    s.push(make_pair(dp[i][j], i));
                }
                else {
                    int top = i-1;
                    while(!s.empty() && s.top().first >= dp[i][j]) {
                        auto val = s.top().first;
                        s.pop();
                        int place = s.empty() ? 0 : s.top().second+1;
                        int len = min(val, top-place+1);
                        maxi = max(maxi, len*len);
                    }
                    
                    s.push(make_pair(dp[i][j], i));
                    
                }
            }
                
            // no need to empty stack since extra 0 in last row will take care of emptying stack 
        }
        return maxi;
    }
};
28300