What is the time complexity of Maximum Matrix Sum?

The optimal solution has a time complexity of O(n*m).

What data structures are used to solve Maximum Matrix Sum?

This problem uses Array, Greedy, Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Matrix Sum?

Insight 1: The core idea is that two negative numbers can be converted to positive numbers by applying the operation to them. Therefore, the goal is to minimize the number of negative numbers to either 0 or 1. Insight 2: The operations can be performed such that all negative signs can be moved around the matrix until only one negative number remains, or all numbers become positive.

Which companies ask LeetCode 1975. Maximum Matrix Sum in interviews?

This problem is frequently asked at Honeywell.

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Maximum Matrix Sum - LeetCode 1975 Solution

Maximum Matrix Sum - Complete Solution Guide

Maximum Matrix Sum is LeetCode problem 1975, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an n x n integer matrix . You can do the following operation any number of times: Choose any two adjacent elements of matrix and multiply each of them by -1 . Two elements are considered adjacent if and only if they share a border . Your goal is to maximize the summation of the matrix's elements. Return the maximum sum of the matrix's elements using the operation mentioned above. Example 1: Input: matrix = [[1,-1],[-1,1]] Output: 4 Explanation: We can follow the following steps to

Detailed Explanation

The problem asks us to maximize the sum of elements in an n x n matrix. We can perform an operation where we choose two adjacent elements (sharing a border) and multiply both by -1. Our goal is to find the maximum possible sum of the matrix elements after performing this operation any number of times.

Solution Approach

The solution counts the number of negative numbers and calculates the sum of absolute values of all numbers in the matrix. If the count of negative numbers is even, we return the sum of absolute values. If the count is odd, we return the sum of absolute values minus twice the minimum absolute value in the matrix. This is because one negative number remains and we want to select the smallest absolute value as the remaining negative number.

Step-by-Step Algorithm

Step 1: Initialize `total_sum` to 0, `neg_count` to 0, and `min_abs_val` to infinity.
Step 2: Iterate through each element of the matrix.
Step 3: If the element is negative, increment `neg_count`.
Step 4: Calculate the absolute value of the current element.
Step 5: Add the absolute value to `total_sum`.
Step 6: Update `min_abs_val` to be the minimum of the current `min_abs_val` and the absolute value of the current element.
Step 7: After iterating through all elements, check if `neg_count` is even.
Step 8: If `neg_count` is even, return `total_sum`.
Step 9: If `neg_count` is odd, return `total_sum - 2 * min_abs_val`.

Key Insights

Insight 1: The core idea is that two negative numbers can be converted to positive numbers by applying the operation to them. Therefore, the goal is to minimize the number of negative numbers to either 0 or 1.
Insight 2: The operations can be performed such that all negative signs can be moved around the matrix until only one negative number remains, or all numbers become positive.
Insight 3: If there's an even number of negative numbers, we can always eliminate all negative signs. If there's an odd number, we're left with one negative number, and to minimize the effect of this negative number, we want to minimize its absolute value.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Matrix.

Companies

Asked at: Honeywell.

Maximum Matrix Sum | LeetCode 1975

Medium

Array Greedy Matrix

Time: O(n*m)Space: O(1)

Solution Code

Available in:

class Solution {
    public long maxMatrixSum(int[][] matrix) {
        long total_sum = 0;
        int neg_count = 0;
        int min_abs_val = Integer.MAX_VALUE;

        for (int[] row : matrix) {
            for (int val : row) {
                if (val < 0) {
                    neg_count++;
                }

                int abs_v = Math.abs(val);
                total_sum += abs_v;
                min_abs_val = Math.min(min_abs_val, abs_v);
            }
        }

        if (neg_count % 2 == 0) {
            return total_sum;
        } else {
            return total_sum - 2 * (long)min_abs_val;
        }
    }
}

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Problem Info

Description

Problem 1975: Maximum Matrix Sum - Detailed explanation and solution approaches.

Topics

Array Greedy Matrix

Companies

Honeywell

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1975

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

9 steps

1
Step 1: Initialize `total_sum` to 0, `neg_count` to 0, and `min_abs_val` to infinity.
2
Step 2: Iterate through each element of the matrix.
3
Step 3: If the element is negative, increment `neg_count`.
4
Step 4: Calculate the absolute value of the current element.
5
Step 5: Add the absolute value to `total_sum`.
6
Step 6: Update `min_abs_val` to be the minimum of the current `min_abs_val` and the absolute value of the current element.
7
Step 7: After iterating through all elements, check if `neg_count` is even.
8
Step 8: If `neg_count` is even, return `total_sum`.
9
Step 9: If `neg_count` is odd, return `total_sum - 2 * min_abs_val`.

💡

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(n*m)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The code's time complexity is directly proportional to the number of elements in the matrix since each element is visited once. The space complexity is constant because the memory used does not depend on the size of the input matrix.

📖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

None - The provided solution is already optimal

There isn't a significantly more efficient algorithm to solve this problem. Dynamic programming is unnecessarily complex.

Time: N/ASpace: N/A

Trade-offs: N/A

✨

Key Insights

1Insight 1: The core idea is that two negative numbers can be converted to positive numbers by applying the operation to them. Therefore, the goal is to minimize the number of negative numbers to either 0 or 1.
2Insight 2: The operations can be performed such that all negative signs can be moved around the matrix until only one negative number remains, or all numbers become positive.
3Insight 3: If there's an even number of negative numbers, we can always eliminate all negative signs. If there's an odd number, we're left with one negative number, and to minimize the effect of this negative number, we want to minimize its absolute value.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: Matrix with all positive numbers. In this case, `neg_count` will be 0 (even), and the code will correctly return the sum of all elements.
Edge case 2: Matrix with all negative numbers. If the size of the matrix is even, then number of negative integers is even and it returns the sum of absolute values. If the size of the matrix is odd, then the code selects the minimum of the absolute values and subtracts it twice from the sum of absolute values.
Edge case 3: Matrix containing zeros. Zeros do not affect the negativity count, and the algorithm correctly handles them because abs(0) = 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the absolute value of the matrix elements before summing. This will lead to an incorrect result if there are negative numbers.

✗Common mistake 2: Failing to account for the case where the number of negative elements is odd. In this case, one negative sign must remain, and the smallest absolute value element should be chosen to carry that negative sign.

✅

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: Optimization problems where pairwise adjustments can lead to a global maximum/minimum, such as resource allocation or network flow problems.

🏦Real-world scenario 2: Image processing, where pixel values might be adjusted to enhance contrast or reduce noise using local neighborhood operations, and the algorithm helps maintain a balance of intensities.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: November 23, 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

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

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 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.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Maximum Matrix Sum"?

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

🤖 Need help with this problem?