<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 {
public:

    vector<int>prevSmaller(vector<int>&input){
        stack<int> st;
         st.push(-1);

    vector<int> ans(input.size()); 
    

    for (int i = 0; i < input.size(); i++) {
        int curr = input[i];

        while ( st.top()!= -1 && input[st.top()] >= curr) {
            st.pop();
        }

        // Small element found
        ans[i] = st.top();
        // Push
        st.push(i);
}
return ans;
}

vector<int>nextSmaller(vector<int>&input){
        stack<int> st;
         st.push(-1);

    vector<int> ans(input.size()); 
    

    for (int i = input.size() - 1; i >= 0; i--) {
        int curr = input[i];

        while (st.top()!= -1 && input[st.top()] >= curr) {
            st.pop();
        }

        // Small element found
        ans[i] = st.top();
        // Push
        st.push(i);
}
return ans;
}

int largestRectangleArea(vector<int>& heights){

    int size = heights.size();
    //step 1: prev Smaller output
    vector<int> prev = prevSmaller(heights);
    //step2: next smaller output
    vector<int> next = nextSmaller(heights);
    //step 2: find max Area
    int maxArea = INT_MIN;

    for(int i = 0 ; i<heights.size();i++){

        int lenght = heights[i];

        if(next[i]== -1){
            next[i] =size;
        }

        int width = next[i]- prev[i] -1;

        int area = lenght*width;
        maxArea = max(area , maxArea);
    }
    return maxArea;

}

    int maximalRectangle(vector<vector<char>>& matrix) {
        vector<vector<int>> v;
        int n = matrix.size();
        int m = matrix[0].size();

        for(int i=0;i<n;i++){
            vector<int>t;
            for(int j=0;j<m;j++){
                t.push_back(matrix[i][j]-'0');
            }
            v.push_back(t);
        }
        int area = largestRectangleArea(v[0]);
        for(int i=1;i<n;i++){
            for(int j=0;j<m;j++){
                //lets update current row with previous values
                if(v[i][j]){
                    v[i][j]+=v[i-1][j];
                }else{
                    v[i][j]=0;
                }
            }
            area = max(area,largestRectangleArea(v[i]));
        }
        return area;
    }
};
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