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. 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'.
Solution :
Solution in C :
int maximalRectangle(char** matrix, int rowSize, int* colSize){
if (rowSize == 0 || *colSize == 0)
return 0;
int* height = (int*)malloc(sizeof(int) * (*colSize));
int* left = (int*)malloc(sizeof(int) * (*colSize));
int* right = (int*)malloc(sizeof(int) * (*colSize));
int count = 0;
for (int i = 0; i < *colSize; i++) {
height[i] = 0;
left[i] = 0;
right[i] = *colSize;
}
for (int i = 0; i < rowSize; i++) {
int currentLeft = 0, currentRight = *colSize;
for (int j = 0; j < *colSize; j++) {
if (matrix[i][j] == '1') {
height[j]++;
left[j] = left[j] > currentLeft ?
left[j] : currentLeft;
} else {
height[j] = 0;
left[j] = 0;
currentLeft = j + 1;
}
}
for (int j = *colSize - 1; j >= 0; j--) {
if(matrix[i][j] == '1')
right[j] = right[j] < currentRight ?
right[j] : currentRight;
else {
right[j] = *colSize;
currentRight = j;
}
}
for (int j = 0; j < *colSize; j++)
count = count > (right[j] - left[j]) * height[j] ?
count : (right[j] - left[j]) * height[j];
}
free(height);
free(left);
free(right);
return count;
}
Solution in C++ :
class Solution {
public:
int largestRectangleArea(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 solve(vector<vector<char>>&mat, int n, int m) {
int maxArea = 0;
vector<int> height(m, 0);
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (mat[i][j] == '1') height[j]++;
else height[j] = 0;
}
int area = largestRectangleArea(height);
maxArea = max(area, maxArea);
}
return maxArea;
}
int maximalRectangle(vector<vector<char>>& matrix) {
int n = matrix.size();
int m = matrix[0].size();
return solve(matrix, n, m);
}
};
Solution in Java :
class Solution {
public int maximalRectangle(char[][] m) {
int h[][]=new int[m.length+1][m[0].length+1];
int b[][]=new int[m.length+1][m[0].length+1];
int res[][]=new int[m.length][m[0].length];
int maxi=Integer.MIN_VALUE;
for(int i=0;i<m.length;i++){
for(int j=0;j<m[0].length;j++){
if(m[i][j]=='0') continue;
else{
h[i+1][j+1]=h[i][j+1]+1;
b[i+1][j+1]=b[i+1][j]+1;
maxi=Math.max(maxi,Math.max(h[i+1][j+1],b[i+1][j+1]));
}
}
}
for(int i=0;i<m.length;i++){
for(int j=0;j<m[0].length;j++){
if(m[i][j]=='0') continue;
if(b[i][j+1]!=0 && b[i+1][j+1]!=0 && h[i+1][j+1]!=0){
int val=find(h,b,i,j,Math.min(b[i][j+1],b[i+1][j+1]));
res[i][j]=val;
maxi=Math.max(maxi,val);
}
else res[i][j]=1;
}
}
return maxi==Integer.MIN_VALUE?0:maxi;
}
public int find(int[][] h,int[][] b,int i,int j,int len){
int mini=Integer.MIN_VALUE;
int v=0,cur=0;
for(int k=1;k<=len;k++){
if(k==1){
v=h[i+1][j+1];
cur=h[i+1][j+1];
}
else{
cur=Math.min(cur,h[i+1][j+1]);
v=k*cur;
}
mini=Math.max(mini,v);
j--;
}
return mini;
}
}
Solution in Python :
class Solution:
def maximalRectangle(self, matrix: List[List[str]]) -> int:
m = len(matrix)
if m == 0:
return 0
n = len(matrix[0])
if n == 0:
return 0
rec = [[0] * n for _ in range(m)]
for i in range(n):
if matrix[0][i] == '1':
rec[0][i] = 1
max_area = 0
for i in range(m):
for j in range(n):
if i > 0 and matrix[i][j] != '0':
rec[i][j] = rec[i - 1][j] + 1
stack = []
heights = rec[i]
for k, h in enumerate(heights):
if not stack or stack[-1][1] < h:
stack.append((k, h))
else:
while stack and stack[-1][1] >= h:
l, p = stack.pop()
max_area = max(max_area, (k - l) * p)
stack.append((l, h))
while stack:
l, p = stack.pop()
max_area = max(max_area, (n - l) * p)
return max_area
View More Similar Problems
Array Pairs
Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .
View Solution →Self Balancing Tree
An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ
View Solution →Array and simple queries
Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty
View Solution →Median Updates
The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o
View Solution →Maximum Element
You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each
View Solution →Balanced Brackets
A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra
View Solution →