Galactic Trade Imbalance — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Galactic Trade Imbalance problem optimally.
O(n)O(1)Problem Description
You are given a sequence of trade transaction values represented as an array of integers. The array indices start at 0. Compute the maximum possible absolute difference between the sum of values at even indices and the sum of values at odd indices. The required output is a single integer representing this absolute difference.
Input format:
- The first line contains a single integer n (1 ≤ n ≤ 10^5), the number of transactions.
- The second line contains n space‑separated integers, each representing the value of a transaction. Each value satisfies -10^9 ≤ value ≤ 10^9.
Output format:
- Output one integer: the absolute difference between the sum of even‑indexed transactions and the sum of odd‑indexed transactions.
The task is straightforward: iterate through the array, accumulate two sums—one for even indices and one for odd indices—and finally output the absolute value of their difference.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Galactic Trade Imbalance"
WHY DOES IT MATTER?
This pattern is a foundational example of 'single-pass accumulation with conditional logic.' It teaches candidates to avoid unnecessary data structures and to leverage deterministic properties of indices (like parity) to optimize space complexity. It is a stepping stone to more complex partitioning problems.
OPTIMIZATION CHALLENGE
The key insight is recognizing that you do not need to store the values at even or odd indices separately. You only need their sums. This reduces space complexity from O(n) to O(1) by using two variables instead of two arrays.
REAL-WORLD CONNECTION
This is analogous to load balancing in distributed systems where requests are routed to servers based on a hash of the request ID (even/odd). Monitoring the 'imbalance' between server loads requires tracking cumulative metrics per partition in real-time without storing all request data, which is exactly what this algorithm does.
In an interview, explicitly state that you are using a single pass to avoid O(n) space. Mention that you are tracking the parity of the index using i % 2 or a boolean flag that toggles each iteration. This demonstrates awareness of both time and space efficiency.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a linear scan where the core challenge is not the summation itself, but the correct partitioning of indices into even and odd sets. A naive approach might involve creating two separate arrays or lists to store values at even and odd indices, then summing them. While this works, it incurs O(n) extra space and unnecessary memory allocation overhead. The optimal paradigm recognizes that the parity of an index (i % 2) is deterministic and can be evaluated on-the-fly during a single pass. By maintaining two running accumulators—one for even indices and one for odd indices—we can compute the required sums in O(1) space.
Interview Questions on This Problem
Q1How would you modify this solution if the array was extremely large and stored in a distributed system, requiring you to compute the difference without loading the entire array into memory?
You would use a streaming approach. Process the data in chunks, maintaining the running sums for even and odd indices as you go. Since the index parity is global, you must track the starting index of each chunk to correctly determine the parity of the first element in that chunk. This allows O(1) memory usage relative to the total array size, only requiring memory for the current chunk and the two accumulators.
Q2If the array values could be negative, does the logic for calculating the absolute difference change? How do you handle integer overflow in languages like Java or C++?
The logic remains identical because we are summing values and then taking the absolute difference. However, integer overflow is a critical concern. In Java, use long for the accumulators instead of int to prevent overflow during summation. In C++, use long long. The final result should be cast back to the required type or handled as a 64-bit integer if the problem constraints allow large sums.
Q3Can you generalize this problem to find the maximum difference between the sum of values at indices divisible by k and the sum of values at indices not divisible by k?
Yes. You would maintain an array of k accumulators, where accumulator[i] stores the sum of values at indices where index % k == i. Iterate through the array once, adding each value to the appropriate accumulator based on its index modulo k. Finally, compute the sum of the first accumulator (for indices divisible by k) and the sum of the remaining k-1 accumulators, then return the absolute difference. This generalizes to O(n) time and O(k) space.
Examples
Input
5 1 2 3 4 5
Output
3
Explanation: Even indices: 0, 2, 4 → values 1, 3, 5 → sum = 9. Odd indices: 1, 3 → values 2, 4 → sum = 6. Absolute difference = |9 - 6| = 3.
Input
4 -1 10 -3 4
Output
18
Explanation: Even indices: 0, 2 → values -1, -3 → sum = -4. Odd indices: 1, 3 → values 10, 4 → sum = 14. Absolute difference = |-4 - 14| = 18.
Input
6 0 0 0 0 0 0
Output
0
Explanation: Both even and odd sums are 0, so the difference is |0 - 0| = 0.
Input
7 5 -2 7 -3 4 1 -6
Output
14
Explanation: Even indices: 0, 2, 4, 6 → values 5, 7, 4, -6 → sum = 10. Odd indices: 1, 3, 5 → values -2, -3, 1 → sum = -4. Absolute difference = |10 - (-4)| = 14.
Constraints
- 1 <= n <= 10^5
- -10^9 <= nums[i] <= 10^9
- The sum of all transaction values fits within a 64‑bit signed integer.
Optimal Approach & Strategy
Iterate through the array once, maintaining two running sums: one for even indices and one for odd indices. Return the absolute difference between the two sums after the loop completes.
Brute Force Approach
Create two separate arrays, one for values at even indices and one for values at odd indices. Sum both arrays and return the absolute difference of the two sums.
Code Solutions
'use strict';
const fs = require('fs');
const input = fs.readFileSync(0, 'utf8').trim().split(/\s+/).map(Number);
let idx = 0;
const n = input[idx++];
let sumEven = 0, sumOdd = 0;
for (let i = 0; i < n; i++) {
const val = input[idx++];
if (i % 2 === 0) sumEven += val;
else sumOdd += val;
}
const result = Math.abs(sumEven - sumOdd);
console.log(result.toString());#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
if(!(cin >> n)) return 0;
long long sumEven = 0, sumOdd = 0;
for(int i = 0; i < n; ++i) {
long long x; cin >> x;
if(i % 2 == 0) sumEven += x;
else sumOdd += x;
}
cout << llabs(sumEven - sumOdd) << '\n';
return 0;
}
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line;
line = br.readLine();
if (line == null || line.isEmpty()) return;
int n = Integer.parseInt(line.trim());
line = br.readLine();
StringTokenizer st = new StringTokenizer(line);
long sumEven = 0, sumOdd = 0;
for (int i = 0; i < n; i++) {
long val = Long.parseLong(st.nextToken());
if (i % 2 == 0) sumEven += val;
else sumOdd += val;
}
System.out.println(Math.abs(sumEven - sumOdd));
}
}
import sys
def main():
data = sys.stdin.read().strip().split()
if not data:
return
n = int(data[0])
arr = list(map(int, data[1:1+n]))
sum_even = sum(arr[i] for i in range(0, n, 2))
sum_odd = sum(arr[i] for i in range(1, n, 2))
print(abs(sum_even - sum_odd))
if __name__ == "__main__":
main()
'use strict';
const fs = require('fs');
const input = fs.readFileSync(0, 'utf8').trim().split(/\s+/).map(Number);
let idx = 0;
const n = input[idx++];
let sumEven = 0, sumOdd = 0;
for (let i = 0; i < n; i++) {
const val = input[idx++];
if (i % 2 === 0) sumEven += val;
else sumOdd += val;
}
const result = Math.abs(sumEven - sumOdd);
console.log(result.toString());
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