# Maximal Rectangle Problem

## Description

LeetCode Problem 85.

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:

``````1
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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:

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Input: matrix = []
Output: 0
``````

Example 3:

``````1
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Input: matrix = [["0"]]
Output: 0
``````

Example 4:

``````1
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Input: matrix = [["1"]]
Output: 1
``````

Example 5:

``````1
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Input: matrix = [["0","0"]]
Output: 0
``````

Constraints:

• rows == matrix.length
• cols == matrix[i].length
• 0 <= row, cols <= 200
• matrix[i][j] is ‘0’ or ‘1’.

## Sample C++ Code

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class Solution {
public:
int maximalRectangle(vector<vector<char>>& matrix) {
int r = matrix.size();
if (r == 0) return 0;
int c = matrix[0].size();

int max_area = 0;
int area, k, tp;
stack<int> s;
vector<int> line;
vector<vector<int>> dp;

for (int i = 0; i < r; i ++) {
line.clear();
for (int j = 0; j < c; j ++) {
if (matrix[i][j] == '0')
line.push_back(0);
else {
if (i > 0)
line.push_back(dp[i-1][j] + 1);
else
line.push_back(matrix[i][j] - '0');
}
}
dp.push_back(line);

k = 0;
while (k < c) {
if ((s.empty()) || (line[s.top()] < line[k])) {
s.push(k);
k ++;
} else {
tp = s.top();
s.pop();
area = line[tp] * (s.empty() ? k : k - s.top() - 1);
if (area > max_area)
max_area = area;
}
}
while (!s.empty()) {
tp = s.top();
s.pop();
area = line[tp] * (s.empty() ? k : k - s.top() - 1);
if (area > max_area)
max_area = area;
}
}
return max_area;
}
};
``````