Alternating Sum Subarray Maximum — Problem Statement & Solution Guide

ArraysMediumAlternating sum subarray

TimeO(n)

SpaceO(1)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Arrays and solve the Alternating Sum Subarray Maximum problem optimally.

TopicArrays

PatternAlternating sum subarray

TimeO(n)

SpaceO(1)

Problem Description

Given an integer array nums, a subarray is any contiguous segment of nums. The alternating sum of a subarray that starts at index l and ends at index r (inclusive) is defined as sum_{k=0}^{r‑l} (-1)^k · nums[l+k]; the first element contributes positively, the second negatively, the third positively, and so on. Your task is to compute the maximum possible alternating sum among all subarrays of nums. Input: the first line contains an integer n (the length of the array). The second line contains n space‑separated integers. Output: a single integer – the largest alternating sum achievable.

DSA Pattern Breakdown

"Alternating Sum Subarray Maximum"

medium

WHY DOES IT MATTER?

Maximum alternating‑sum subarrays appear in financial signal processing, error‑correction codes, and any domain where gains and losses alternate; mastering the sign‑flip reduction teaches you to convert a seemingly exotic objective into a classic one, a reusable pattern for many interview problems.

OPTIMIZATION CHALLENGE

The breakthrough is recognizing that the alternating sign pattern is deterministic and can be baked into the data itself, turning a O(n^2) search into two linear Kadane passes. This eliminates the need for nested loops or prefix‑difference tables.

REAL-WORLD CONNECTION

Think of a power‑grid where generators and loads alternate along a line; the net effective power is the alternating sum of capacities. By re‑labeling loads as negative, the net power becomes a simple total, just like converting the alternating sum to a normal sum for easier analysis.

When you spot a fixed sign pattern, immediately consider a parity‑based transformation; it often collapses the problem to a well‑known O(n) algorithm like Kadane, saving you minutes in an interview.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The alternating sum of a subarray can be expressed as a linear combination of the original elements with coefficients that flip sign every position. A naïve solution would enumerate every possible (l,r) pair, compute the alternating sum in O(r‑l) time and keep the maximum, leading to O(n^3) overall – impossible for n up to 10^5 or more. The key insight is to view the sign pattern as a pre‑computed transformation: if we multiply every element at an odd index by –1, the alternating sum of any subarray that starts at an even index becomes the ordinary sum of the transformed segment. Likewise, flipping the sign of every element at an even index (or equivalently negating the whole transformed array) captures subarrays that start at an odd index. Thus the problem reduces to finding the maximum subarray sum (Kadane’s algorithm) on two linear‑time scans, yielding an O(n) solution with O(1) extra space.

Interview Questions on This Problem

Q1How would you adapt Kadane’s algorithm to compute the maximum alternating sum subarray in a single pass?

Maintain two DP variables: bestPos – the maximum alternating sum of a subarray ending at i with a positive sign on nums[i]; bestNeg – the same but with a negative sign on nums[i]. Update bestPos = max(nums[i], bestNeg + nums[i]) and bestNeg = max(-nums[i], bestPos - nums[i]) (using previous bestPos). The answer is the maximum bestPos seen.

Q2Why does the “sign‑flip” transformation work for subarrays that start at an odd index?

When a subarray starts at an odd index, the first element should be added positively, but in the original array its index parity is odd, so we need the opposite sign pattern. Negating the whole transformed array (or swapping the parity of the sign flip) flips the sign of every term, turning the alternating sum into a regular sum again, allowing Kadane to capture those cases.

Q3Can you solve the problem in O(1) extra space without storing the transformed array? Explain.

Yes. While scanning the original array, keep two running sums: curEven and curOdd, representing the best alternating sum ending at the current position assuming the subarray started at an even or odd index respectively. Update them using the recurrence relations above; no auxiliary array is required, only a few scalar variables.

Examples

Example 1

Input

5
4 -1 2 -3 5

Output

Explanation: All subarrays are examined. The whole array gives 4-(-1)+2-(-3)+5 = 15, which is larger than any shorter subarray (e.g., [2,-3,5] yields 10, [4,-1] yields 5). Hence the maximum alternating sum is 15.

