# 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])

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

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```
```

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