LeetCode #1292 — MEDIUM

Maximum Side Length of a Square with Sum Less than or Equal to Threshold

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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 or equal to 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

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n <= 300
0 <= mat[i][j] <= 10⁴
0 <= threshold <= 10⁵

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]]
4

Example 2

[[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2]]
1

Step 02

Core Insight

What unlocks the optimal approach

Store prefix sum of all grids in another 2D array.
Try all possible solutions and if you cannot find one return 0.
If x is a valid answer then any y < x is also valid answer. Use binary search to find answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1292: Maximum Side Length of a Square with Sum Less than or Equal to Threshold
class Solution {
    private int m;
    private int n;
    private int threshold;
    private int[][] s;

    public int maxSideLength(int[][] mat, int threshold) {
        m = mat.length;
        n = mat[0].length;
        this.threshold = threshold;
        s = new int[m + 1][n + 1];
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + mat[i - 1][j - 1];
            }
        }
        int l = 0, r = Math.min(m, n);
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (check(mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }

    private boolean check(int k) {
        for (int i = 0; i < m - k + 1; ++i) {
            for (int j = 0; j < n - k + 1; ++j) {
                if (s[i + k][j + k] - s[i][j + k] - s[i + k][j] + s[i][j] <= threshold) {
                    return true;
                }
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #1292: Maximum Side Length of a Square with Sum Less than or Equal to Threshold
func maxSideLength(mat [][]int, threshold int) int {
	m, n := len(mat), len(mat[0])
	s := make([][]int, m+1)
	for i := range s {
		s[i] = make([]int, n+1)
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + mat[i-1][j-1]
		}
	}
	check := func(k int) bool {
		for i := 0; i < m-k+1; i++ {
			for j := 0; j < n-k+1; j++ {
				if s[i+k][j+k]-s[i][j+k]-s[i+k][j]+s[i][j] <= threshold {
					return true
				}
			}
		}
		return false
	}
	l, r := 0, min(m, n)
	for l < r {
		mid := (l + r + 1) >> 1
		if check(mid) {
			l = mid
		} else {
			r = mid - 1
		}
	}
	return l
}

# Accepted solution for LeetCode #1292: Maximum Side Length of a Square with Sum Less than or Equal to Threshold
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

// Accepted solution for LeetCode #1292: Maximum Side Length of a Square with Sum Less than or Equal to Threshold
impl Solution {
    pub fn max_side_length(mat: Vec<Vec<i32>>, threshold: i32) -> i32 {
        let m = mat.len();
        let n = mat[0].len();
        let mut s = vec![vec![0; n + 1]; m + 1];
        for i in 1..=m {
            for j in 1..=n {
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + mat[i - 1][j - 1];
            }
        }
        let check = |k: usize| -> bool {
            for i in 0..=m - k {
                for j in 0..=n - k {
                    if s[i + k][j + k] - s[i][j + k] - s[i + k][j] + s[i][j] <= threshold {
                        return true;
                    }
                }
            }
            false
        };
        let mut l = 0usize;
        let mut r = m.min(n);
        while l < r {
            let mid = (l + r + 1) >> 1;
            if check(mid) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        l as i32
    }
}

// Accepted solution for LeetCode #1292: Maximum Side Length of a Square with Sum Less than or Equal to Threshold
function maxSideLength(mat: number[][], threshold: number): number {
    const m = mat.length;
    const n = mat[0].length;
    const s: number[][] = Array(m + 1)
        .fill(0)
        .map(() => Array(n + 1).fill(0));
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + mat[i - 1][j - 1];
        }
    }
    const check = (k: number): boolean => {
        for (let i = 0; i < m - k + 1; ++i) {
            for (let j = 0; j < n - k + 1; ++j) {
                if (s[i + k][j + k] - s[i + k][j] - s[i][j + k] + s[i][j] <= threshold) {
                    return true;
                }
            }
        }
        return false;
    };

    let l = 0;
    let r = Math.min(m, n);
    while (l < r) {
        const mid = (l + r + 1) >> 1;
        if (check(mid)) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }
    return l;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n × log \min(m, n)

Space

O(m × n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.