Maximum Trade Value Optimization — Problem Statement & Solution Guide

ArraysMediumMixed

TimeO(n)

SpaceO(1)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Arrays and solve the Maximum Trade Value Optimization problem optimally.

TopicArrays

PatternMixed

TimeO(n)

SpaceO(1)

Problem Description

Given an integer array tradeValues, you may independently decide for each position i to either retain its original value tradeValues[i] or replace it with the value (sum of all elements before i) − (sum of all elements after i). After making a choice for every index, the array is summed. Return the greatest possible total sum achievable by optimal selections.

DSA Pattern Breakdown

"Maximum Trade Value Optimization"

medium

WHY DOES IT MATTER?

This pattern exemplifies the power of prefix‑sum transformations to turn a seemingly global decision into independent local choices, a technique that appears in many optimization and range‑query problems.

OPTIMIZATION CHALLENGE

The key insight is algebraically rewriting the replacement expression so that it depends only on a running prefix and a constant total, collapsing an O(n²) dependency into O(1) per index.

REAL-WORLD CONNECTION

Think of a distributed ledger where each node can either keep its own transaction amount or replace it with the net balance of all preceding versus succeeding nodes; computing the optimal ledger state efficiently mirrors the prefix‑sum trick used here.

During an interview, write down the formula for the alternative value, simplify it on the whiteboard, and immediately point out that you only need total and a running prefix – that signals you can achieve linear time.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to a per‑index decision: keep the original element or replace it with a value derived from the sums of elements before and after that index. By expanding the replacement expression, prefix‑sum(i‑1) − suffix‑sum(i+1) can be rewritten as 2*prefix + a[i] − total, where total is the sum of the whole array and prefix is the sum of elements before i. This transformation shows that the alternative value for each position can be computed in O(1) once we know the running prefix and the overall total, eliminating the need for nested loops. A naive O(n²) approach would recompute prefix and suffix for every i, which quickly becomes infeasible for large n (e.g., 10⁵ or more). The optimal paradigm is a single linear scan that maintains a running prefix sum while the total sum is pre‑computed, allowing us to evaluate max(original, alternative) for each index in constant time, yielding an overall O(n) solution with O(1) extra space.

Interview Questions on This Problem

Q1How would you compute the alternative value for each index without recomputing prefix and suffix sums each time?

Pre‑compute the total sum of the array, then iterate once while maintaining a running prefix sum. For index i the alternative is 2*prefix + a[i] − total, which is O(1) per element.

Q2If the array contains up to 10⁶ elements and values up to 10⁹, what data type should you use for the accumulated sums and why?

Use a 64‑bit signed integer (long long in C++, long in Java, or Python's int which is arbitrary precision) because the total sum can exceed 32‑bit limits (10⁶ * 10⁹ ≈ 10¹⁵).

Q3Can the optimal total be negative? How would you handle that case in your implementation?

Yes, if every alternative and original value is negative the maximum sum will be negative. The algorithm still works because we always take max(original, alternative) and accumulate the result; no special case is needed beyond using a signed type.

Examples

Example 1

Input

[3,-1,2]

Output

Explanation: i=0: before=0 after=1 diff=-1 keep 3; i=1: before=3 after=2 diff=1 replace -1 with 1; i=2: before=2 after=0 diff=2 keep 2; total=3+1+2=6.

Example 2

Input

[-5,4,-2,7]

Output

Explanation: All diffs are worse than original values, so keep every element. Sum = -5+4-2+7=4.

Example 3

Input

[10,1,1,1,1]

Output

Explanation: i0 keep 10; i1 diff=7 replace 1; i2 diff=9 replace 1; i3 diff=11 replace 1; i4 diff=13 replace 1; total=10+7+9+11+13=50.

Constraints

1 <= tradeValues.length <= 100000
-1000000000 <= tradeValues[i] <= 1000000000
Result fits in 64‑bit signed integer

Optimal Approach & Strategy

First compute the total sum, then scan once while maintaining a prefix sum, compute the alternative as 2*prefix + a[i] − total, and add the larger of the two to the answer, achieving O(n) time and O(1) space.

Brute Force Approach

For each index recompute the sum of elements before and after it, calculate the alternative value, compare with the original, and sum the best choice; this requires O(n²) time.

Code Solutions

Language:

JavaScript Solution

Time: O(n)

const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(data.length===0){process.exit(0);} 
let pos=0;
const n=data[pos++];
const arr=data.slice(pos,pos+n);
function maxTradeValue(tradeValues){
    let total=0n;
    for(const v of tradeValues) total+=BigInt(v);
    let prefix=0n;
    let ans=total;
    for(const v of tradeValues){
        const diff=2n*prefix - total;
        if(diff>0n) ans+=diff;
        prefix+=BigInt(v);
    }
    return ans;
}
console.log(maxTradeValue(arr).toString());

#include <bits/stdc++.h>
using namespace std;
long long maxTradeValue(const vector<int>& tradeValues){
    long long total=0; for(int v:tradeValues) total+=v;
    long long prefix=0, ans=total;
    for(size_t i=0;i<tradeValues.size();++i){
        long long diff=2*prefix - total; // gain if we replace at i
        if(diff>0) ans+=diff;
        prefix+=tradeValues[i];
    }
    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<<maxTradeValue(a);
    return 0;
}

import java.io.*;
import java.util.*;
public class Main {
    public static long maxTradeValue(int[] tradeValues){
        long total=0L;
        for(int v:tradeValues) total+=v;
        long prefix=0L;
        long ans=total;
        for(int v:tradeValues){
            long diff=2L*prefix - total;
            if(diff>0) ans+=diff;
            prefix+=v;
        }
        return ans;
    }
    public static void main(String[] args) throws Exception{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        String line=br.readLine();
        if(line==null||line.isEmpty()) return;
        int n=Integer.parseInt(line.trim());
        int[] arr=new int[n];
        StringTokenizer st=new StringTokenizer(br.readLine());
        for(int i=0;i<n;i++) arr[i]=Integer.parseInt(st.nextToken());
        System.out.println(maxTradeValue(arr));
    }
}

import sys
def max_trade_value(tradeValues):
    total=sum(tradeValues)
    prefix=0
    ans=total
    for v in tradeValues:
        diff=2*prefix - total
        if diff>0:
            ans+=diff
        prefix+=v
    return ans

def main():
    data=sys.stdin.read().strip().split()
    if not data:
        return
    n=int(data[0])
    arr=list(map(int,data[1:1+n]))
    print(max_trade_value(arr))
if __name__=="__main__":
    main()

const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(data.length===0){process.exit(0);} 
let pos=0;
const n=data[pos++];
const arr=data.slice(pos,pos+n);
function maxTradeValue(tradeValues){
    let total=0n;
    for(const v of tradeValues) total+=BigInt(v);
    let prefix=0n;
    let ans=total;
    for(const v of tradeValues){
        const diff=2n*prefix - total;
        if(diff>0n) ans+=diff;
        prefix+=BigInt(v);
    }
    return ans;
}
console.log(maxTradeValue(arr).toString());

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