What is the time complexity of Maximize Sum Of Array After K Negations?

The optimal solution has a time complexity of O(nlogn).

What data structures are used to solve Maximize Sum Of Array After K Negations?

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

What are the key insights for solving Maximize Sum Of Array After K Negations?

**Prioritize flipping negative numbers:** Maximizing the array sum fundamentally involves making numbers as large (positive) as possible. The most impactful way to achieve this is to convert existing negative values into their positive counterparts. Sorting the `nums` array allows us to systematically address the most negative numbers first, as these yield the largest absolute increase in sum when flipped. **Strategically minimize harm for leftover negations:** Once all numbers in the array are non-negative (or `k` operations run out before all negatives are flipped), any further negation will inevitably reduce the total sum. To maximize the final sum, we must strategically minimize this reduction by repeatedly targeting the number with the smallest absolute value. Flipping `1` to `-1` causes less sum reduction than flipping `100` to `-100`.

Which companies ask LeetCode 1005. Maximize Sum Of Array After K Negations in interviews?

This problem is frequently asked at Druva.

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Maximize Sum Of Array After K Negations - LeetCode 1005 Solution

Maximize Sum Of Array After K Negations - Complete Solution Guide

Maximize Sum Of Array After K Negations is LeetCode problem 1005, a Easy 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 integer array nums and an integer k , modify the array in the following way: choose an index i and replace nums[i] with -nums[i] . You should apply this process exactly k times. You may choose the same index i multiple times. Return the largest possible sum of the array after modifying it in this way . Example 1: Input: nums = [4,2,3], k = 1 Output: 5 Explanation: Choose index 1 and nums becomes [4,-2,3]. Example 2: Input: nums = [3,-1,0,2], k = 3 Output: 6 Explanation: Choose indices (

Detailed Explanation

This problem asks us to manipulate an array `nums` by repeatedly choosing an index `i` and replacing `nums[i]` with `-nums[i]`. We must perform this operation exactly `k` times, and we're allowed to pick the same index multiple times. Our goal is to find the largest possible sum of the array elements after all `k` negations are complete. The core challenge lies in strategically applying these `k` negations to maximize the final sum, especially considering the implications of choosing the same index more than once.

Solution Approach

The most intuitive way to maximize a sum is to make all numbers as positive as possible. Our solution employs a greedy strategy: first, we sort the `nums` array to easily identify the most negative numbers. We then iterate through the array, flipping every negative number to its positive counterpart using one of our `k` operations. This step always increases the total sum. If we exhaust `k` or run out of negative numbers, we proceed to the second phase. At this point, all numbers in the array are non-negative. If we still have `k` operations remaining, any further negation will convert a positive number to a negative one, thus *decreasing* the total sum. To minimize this decrease, we must target the number with the smallest absolute value. The crucial insight here is the parity of the remaining `k`: if `k` is even, we can perform `k/2` pairs of negations on the smallest number (e.g., `5 -> -5 -> 5`), effectively canceling out and causing no net change to the sum. If `k` is odd, we perform `(k-1)/2` such pairs, and then one final effective negation on the smallest number, turning it negative. Therefore, after flipping all initial negatives, we check if `k` is still positive and odd. If so, we re-sort the array (because the initial flips might have changed the relative order of numbers, e.g., -10 becoming 10 might be larger than 5, but after sorting, 5 would be `nums[0]`), then flip the smallest element (`nums[0]`) to its negative form. Finally, we sum the modified array.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order. This allows efficient access to the smallest elements.
Step 2: Iterate through the array. For each negative element encountered, negate it and decrement `k`. This step continues until either all negative numbers have been processed or `k` becomes 0.
Step 3: If `k` is still greater than 0 after Step 2, it means we've negated all negative numbers. If `k` is odd, negating the smallest element (which is now non-negative) will increase the sum (or keep it the same if its 0).
Step 4: Calculate the sum of the modified array and return the result.

Key Insights

**Prioritize flipping negative numbers:** Maximizing the array sum fundamentally involves making numbers as large (positive) as possible. The most impactful way to achieve this is to convert existing negative values into their positive counterparts. Sorting the `nums` array allows us to systematically address the most negative numbers first, as these yield the largest absolute increase in sum when flipped.
**Strategically minimize harm for leftover negations:** Once all numbers in the array are non-negative (or `k` operations run out before all negatives are flipped), any further negation will inevitably reduce the total sum. To maximize the final sum, we must strategically minimize this reduction by repeatedly targeting the number with the smallest absolute value. Flipping `1` to `-1` causes less sum reduction than flipping `100` to `-100`.
**Leveraging `k`'s parity for 'undoing' flips:** The problem explicitly states we can choose the same index multiple times. This is critical: two negations on a number cancel each other out (e.g., `x -> -x -> x`). If, after handling all initial negative numbers, `k` operations are still remaining and `k` is *even*, we can pair up these negations on the current smallest absolute value number, resulting in no net change to the sum. Only if `k` is *odd* do we need to perform one final, effective flip on the smallest number, turning it negative to complete the `k` operations.

