<p>Given a <code>rows x cols</code>&nbsp;binary <code>matrix</code> filled with <code>0</code>&#39;s and <code>1</code>&#39;s, find the largest rectangle containing only <code>1</code>&#39;s and return <em>its area</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/14/maximal.jpg" style="width: 402px; height: 322px;" />
<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> 6
<strong>Explanation:</strong> The maximal rectangle is shown in the above picture.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[&quot;0&quot;]]
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[&quot;1&quot;]]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>rows == matrix.length</code></li>
	<li><code>cols == matrix[i].length</code></li>
	<li><code>1 &lt;= row, cols &lt;= 200</code></li>
	<li><code>matrix[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution {
private:
    vector<int> nextSmallerElementIndex(int* arr,int n){
        stack<int> st;
        st.push(-1);
        vector<int> ans(n);
        
        for(int i=n-1;i>=0;i--){
            if(st.top()!=-1 && arr[st.top()]<arr[i]){
                ans[i]=st.top();
                st.push(i);
            }else{
                while(st.top()!=-1 && arr[st.top()]>=arr[i]){
                    st.pop();
                }
                ans[i]=st.top();
                st.push(i);
            }
        }
        return ans;
    }
    vector<int> prevSmallerElementIndex(int* arr,int n){
        stack<int> st;
        st.push(-1);
        vector<int> ans(n);
        
        for(int i=0;i<n;i++){
            if(st.top()!=-1 && arr[st.top()]<arr[i]){
                ans[i]=st.top();
                st.push(i);
            }else{
                while(st.top()!=-1 && arr[st.top()]>=arr[i]){
                    st.pop();
                }
                ans[i]=st.top();
                st.push(i);
            }
        }
        return ans;
    }
      int largestRectangleArea(int* heights,int n) {

        //2nd method - using stack
        
        vector<int> prev(n);
        prev=prevSmallerElementIndex(heights,n);
        
        vector<int> next(n);
        next=nextSmallerElementIndex(heights,n);
        
        int ans=INT_MIN;
        for(int i=0;i<n;i++){
            int l=heights[i];
            
            if(next[i]==-1){
                next[i]=n;
            }
            
            int b=next[i]-prev[i]-1;
            
            int area=l*b;
            ans=max(ans,area);
        }
        return ans;
    }
  public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        
        int n=matrix.size();
        int m=matrix[0].size();
        
        int M[200][200];
        
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                M[i][j]=matrix[i][j]-'0';
            }
        }
        
        int ans=largestRectangleArea(M[0],m);
        
        for(int i=1;i<n;i++){
            for(int j=0;j<m;j++){
                if(M[i][j]!=0){
                    M[i][j]=M[i][j]+M[i-1][j];
                }
            }
            ans=max(ans,largestRectangleArea(M[i],m));
        }
        return ans;
    }
};
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