Alternating Sum Maximization — 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 Maximization problem optimally.

TopicArrays

PatternAlternating sum subarray

TimeO(n)

SpaceO(1)

Problem Description

Given an integer array values, choose any non‑empty contiguous subarray and compute its alternating sum, where the first element is added, the second subtracted, the third added, and so on (i.e., for subarray values[l..r] the sum is values[l]−values[l+1]+values[l+2]−… ). Return the maximum possible alternating sum among all subarrays. The algorithm must run in linear time relative to the array length.

DSA Pattern Breakdown

"Alternating Sum Maximization"

medium

WHY DOES IT MATTER?

Understanding how to transform a problem with alternating signs into a standard maximum subarray problem teaches the powerful technique of sign‑normalization, which appears in finance (profit‑loss streams) and signal processing.

OPTIMIZATION CHALLENGE

The key insight is that the sign of each element is deterministic once the subarray’s start parity is fixed, allowing us to pre‑compute two linear‑time scans instead of quadratic enumeration.

REAL-WORLD CONNECTION

Think of a stock trader who alternately buys and sells each day; the net profit is an alternating sum of daily price changes. Optimizing the trading window is analogous to finding the best alternating‑sign subarray.

When coding, keep two variables (pos, neg) representing the best alternating sum ending at the current index with a '+' or '-' sign; update them in place and track the global maximum from the '+' variable.

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 the dot product of the subarray with a sign pattern (+,‑, +,‑,…). By pre‑multiplying the original array with two possible sign patterns—one assuming the subarray starts with a plus and the other assuming it starts with a minus—we reduce the problem to finding the maximum subarray sum for each transformed array. This is exactly the classic Kadane’s algorithm, which runs in linear time. A naïve solution would enumerate every possible (l,r) pair and compute the alternating sum in O(n^2) time, which quickly becomes infeasible for large n because the number of subarrays grows quadratically. The optimal paradigm leverages dynamic programming: at each index we maintain the best alternating sum ending at that position for both parity states, updating them in O(1) and thus achieving overall O(n) time with O(1) extra space.

Interview Questions on This Problem

Q1How would you modify Kadane’s algorithm to handle the alternating sign requirement?

Create two running sums: one for subarrays that would end with a '+' sign and one for those ending with a '-' sign. At each element, update them by either extending the previous sum (flipping the sign) or starting fresh, then keep the global maximum of the '+'‑ending sum.

Q2Why does the answer never come from a subarray that starts with a negative contribution?

If a subarray starts with a negative contribution, dropping that first element yields a strictly larger alternating sum because the sign pattern flips, turning the former negative into a positive contribution; therefore the optimal subarray always begins with a '+' in its local sign context.

Q3Can this problem be solved with a segment tree? If so, what would be the node information?

Yes; each node would store four values: max prefix sum when the segment starts with '+', max prefix when it starts with '-', max suffix for both start signs, and the overall max alternating sum inside the segment. Merging two children requires handling sign flips between segments.

Examples

Example 1

Input

[4,-1,2,3]

Output

Explanation: All subarrays are examined. The subarray [4,-1,2] yields 4-(-1)+2=7, which is larger than any other alternating sum (e.g., [4,-1]=5, [4,-1,2,3]=4, [2]=2, etc.). Hence the answer is 7.

Example 2

Input

[-5,6,-2,1,-4]

Output

Explanation: Starting at index 1 gives the best result. For subarray [6,-2,1,-4] the alternating sum is 6-(-2)+1-(-4)=6+2+1+4=13, which exceeds all other possibilities (e.g., single 6 gives 6, [6,-2,1] gives 9). Thus the maximum is 13.

Example 3

Input

[10,20,30]

Output

Explanation: The single element subarray [30] yields an alternating sum of 30, which is greater than any longer subarray (e.g., [10,20,30] gives 10-20+30=20). Therefore the maximum alternating sum is 30.

Constraints

1 <= values.length <= 100000
-1000000000 <= values[i] <= 1000000000
Time complexity O(n)
O(1) additional memory

Optimal Approach & Strategy

Maintain two running sums for subarrays ending at the current index—one assuming the next sign is '+' and the other '-'. Update them in O(1) per element and track the global maximum, achieving O(n) time and O(1) space.

