<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 {
    vector<int> nextSmallIndex(vector<int>&arr)
    {
        stack<int>st;
        int n = arr.size();
        vector<int>ans(n);

        for(int i=n-1;i>=0;i--)
        {
            while(!st.empty() && arr[st.top()] >= arr[i])
            {
                st.pop();
            }

            if(st.empty())
            {
                ans[i] = n;
            }
            else
            {
                ans[i] = st.top();
            }

            st.push(i);
        }
        return ans;
    }

    vector<int> prevSmallIndex(vector<int>&arr)
    {
        stack<int>st;
        int n = arr.size();
        vector<int>ans(n);

        for(int i=0;i<n;i++)
        {
            while(!st.empty() && arr[st.top()] > arr[i])
            {
                st.pop();
            }

            if(st.empty())
            {
                ans[i] = -1;
            }
            else
            {
                ans[i] = st.top();
            }

            st.push(i);
        }
        return ans;
    }


    int largestRectangleArea(vector<int>&arr)
    {
        vector<int>nextSmall = nextSmallIndex(arr);
        vector<int>prevSmall = prevSmallIndex(arr);

        int n = arr.size();
        int ans = INT_MIN;
        for(int i=0;i<n;i++)
        {
            int curr = ( nextSmall[i] - prevSmall[i] - 1) * arr[i];

            ans = max(ans , curr);
        }

        return ans;
    }

public:
    int maximalRectangle(vector<vector<char>>& matrix) 
    {
       int n = matrix.size();
       int m = matrix[0].size();
    
       int ans = INT_MIN;
       vector<int>CurrArr(m , 0);

       for(int i=0;i<n;i++)
       {
            for(int j=0;j<m;j++)
            {
                if(matrix[i][j] == '1')
                {
                    CurrArr[j]++;
                }
                else
                {
                    CurrArr[j] = 0;
                }
            }

            ans = max(ans , largestRectangleArea(CurrArr));
       }

       return ans;
    }
};
21254