What is the time complexity of Max Sum of Rectangle No Larger Than K?

The optimal solution has a time complexity of O(min(m, n)^2 * max(m, n) * log(max(m, n))).

What is the space complexity of Max Sum of Rectangle No Larger Than K?

The space complexity is O(max(m, n)).

What data structures are used to solve Max Sum of Rectangle No Larger Than K?

This problem uses Array, Binary Search, Matrix, Prefix Sum, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Max Sum of Rectangle No Larger Than K?

Insight 1: The core idea is to iterate through all possible pairs of rows (i, j) in the matrix. For each row pair, calculate the sum of elements in each column between these rows, creating a 1D array. Insight 2: The problem then reduces to finding the maximum subarray sum in the 1D array, which is no larger than k. This can be efficiently solved using prefix sums and a binary search (or a balanced tree) to find a suitable prefix sum.

Which companies ask LeetCode 363. Max Sum of Rectangle No Larger Than K in interviews?

This problem is frequently asked at DE Shaw.

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Max Sum of Rectangle No Larger Than K - LeetCode 363 Solution

Max Sum of Rectangle No Larger Than K - Complete Solution Guide

Max Sum of Rectangle No Larger Than K is LeetCode problem 363, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks us to find the maximum sum of any rectangle within a given m x n matrix, such that this sum is no larger than a given integer k. The problem guarantees that at least one such rectangle exists. In simpler terms, we want to find a rectangle in the matrix whose sum is as close to k as possible, without exceeding it. The inputs are the 2D matrix and the integer k. The output is the maximum sum of a rectangle that satisfies the condition.

Solution Approach

The solution iterates through all possible pairs of rows (or columns if transposed) in the matrix. For each pair, it creates a 1D array representing the sum of elements between those rows/columns. Then, it uses a binary search approach (implemented with bisect in Python, TreeSet in Java, set in C++, and custom binary search in C) to find the maximum subarray sum within this 1D array that is no larger than k. By comparing all these maximum sums, it determines the overall maximum sum across the entire matrix.

Step-by-Step Algorithm

Step 1: Determine if the matrix should be transposed (if m > n). This optimizes performance when rows greatly exceed columns.
Step 2: Iterate through all possible starting rows (i) from 0 to rows - 1.
Step 3: For each starting row i, iterate through all possible ending rows (j) from i to rows - 1.
Step 4: Inside the inner loop (j), create or update a 1D array (current_1d_array) representing the column sums between rows i and j.
Step 5: Call the `find_max_sum_le_k_1d` function (or its equivalent) to find the maximum subarray sum in the 1D array that is no larger than k.
Step 6: Update the overall `max_val` if the current subarray's maximum sum is greater.
Step 7: If `max_val` equals k, return k directly (as it cannot be improved further).
Step 8: After iterating through all possible row pairs, return the `max_val` found.

Key Insights

Insight 1: The core idea is to iterate through all possible pairs of rows (i, j) in the matrix. For each row pair, calculate the sum of elements in each column between these rows, creating a 1D array.
Insight 2: The problem then reduces to finding the maximum subarray sum in the 1D array, which is no larger than k. This can be efficiently solved using prefix sums and a binary search (or a balanced tree) to find a suitable prefix sum.
Insight 3: Consider the case where the number of rows is much larger than the number of columns. Transposing the matrix and processing it that way will improve performance, as it reduces the outer loop iterations.

Complexity Analysis

Time Complexity: O(min(m, n)^2 * max(m, n) * log(max(m, n)))

Space Complexity: O(max(m, n))

Topics

This problem involves: Array, Binary Search, Matrix, Prefix Sum, Ordered Set.

Companies

Asked at: DE Shaw.

Max Sum of Rectangle No Larger Than K | LeetCode 363

Hard

Array Binary Search Matrix Prefix Sum Ordered Set

Time: O(min(m, n)^2 * max(m, n) * log(max(m, n)))Space: O(max(m, n))

Solution Code

Available in:

import java.util.TreeSet;

class Solution {
    public int maxSumSubmatrix(int[][] matrix, int k) {
        int m = matrix.length;
        int n = matrix[0].length;

        boolean isTransposed = m > n;
        int rows = isTransposed ? n : m;
        int cols = isTransposed ? m : n;

        int maxVal = Integer.MIN_VALUE;

        for (int i = 0; i < rows; i++) {
            int[] current1DArray = new int[cols];
            for (int j = i; j < rows; j++) {
                for (int l = 0; l < cols; l++) {
                    if (isTransposed) {
                        current1DArray[l] += matrix[l][j];
                    } else {
                        current1DArray[l] += matrix[j][l];
                    }
                }

                int currentMax = findMaxSumLeK1D(current1DArray, k);

                if (currentMax > maxVal) {
                    maxVal = currentMax;
                }

                if (maxVal == k) {
                    return k;
                }
            }
        }

        return maxVal;
    }

    private int findMaxSumLeK1D(int[] arr, int kVal) {
        int maxSum = Integer.MIN_VALUE;
        int prefixSum = 0;
        TreeSet<Integer> sortedPrefixSums = new TreeSet<>();
        sortedPrefixSums.add(0);

        for (int num : arr) {
            prefixSum += num;
            Integer target = prefixSum - kVal;
            Integer ceiling = sortedPrefixSums.ceiling(target);

            if (ceiling != null) {
                maxSum = Math.max(maxSum, prefixSum - ceiling);
            }

            sortedPrefixSums.add(prefixSum);
        }

        return maxSum;
    }
}

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Problem Info

Description

Problem 363: Max Sum of Rectangle No Larger Than K - Detailed explanation and solution approaches.

