Largest Rectangle Submatrix - Amazon Top Interview Questions
Problem Statement :
Given a two-dimensional integer matrix consisting only of 1s and 0s, return the area of the largest rectangle containing only 1s.
Constraints
0 ≤ n, m ≤ 250 where n and m are the number of rows and columns in matrix
Example 1
Input
matrix = [
[1, 0, 0, 0],
[1, 0, 1, 1],
[1, 0, 1, 1],
[0, 1, 0, 0]
]
Output
4
Explanation
The biggest rectangle here is the 2 by 2 square of 1s on the right.
Example 2
Input
matrix = [
[1, 0, 0, 0, 0],
[0, 0, 1, 1, 0],
[0, 1, 1, 0, 0],
[0, 0, 0, 0, 0],
[1, 1, 0, 0, 1],
[1, 1, 0, 0, 1]
]
Output
4
Example 3
Input
matrix = [
[1, 1, 1, 1],
[1, 1, 1, 1],
[1, 1, 1, 1],
[0, 0, 0, 0]
]
Output
12
Example 4
Input
matrix = [
[1, 1, 1, 1],
[1, 0, 0, 1],
[1, 1, 1, 1],
[1, 1, 1, 1]
]
Output
8
Solution :
Solution in C++ :
int solve(vector<vector<int>>& matrix) {
if (!matrix.size() || !matrix[0].size()) return 0;
int i, j, m = matrix.size(), n = matrix[0].size(), ret = 0;
vector<int> height(n + 1, 0);
for (i = 0; i < m; i++) {
vector<int> st;
for (j = 0; j <= n; j++) {
if (j < n) height[j] = (matrix[i][j] ? height[j] + 1 : 0);
while (st.size() && height[st.back()] >= height[j]) {
int h = height[st.back()];
st.pop_back();
int w = (st.size() ? j - st.back() - 1 : j);
ret = max(ret, w * h);
}
st.push_back(j);
}
}
return ret;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] matrix) {
final int C = matrix[0].length;
int[] dp = new int[C];
int res = 0;
Stack<int[]> stack = new Stack<>();
for (int[] row : matrix) {
for (int c = 0; c != C; c++) {
dp[c] = row[c] == 0 ? 0 : dp[c] + 1;
int earliest = c;
while (stack.isEmpty() == false && stack.peek()[1] >= dp[c]) {
int[] prev = stack.pop();
earliest = prev[0];
res = Math.max(res, (c - prev[0]) * prev[1]);
}
stack.push(new int[] {earliest, dp[c]});
}
while (stack.isEmpty() == false) {
int[] prev = stack.pop();
res = Math.max(res, (C - prev[0]) * prev[1]);
}
}
return res;
}
}
Solution in Python :
class Solution:
def solve(self, matrix):
M, N = len(matrix), len(matrix[0])
skyline = [0] * N
area = 0
for row in matrix:
skyline = self.update_skyline(row, skyline)
area = max(area, self.largest_rect_area(skyline))
return area
def update_skyline(self, row, prev_skyline):
new_skyline = []
for nonzero, prev_height in zip(row, prev_skyline):
new_skyline.append(prev_height + 1 if nonzero else 0)
return new_skyline
def largest_rect_area(self, skyline):
stack = []
area = 0
for i, height in enumerate(skyline + [0]):
earliest_j = i
while stack and height <= stack[-1][1]:
j, other_height = stack.pop()
area = max(area, (i - j) * other_height)
earliest_j = min(earliest_j, j)
stack.append((earliest_j, height))
return area
View More Similar Problems
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 →Equal Stacks
ou have three stacks of cylinders where each cylinder has the same diameter, but they may vary in height. You can change the height of a stack by removing and discarding its topmost cylinder any number of times. Find the maximum possible height of the stacks such that all of the stacks are exactly the same height. This means you must remove zero or more cylinders from the top of zero or more of
View Solution →Game of Two Stacks
Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B = [b0, b1, . . . , b m-1] where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game: In each move, Nick can remove one integer from the top of either stack A or stack B. Nick keeps a running sum of the integers he removes from the two stacks. Nick is disqualified f
View Solution →Largest Rectangle
Skyline Real Estate Developers is planning to demolish a number of old, unoccupied buildings and construct a shopping mall in their place. Your task is to find the largest solid area in which the mall can be constructed. There are a number of buildings in a certain two-dimensional landscape. Each building has a height, given by . If you join adjacent buildings, they will form a solid rectangle
View Solution →Simple Text Editor
In this challenge, you must implement a simple text editor. Initially, your editor contains an empty string, S. You must perform Q operations of the following 4 types: 1. append(W) - Append W string to the end of S. 2 . delete( k ) - Delete the last k characters of S. 3 .print( k ) - Print the kth character of S. 4 . undo( ) - Undo the last (not previously undone) operation of type 1 or 2,
View Solution →Poisonous Plants
There are a number of plants in a garden. Each of the plants has been treated with some amount of pesticide. After each day, if any plant has more pesticide than the plant on its left, being weaker than the left one, it dies. You are given the initial values of the pesticide in each of the plants. Determine the number of days after which no plant dies, i.e. the time after which there is no plan
View Solution →