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# Maximal Rectangle LeetCode Solution

## Problem – Maximal Rectangle LeetCode Solution

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'`.

## Maximal Rectangle LeetCode Solution in Python

``````def maximalRectangle(self, matrix):
if not matrix or not matrix:
return 0
n = len(matrix)
height =  * (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
``````

## Maximal Rectangle LeetCode Solution in Java

``````public class Solution {
public int maximalRectangle(char[][] matrix) {
if(matrix == null || matrix.length == 0 || matrix.length == 0) return 0;

int[] height = new int[matrix.length];
for(int i = 0; i < matrix.length; i ++){
if(matrix[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.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;
}
``````

## Maximal Rectangle LeetCode Solution in C++

``````class Solution {
public:
int maximalRectangle(vector<vector<char>>& M) {
if(!size(M)) return 0;
int ans = 0, m = size(M), n = size(M);
vector<vector<short>> dp(m+1, vector<short>(n+1));
for(int i = m-1; ~i; i--)
for(int j = n-1; ~j; j--)
dp[i][j] = M[i][j] == '1' ? dp[i][j+1] + 1 : 0;

for(int i = 0; i < m; i++)
for(int j = 0; j < n; j++)
for(int row = i, colLen = n; row < m && M[row][j] == '1'; row++)
ans = max(ans, (row-i+1) * (colLen = min(colLen, dp[row][j]*1)));

return ans;
}
};
``````
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