Square Submatrix Sum Below Target - Google Top Interview Questions
Problem Statement :
You are given a two-dimensional list of non-negative integers matrix and a non-negative integer target. Return the area of the largest square sub-matrix whose sum is less than or equal to target.
If there's no such matrix, return 0.
Constraints
1 ≤ n, m ≤ 250 where n and m are the number of rows and columns in matrix
Example 1
Input
matrix = [
[3, 2, 1],
[2, 0, 7],
[1, 1, 9]
]
target = 10
Output
4
Explanation
This sub-matrix sums to 10:
[2, 1]
[0, 7]
Solution :
Solution in C++ :
int dp[255][255];
int solve(vector<vector<int>>& matrix, int target) {
int r = matrix.size();
int c = matrix[0].size();
int ret = 0;
for (int i = 1; i <= r; i++) {
for (int j = 1; j <= c; j++) {
dp[i][j] = matrix[i - 1][j - 1] + dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1];
while (ret + 1 <= min(i, j) && dp[i][j] - dp[i - ret - 1][j] - dp[i][j - ret - 1] +
dp[i - ret - 1][j - ret - 1] <=
target)
ret++;
}
}
return ret * ret;
}
Solution in Java :
import java.util.*;
class Solution {
int[][] dp;
int[][] matrix;
public int area(int r1, int c1, int r2, int c2) {
int m = matrix.length, n = matrix[0].length;
if (r2 >= m || c2 >= n)
return Integer.MAX_VALUE;
return dp[r2 + 1][c2 + 1] + dp[r1][c1] - dp[r2 + 1][c1] - dp[r1][c2 + 1];
}
public boolean ok(int mid, int target) {
int m = matrix.length, n = matrix[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (area(i, j, i + mid - 1, j + mid - 1) <= target)
return true;
}
}
return false;
}
public int solve(int[][] mm, int target) {
this.matrix = mm;
int m = matrix.length;
int n = matrix[0].length;
dp = new int[m + 1][n + 1];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
dp[i + 1][j + 1] = matrix[i][j] + dp[i][j + 1] + dp[i + 1][j] - dp[i][j];
}
}
int res = 0;
int left = 1;
int right = Math.min(m, n);
while (left <= right) {
int mid = left + (right - left) / 2;
if (ok(mid, target)) {
res = mid * mid;
left = mid + 1;
} else {
right = mid - 1;
}
}
return res;
}
}
Solution in Python :
class Solution:
def solve(self, matrix, target):
n, m = len(matrix), len(matrix[0])
def prefix_sum_with_padding():
prefix = [[0 for i in range(m + 1)] for j in range(n + 1)]
for i in range(1, n + 1):
for j in range(1, m + 1):
prefix[i][j] = matrix[i - 1][j - 1]
prefix[i][j] += prefix[i][j - 1]
prefix[i][j] += prefix[i - 1][j]
prefix[i][j] -= prefix[i - 1][j - 1]
return prefix
prefix = prefix_sum_with_padding()
def found_less_equal(side):
for i in range(side, n + 1):
for j in range(side, m + 1):
cur_sum = (
prefix[i][j]
- prefix[i - side][j]
- prefix[i][j - side]
+ prefix[i - side][j - side]
)
if cur_sum <= target:
return True
return False
def binary_search_for_side():
l, r = 0, max(n, m)
while l <= r:
mid = (l + r) // 2
if found_less_equal(mid):
l = mid + 1
else:
r = mid - 1
return r
side = binary_search_for_side()
return side * side
View More Similar Problems
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 →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 →