Brute Force Approach

Enumerate every possible subarray, compute its alternating sum by iterating through its elements, and keep the maximum; this requires O(n^2) time. The double loop makes it impractical for large inputs.

Code Solutions

Language:

JavaScript Solution

Time: O(n)

const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let pos=0;
const n = data[pos++]||0;
const values = data.slice(pos,pos+n);
function maxAlternatingSum(arr){
    if(arr.length===0) return 0;
    let dpPlus = arr[0]; // '+' sign at current end
    let dpMinus = Number.NEGATIVE_INFINITY; // '-' sign at current end
    let ans = dpPlus;
    for(let i=1;i<arr.length;i++){
        const v = arr[i];
        const newPlus = Math.max(v, dpMinus + v);
        const newMinus = Math.max(-v, dpPlus - v);
        dpPlus = newPlus;
        dpMinus = newMinus;
        if(dpPlus>ans) ans=dpPlus;
    }
    return ans;
}
if(n!==0) console.log(maxAlternatingSum(values));

#include <bits/stdc++.h>
using namespace std;
long long maxAlternatingSum(const vector<int>& values){
    int n=values.size();
    long long dpPlus=values[0]; // ending at i with '+' sign
    long long dpMinus=LLONG_MIN; // ending at i with '-' sign (invalid for first element)
    long long ans=dpPlus;
    for(int i=1;i<n;++i){
        long long v=values[i];
        long long newPlus=max(v, dpMinus+v);
        long long newMinus=max(-v, dpPlus - v);
        dpPlus=newPlus;
        dpMinus=newMinus;
        ans=max(ans,dpPlus);
    }
    return ans;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n; if(!(cin>>n)) return 0;
    vector<int> a(n);
    for(int i=0;i<n;++i)cin>>a[i];
    cout<<maxAlternatingSum(a);
    return 0;
}

import java.util.*;
public class Main {
    public static long maxAlternatingSum(int[] values){
        if(values.length==0) return 0;
        long dpPlus = values[0]; // '+' sign at current end
        long dpMinus = Long.MIN_VALUE; // '-' sign at current end (invalid initially)
        long ans = dpPlus;
        for(int i=1;i<values.length;i++){
            long v = values[i];
            long newPlus = Math.max(v, dpMinus + v);
            long newMinus = Math.max(-v, dpPlus - v);
            dpPlus = newPlus;
            dpMinus = newMinus;
            if(dpPlus>ans) ans = dpPlus;
        }
        return ans;
    }
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        if(!sc.hasNextInt()) return;
        int n = sc.nextInt();
        int[] arr = new int[n];
        for(int i=0;i<n;i++) arr[i]=sc.nextInt();
        System.out.print(maxAlternatingSum(arr));
    }
}

import sys
def maxAlternatingSum(values):
    if not values:
        return 0
    dp_plus = values[0]          # ending here with '+' sign
    dp_minus = float('-inf')     # ending here with '-' sign (invalid for first element)
    ans = dp_plus
    for v in values[1:]:
        new_plus = max(v, dp_minus + v)
        new_minus = max(-v, dp_plus - v)
        dp_plus, dp_minus = new_plus, new_minus
        if dp_plus > ans:
            ans = dp_plus
    return ans

data = sys.stdin.read().strip().split()
if not data:
    sys.exit()
n = int(data[0])
arr = list(map(int, data[1:1+n]))
print(maxAlternatingSum(arr))

const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let pos=0;
const n = data[pos++]||0;
const values = data.slice(pos,pos+n);
function maxAlternatingSum(arr){
    if(arr.length===0) return 0;
    let dpPlus = arr[0]; // '+' sign at current end
    let dpMinus = Number.NEGATIVE_INFINITY; // '-' sign at current end
    let ans = dpPlus;
    for(let i=1;i<arr.length;i++){
        const v = arr[i];
        const newPlus = Math.max(v, dpMinus + v);
        const newMinus = Math.max(-v, dpPlus - v);
        dpPlus = newPlus;
        dpMinus = newMinus;
        if(dpPlus>ans) ans=dpPlus;
    }
    return ans;
}
if(n!==0) console.log(maxAlternatingSum(values));

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