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
C++
class Solution {
public:
int rectangles_in_histogram(vector<int> &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<vector<char>>& matrix) {
int m = matrix.size();
int n = matrix[0].size();
vector<vector<int>> hist(m, vector<int>(n,0));
for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
int count = 0;
for(int k=i; k>=0; k--){
if(matrix[k][j] == '1')
count++;
else
break;
}
hist[i][j] = count;
}
}
int maxi = 0;
for(auto &it: hist){
vector<int> temp(it);
int ans = rectangles_in_histogram(it);
maxi = max(maxi, ans);
}
return maxi;
}
};