Example 2

Input

5
-2 3 -5 7 -1

Output

Explanation: The subarray [3,-5,7,-1] (indices 1‑4) yields 3-(-5)+7-(-1)=3+5+7+1=16, which exceeds all other candidates such as the whole array (-18) or single element 7.

Example 3

Input

4
1 2 3 4

Output

Explanation: Single‑element subarrays give sums 1,2,3,4. Any longer subarray alternates signs and reduces the total (e.g., [1,2,3] → 1-2+3=2, [1,2] → -1). The largest value is the single element 4.

Constraints

1 <= nums.length <= 100000
-1000000000 <= nums[i] <= 1000000000
Solution must run in O(n) time and O(1) extra space

Optimal Approach & Strategy

Transform the array by flipping signs on odd indices (and also its negation) and apply Kadane’s algorithm twice, achieving O(n) time and O(1) extra space.

Brute Force Approach

Enumerate all O(n^2) subarrays, compute each alternating sum in O(length) and keep the maximum, resulting in O(n^3) time.

Code Solutions

Language:

JavaScript Solution

Time: O(n)

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxAlternatingSum = function(nums) {
    if (nums.length === 0) return 0;
    
    let pos = nums[0];
    let neg = -nums[0];
    let ans = nums[0];
    
    for (let i = 1; i < nums.length; i++) {
        let new_pos = Math.max(nums[i], neg + nums[i]);
        let new_neg = Math.max(-nums[i], pos - nums[i]);
        
        pos = new_pos;
        neg = new_neg;
        
        ans = Math.max(ans, Math.max(pos, neg));
    }
    
    return ans;
};

class Solution {
public:
    long long maxAlternatingSum(vector<int>& nums) {
        if (nums.empty()) return 0;
        
        // pos: Max alternating sum of a subarray ending at current index where the current element has a positive sign
        // neg: Max alternating sum of a subarray ending at current index where the current element has a negative sign
        long long pos = nums[0];
        long long neg = -nums[0];
        long long ans = nums[0];
        
        for (int i = 1; i < nums.size(); ++i) {
            // If we append nums[i] to a subarray ending at i-1:
            // - If the previous sign was negative, the new sign is positive. New pos = neg + nums[i]
            // - If the previous sign was positive, the new sign is negative. New neg = pos - nums[i]
            // We can also start a new subarray at i:
            // - New pos = nums[i]
            // - New neg = -nums[i]
            
            long long new_pos = max(static_cast<long long>(nums[i]), neg + nums[i]);
            long long new_neg = max(static_cast<long long>(-nums[i]), pos - nums[i]);
            
            pos = new_pos;
            neg = new_neg;
            
            // The maximum alternating sum for any subarray ending at i is max(pos, neg)
            // But since we want the maximum over all subarrays, we update ans
            ans = max(ans, max(pos, neg));
        }
        
        return ans;
    }
};

class Solution {
    public long maxAlternatingSum(int[] nums) {
        if (nums.length == 0) return 0;
        
        long pos = nums[0];
        long neg = -nums[0];
        long ans = nums[0];
        
        for (int i = 1; i < nums.length; i++) {
            long new_pos = Math.max((long)nums[i], neg + nums[i]);
            long new_neg = Math.max((long)-nums[i], pos - nums[i]);
            
            pos = new_pos;
            neg = new_neg;
            
            ans = Math.max(ans, Math.max(pos, neg));
        }
        
        return ans;
    }
}

class Solution:
    def maxAlternatingSum(self, nums: List[int]) -> int:
        if not nums:
            return 0
            
        pos = nums[0]
        neg = -nums[0]
        ans = nums[0]
        
        for i in range(1, len(nums)):
            new_pos = max(nums[i], neg + nums[i])
            new_neg = max(-nums[i], pos - nums[i])
            
            pos = new_pos
            neg = new_neg
            
            ans = max(ans, max(pos, neg))
        
        return ans

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxAlternatingSum = function(nums) {
    if (nums.length === 0) return 0;
    
    let pos = nums[0];
    let neg = -nums[0];
    let ans = nums[0];
    
    for (let i = 1; i < nums.length; i++) {
        let new_pos = Math.max(nums[i], neg + nums[i]);
        let new_neg = Math.max(-nums[i], pos - nums[i]);
        
        pos = new_pos;
        neg = new_neg;
        
        ans = Math.max(ans, Math.max(pos, neg));
    }
    
    return ans;
};

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