Topics

Array Binary Search Matrix Prefix Sum Ordered Set

Companies

DE Shaw

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:363

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Determine if the matrix should be transposed (if m > n). This optimizes performance when rows greatly exceed columns.
2
Step 2: Iterate through all possible starting rows (i) from 0 to rows - 1.
3
Step 3: For each starting row i, iterate through all possible ending rows (j) from i to rows - 1.
4
Step 4: Inside the inner loop (j), create or update a 1D array (current_1d_array) representing the column sums between rows i and j.
5
Step 5: Call the `find_max_sum_le_k_1d` function (or its equivalent) to find the maximum subarray sum in the 1D array that is no larger than k.
6
Step 6: Update the overall `max_val` if the current subarray's maximum sum is greater.
7
Step 7: If `max_val` equals k, return k directly (as it cannot be improved further).
8
Step 8: After iterating through all possible row pairs, return the `max_val` found.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(min(m, n)^2 * max(m, n) * log(max(m, n)))

💾

Space Complexity

O(max(m, n))

📝

Why?

The algorithm iterates through all possible pairs of rows/columns, which takes O(min(m, n)^2). For each pair, it needs to sum the columns/rows, taking O(max(m, n)). The `find_max_sum_le_k_1d` function uses binary search on a sorted set of prefix sums of length O(max(m, n)), which takes O(log(max(m, n))). Therefore, the overall time complexity is O(min(m, n)^2 * max(m, n) * log(max(m, n))). The space complexity comes from the 1D array of column/row sums, which has a size of O(max(m, n)), and from the sorted set or TreeSet/TreeSet, which also stores at most max(m, n) prefix sums.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Kadane's Algorithm with Binary Search

An alternative approach uses Kadane's algorithm to find the maximum subarray sum in the 1D array. Kadane's algorithm can find the maximum subarray sum if there is no k constraint. By combining it with Binary Search and prefix sum, we can find the maximum sum subarray no larger than k.

Time: O(min(m,n)^2 * max(m,n) * log(max(m,n)))Space: O(max(m,n))

Trade-offs: Kadane's algorithm can potentially be more efficient in finding the maximum subarray sum if it wasn't for the k constraint. The binary search ensures that the maximum subarray no larger than k is found. It might be marginally faster in some specific cases but has the same asymptotic complexity.

✨

Key Insights

1Insight 1: The core idea is to iterate through all possible pairs of rows (i, j) in the matrix. For each row pair, calculate the sum of elements in each column between these rows, creating a 1D array.
2Insight 2: The problem then reduces to finding the maximum subarray sum in the 1D array, which is no larger than k. This can be efficiently solved using prefix sums and a binary search (or a balanced tree) to find a suitable prefix sum.
3Insight 3: Consider the case where the number of rows is much larger than the number of columns. Transposing the matrix and processing it that way will improve performance, as it reduces the outer loop iterations.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: The matrix can contain negative numbers. The algorithm handles this correctly by using INT_MIN (or equivalent) as the initial maximum sum and by correctly calculating prefix sums that may decrease as well as increase.
Edge case 2: The value of k can be negative. The algorithm functions correctly whether k is positive, negative, or zero.
Edge case 3: A zero-sized matrix should be handled. However, the constraints state that m and n are both at least 1, so this case does not need to be specifically addressed in the solution.
Edge case 4: When a rectangle sums to exactly k, we can return k immediately to save time. This is included in the code.
Edge case 5: Large negative values that when summed in the prefix sums exceed the integer limits. The problem constraints do not guarantee this but it can cause an overflow issue which would lead to inaccurate results.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle negative numbers correctly in the prefix sum calculation and initial maximum sum.

✗Common mistake 2: Not considering the case where the number of rows is much larger than the number of columns, leading to inefficient computation. Transposing can improve performance.

✗Common mistake 3: Incorrectly implementing the binary search or insort (or TreeSet/set operations), which can lead to incorrect results or timeouts.

✗Common mistake 4: Integer overflow with the sum of the matrix elements and the prefix sums. The constraints state -100 <= matrix[i][j] <= 100 and -10^5 <= k <= 10^5. The maximum size of the submatrix would be 100 * 100 elements which would be 100 * 100 * 100 = 10^6 or one million which is still fine with standard integers.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Financial analysis, where you might want to find the period of maximum profit within a certain budget or risk tolerance (represented by k).

🏦Real-world scenario 2: Image processing, where you might want to identify regions with the highest intensity values within a certain limit to avoid overexposure or noise.

🎮Real-world scenario 3: Data mining, where you want to identify clusters of high-value data points subject to constraints on resources or storage.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Max Sum of Rectangle No Larger Than K"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?