Is Compute Alternating Sum easy or hard?

Compute Alternating Sum is considered an Easy problem. The solution involves basic array traversal and simple arithmetic, making it a common introductory exercise for practicing simulation and loop logic.

Compute Alternating Sum Python/Java solution

The implementation is identical across languages: loop through the array and alternate between adding and subtracting elements. Python, Java, C++, Go, and TypeScript versions typically use either an index parity check or a sign variable to maintain the alternating pattern.

How to solve Compute Alternating Sum in O(n)?

Iterate through the array once while maintaining a running total. Add the value when the index is even and subtract it when the index is odd, or multiply each element by a sign variable that alternates between +1 and -1. This single traversal computes the result in O(n) time and constant space.

What is the best approach for Compute Alternating Sum?

The best approach is a single-pass simulation that alternates between addition and subtraction while iterating through the array. You can either check index parity (even add, odd subtract) or maintain a sign variable that flips each iteration. Both versions run in O(n) time with O(1) extra space.

Is Compute Alternating Sum asked at Google/Amazon/Meta?

Alternating sum style problems commonly appear as warm-up or screening questions at large tech companies. They test array traversal, simple simulation logic, and clean coding under time constraints rather than complex algorithms.

What data structure is used in Compute Alternating Sum?

The problem primarily uses an array as the input structure. The solution only needs a few scalar variables such as a running total and possibly a sign indicator, so no additional data structures are required.

What is the time complexity of Compute Alternating Sum?

The time complexity is O(n) because the algorithm processes each element of the array exactly once. No nested loops or additional data structures are required, so the space complexity stays O(1).

3701. Compute Alternating Sum - Practice Online

You are given an integer array nums.

The alternating sum of nums is the value obtained by adding elements at even indices and subtracting elements at odd indices. That is, nums[0] - nums[1] + nums[2] - nums[3]...

Return an integer denoting the alternating sum of nums.

Example 1:

Input: nums = [1,3,5,7]

Output: -4

Explanation:

Elements at even indices are nums[0] = 1 and nums[2] = 5 because 0 and 2 are even numbers.
Elements at odd indices are nums[1] = 3 and nums[3] = 7 because 1 and 3 are odd numbers.
The alternating sum is nums[0] - nums[1] + nums[2] - nums[3] = 1 - 3 + 5 - 7 = -4.

Example 2:

Input: nums = [100]

Output: 100

Explanation:

The only element at even indices is nums[0] = 100 because 0 is an even number.
There are no elements on odd indices.
The alternating sum is nums[0] = 100.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100