Complexity Analysis

Time Complexity: O(nlogn)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Sorting.

Companies

Asked at: Druva.

Maximize Sum Of Array After K Negations | LeetCode 1005

Easy

Array Greedy Sorting

Time: O(nlogn)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int largestSumAfterKNegations(int[] nums, int k) {
        Arrays.sort(nums);
        int n = nums.length;
        for (int i = 0; i < n && k > 0; ++i) {
            if (nums[i] < 0) {
                nums[i] = -nums[i];
                k--;
            }
        }
        if (k % 2 == 1) {
            Arrays.sort(nums);
            nums[0] = -nums[0];
        }
        int sum = 0;
        for (int num : nums) {
            sum += num;
        }
        return sum;
    }
}

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Problem Info

Description

Problem 1005: Maximize Sum Of Array After K Negations - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting

Companies

Druva

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1005

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

4 steps

1
Step 1: Sort the input array `nums` in ascending order. This allows efficient access to the smallest elements.
2
Step 2: Iterate through the array. For each negative element encountered, negate it and decrement `k`. This step continues until either all negative numbers have been processed or `k` becomes 0.
3
Step 3: If `k` is still greater than 0 after Step 2, it means we've negated all negative numbers. If `k` is odd, negating the smallest element (which is now non-negative) will increase the sum (or keep it the same if its 0).
4
Step 4: Calculate the sum of the modified array and return the result.

💡

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(nlogn)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is the sorting of the array, which takes O(n log n) time using efficient sorting algorithms like merge sort or quicksort. The rest of the operations (iteration, negation, summation) take linear O(n) time. The space complexity is O(1) because the algorithm works in-place; it modifies the original array without requiring significant extra space.

📖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

Using a Min-Heap

Instead of sorting, you could use a min-heap data structure. This allows you to efficiently find and remove the smallest element repeatedly.

Time: O(k log n)Space: O(n)

Trade-offs: This approach is more efficient if `k` is much smaller than `n`. Sorting is more efficient for larger `k` values.

✨

Key Insights

1**Prioritize flipping negative numbers:** Maximizing the array sum fundamentally involves making numbers as large (positive) as possible. The most impactful way to achieve this is to convert existing negative values into their positive counterparts. Sorting the `nums` array allows us to systematically address the most negative numbers first, as these yield the largest absolute increase in sum when flipped.
2**Strategically minimize harm for leftover negations:** Once all numbers in the array are non-negative (or `k` operations run out before all negatives are flipped), any further negation will inevitably reduce the total sum. To maximize the final sum, we must strategically minimize this reduction by repeatedly targeting the number with the smallest absolute value. Flipping `1` to `-1` causes less sum reduction than flipping `100` to `-100`.
3**Leveraging `k`'s parity for 'undoing' flips:** The problem explicitly states we can choose the same index multiple times. This is critical: two negations on a number cancel each other out (e.g., `x -> -x -> x`). If, after handling all initial negative numbers, `k` operations are still remaining and `k` is *even*, we can pair up these negations on the current smallest absolute value number, resulting in no net change to the sum. Only if `k` is *odd* do we need to perform one final, effective flip on the smallest number, turning it negative to complete the `k` operations.

📚

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.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: All elements in `nums` are positive. In this case, the algorithm will correctly identify the smallest element and negate it if `k` is odd.
Edge case 2: `k` is larger than the number of elements. The algorithm handles this by negating all elements and then potentially negating the smallest element again if `k` is odd after that.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where `k` is odd after negating all negative numbers. This can lead to an incorrect final sum.

✗Common mistake 2: Inefficiently finding the minimum element repeatedly without sorting or using a min-heap. This results in a time complexity of O(n*k), significantly worse than the optimal O(n log n).

✅

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: Optimizing resource allocation. Imagine you have a set of tasks with varying costs (positive or negative representing profit/loss). You want to choose `k` tasks to maximize the overall profit, negating the costs of some tasks as needed.

🏦Real-world scenario 2: Investment portfolio optimization. You might have a portfolio with investments yielding varying returns. You can perform a limited number of trades (k) to adjust your positions and maximize total returns by 'negating' (selling) some assets.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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: September 14, 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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Maximize Sum Of Array After K Negations"?

For this easy-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.

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