LeetCode #363 — HARD

Max Sum of Rectangle No Larger Than K

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

Given an m x n matrix matrix and an integer k, return the max sum of a rectangle in the matrix such that its sum is no larger than k.

It is guaranteed that there will be a rectangle with a sum no larger than k.

Example 1:

Input: matrix = [[1,0,1],[0,-2,3]], k = 2
Output: 2
Explanation: Because the sum of the blue rectangle [[0, 1], [-2, 3]] is 2, and 2 is the max number no larger than k (k = 2).

Example 2:

Input: matrix = [[2,2,-1]], k = 3
Output: 3

Constraints:

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 100
  • -100 <= matrix[i][j] <= 100
  • -105 <= k <= 105

Follow up: What if the number of rows is much larger than the number of columns?

Patterns Used
🔍Modified Binary Search→📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Given an m x n matrix matrix and an integer k, return the max sum of a rectangle in the matrix such that its sum is no larger than k. It is guaranteed that there will be a rectangle with a sum no larger than k.

Baseline thinking

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

Pattern signal: Array · Binary Search · Segment Tree

Example 1

[[1,0,1],[0,-2,3]]
2

Example 2

[[2,2,-1]]
3
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #363: Max Sum of Rectangle No Larger Than K
class Solution {
    public int maxSumSubmatrix(int[][] matrix, int k) {
        int m = matrix.length;
        int n = matrix[0].length;
        final int inf = 1 << 30;
        int ans = -inf;
        for (int i = 0; i < m; ++i) {
            int[] nums = new int[n];
            for (int j = i; j < m; ++j) {
                for (int h = 0; h < n; ++h) {
                    nums[h] += matrix[j][h];
                }
                int s = 0;
                TreeSet<Integer> ts = new TreeSet<>();
                ts.add(0);
                for (int x : nums) {
                    s += x;
                    Integer y = ts.ceiling(s - k);
                    if (y != null) {
                        ans = Math.max(ans, s - y);
                    }
                    ts.add(s);
                }
            }
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(m^2 × n × log n)
Space
O(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.

Related Problems
#315Count of Smaller Numbers After Selfhard→#327Count of Range Sumhard→#456132 Patternmedium→#493Reverse Pairshard→#497Random Point in Non-overlapping Rectanglesmedium→