Leetcode #1292: Maximum Side Length of a Square with Sum Less than or Equal to Threshold
In this guide, we solve Leetcode #1292 Maximum Side Length of a Square with Sum Less than or Equal to Threshold in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Binary Search, Matrix, Prefix Sum
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
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.
Python Solution
class Solution:
def maxSideLength(self, mat: List[List[int]], threshold: int) -> int:
def check(k: int) -> bool:
for i in range(m - k + 1):
for j in range(n - k + 1):
v = s[i + k][j + k] - s[i][j + k] - s[i + k][j] + s[i][j]
if v <= threshold:
return True
return False
m, n = len(mat), len(mat[0])
s = [[0] * (n + 1) for _ in range(m + 1)]
for i, row in enumerate(mat, 1):
for j, x in enumerate(row, 1):
s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + x
l, r = 0, min(m, n)
while l < r:
mid = (l + r + 1) >> 1
if check(mid):
l = mid
else:
r = mid - 1
return l
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.