# 85. Maximal Rectangle
## Problem Statement
Given a `rows x cols` binary `matrix` filled with `0`'s and `1`'s, find the largest rectangle containing only `1`'s and return *its area*.
**Example 1:**
Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
Output: 6
Explanation: The maximal rectangle is shown in the above picture.
**Example 2:**
Input: matrix = [["0"]]
Output: 0
**Example 3:**
Input: matrix = [["1"]]
Output: 1
**Constraints:**
* `rows == matrix.length`
* `cols == matrix[i].length`
* `1 <= row, cols <= 200`
* `matrix[i][j]` is `'0'` or `'1'`.
## Intuition
```
Basically, Just make a count of buildings from
that point
Converting the problem to Largest rectangle in Histogram
For eg->
For above figure the values would be
1 0 1 0 0
2 0 2 1 1 so on
Basically on those row arrays, we apply histogram logic
```
### Links
### Video Links
### Approach 1:
```
Monotonic Stack
```
{% code title="C++" lineNumbers="true" %}
```cpp
class Solution {
public:
int rectangles_in_histogram(vector &histo){
stack < int > st;
int maxA = 0;
int n = histo.size();
for (int i = 0; i <= n; i++) {
while (!st.empty() && (i == n || histo[st.top()] >= histo[i])) {
int height = histo[st.top()];
st.pop();
int width;
if (st.empty())
width = i;
else
width = i - st.top() - 1;
maxA = max(maxA, width * height);
}
st.push(i);
}
return maxA;
}
int maximalRectangle(vector>& matrix) {
int m = matrix.size();
int n = matrix[0].size();
vector> hist(m, vector(n,0));
for(int i=0; i=0; k--){
if(matrix[k][j] == '1')
count++;
else
break;
}
hist[i][j] = count;
}
}
int maxi = 0;
for(auto &it: hist){
vector temp(it);
int ans = rectangles_in_histogram(it);
maxi = max(maxi, ans);
}
return maxi;
}
};
```
{% endcode %}
### Approach 2:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Approach 3:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Approach 4:
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Similar Problems
###
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