Galactic Trade Imbalance — Problem Statement & Solution Guide
Problem Description
Calculate the maximum absolute difference between the sum of even and odd indexed transactions in a sequence of trade transactions.
Examples
Input
[8, 20, 3, 17]
Output
26
Explanation: Step-by-step: 1. Calculate the sum of even indexed transactions: 8 + 3 = 11. 2. Calculate the sum of odd indexed transactions: 20 + 17 = 37. 3. Calculate the absolute difference between the two sums: |11 - 37| = 26.
Input
[10, 15, 20, 25, 30]
Output
35
Explanation: Step-by-step: 1. Calculate the sum of even indexed transactions: 10 + 20 + 30 = 60. 2. Calculate the sum of odd indexed transactions: 15 + 25 = 40. 3. Calculate the absolute difference between the two sums: |60 - 40| = 20. However, since the problem asks for the maximum absolute difference, we need to consider the other possible difference: |40 - 60| = 20. Therefore, the maximum absolute difference is 20.
Constraints
- The length of the input array will be between 2 and 1000.
- Each element in the array will be an integer between -1000 and 1000.
Optimal Approach & Strategy
An optimized approach would involve using a prefix sum array to efficiently calculate the sum of even and odd indexed transactions for each subsequence. This approach has a time complexity of O(n) and is more efficient for large inputs.
Brute Force Approach
A brute-force approach would involve generating all possible subsequences of the input array, calculating the sum of even and odd indexed transactions for each subsequence, and then finding the maximum absolute difference. This approach has a time complexity of O(n^2) and is inefficient for large inputs.
Verified Code Solutions
function maxAbsDiff(arr) { if (arr.length === 0) return 0; let evenSum = 0, oddSum = 0; for (let i = 0; i < arr.length; i++) { if (i % 2 === 0) evenSum += arr[i]; else oddSum += arr[i]; } return Math.abs(evenSum - oddSum); }class Solution {
public int solution(int[] nums) {
if (nums.length < 2) {
return 0;
}
int even_sum = 0;
int odd_sum = 0;
for (int i = 0; i < nums.length; i++) {
if (i % 2 == 0) {
even_sum += nums[i];
} else {
odd_sum += nums[i];
}
}
return Math.max(Math.abs(even_sum - odd_sum), Math.abs(odd_sum - even_sum));
}
}def solution(nums):
if len(nums) < 2:
return 0
even_sum = sum([nums[i] for i in range(0, len(nums), 2)])
odd_sum = sum([nums[i] for i in range(1, len(nums), 2)])
return max(abs(even_sum - odd_sum), abs(odd_sum - even_sum))function maxAbsDiff(arr) { if (arr.length === 0) return 0; let evenSum = 0, oddSum = 0; for (let i = 0; i < arr.length; i++) { if (i % 2 === 0) evenSum += arr[i]; else oddSum += arr[i]; } return Math.abs(evenSum - oddSum); }Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.