# Maximal Rectangle – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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:6Explanation: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'`

.

# C++ Maximal Rectangle LeetCode Solution

```
``````
class Solution {
public:
int maximalRectangle(vector<vector<char>>& M) {
if(!size(M)) return 0;
int ans = 0, m = size(M), n = size(M[0]);
for(int start_i = 0; start_i < m; start_i++)
for(int start_j = 0; start_j < n; start_j++)
for(int end_i = start_i; end_i < m; end_i++)
for(int end_j = start_j; end_j < n; end_j++) {
bool allOnes = true;
for(int i = start_i; i <= end_i && allOnes; i++)
for(int j = start_j; j <= end_j && allOnes; j++)
if(M[i][j] != '1') allOnes = false;
ans = max(ans, allOnes * (end_i - start_i + 1) * (end_j - start_j + 1));
}
return ans;
}
};
```

# Java Maximal Rectangle LeetCode Solution

```
``````
public class Solution {
public int maximalRectangle(char[][] matrix) {
if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0;
int[] height = new int[matrix[0].length];
for(int i = 0; i < matrix[0].length; i ++){
if(matrix[0][i] == '1') height[i] = 1;
}
int result = largestInLine(height);
for(int i = 1; i < matrix.length; i ++){
resetHeight(matrix, height, i);
result = Math.max(result, largestInLine(height));
}
return result;
}
private void resetHeight(char[][] matrix, int[] height, int idx){
for(int i = 0; i < matrix[0].length; i ++){
if(matrix[idx][i] == '1') height[i] += 1;
else height[i] = 0;
}
}
public int largestInLine(int[] height) {
if(height == null || height.length == 0) return 0;
int len = height.length;
Stack<Integer> s = new Stack<Integer>();
int maxArea = 0;
for(int i = 0; i <= len; i++){
int h = (i == len ? 0 : height[i]);
if(s.isEmpty() || h >= height[s.peek()]){
s.push(i);
}else{
int tp = s.pop();
maxArea = Math.max(maxArea, height[tp] * (s.isEmpty() ? i : i - 1 - s.peek()));
i--;
}
}
return maxArea;
}
```

}

# Python 3 Maximal Rectangle LeetCode Solution

```
``````
def maximalRectangle(self, matrix):
if not matrix or not matrix[0]:
return 0
n = len(matrix[0])
height = [0] * (n + 1)
ans = 0
for row in matrix:
for i in xrange(n):
height[i] = height[i] + 1 if row[i] == '1' else 0
stack = [-1]
for i in xrange(n + 1):
while height[i] < height[stack[-1]]:
h = height[stack.pop()]
w = i - 1 - stack[-1]
ans = max(ans, h * w)
stack.append(i)
return ans
```

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