Given a m x n matrix mat and an integer threshold. Return the maximum side-length of a square with a sum less than or equal to threshold or return 0 if there is no such square.
 
Example 1:

Input: mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]], threshold = 4
Output: 2
Explanation: The maximum side length of square with sum less than 4 is 2 as shown.

Example 2:
Input: mat = [[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2]], threshold = 1
Output: 0

Example 3:
Input: mat = [[1,1,1,1],[1,0,0,0],[1,0,0,0],[1,0,0,0]], threshold = 6
Output: 3

Example 4:
Input: mat = [[18,70],[61,1],[25,85],[14,40],[11,96],[97,96],[63,45]], threshold = 40184
Output: 2

 
Constraints:

1 <= m, n <= 300
m == mat.length
n == mat[i].length
0 <= mat[i][j] <= 10000
0 <= threshold <= 10^5

Source: leetcode | Difficulty: MEDIUM_HARD
Tags: Matrices

Sorting

Data structures

Range queries

Starter code: class Solution:
    def maxSideLength(self, mat: List[List[int]], threshold: int) -> int:
        

Solutions:
class Solution:

	def maxSideLength(self, mat: List[List[int]], threshold: int) -> int:
		dp = [[0 for _ in range(len(mat[0]) + 1)] for r in range(len(mat) + 1)]
		for r in range(1, len(mat) + 1):
			for c in range(1, len(mat[r - 1]) + 1):
				dp[r][c] += mat[r - 1][c - 1]
				if not r and (not c):
					continue
				elif not r:
					dp[r][c] += dp[r][c - 1]
					continue
				elif not c:
					dp[r][c] += dp[r - 1][c]
					continue
				dp[r][c] += dp[r][c - 1] + dp[r - 1][c] - dp[r - 1][c - 1]
		highest = -1
		for r in range(1, len(dp)):
			r0 = r1 = r
			c0 = c1 = 1
			while r1 < len(dp) and c1 < len(dp[0]):
				result = dp[r1][c1] + dp[r0 - 1][c0 - 1] - dp[r1][c0 - 1] - dp[r0 - 1][c1]
				if result <= threshold:
					highest = max(r1 - r0, highest)
					r1 += 1
					c1 += 1
				else:
					r1 -= 1
					c0 += 1
				r1 = max(r0 + 1, r1)
				c1 = max(c0 + 1, c1)
		return highest + 1