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Compute Alternating Sum - Solution & Explanation

EasyArraySimulation5 min read
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Problem Statement

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

Approach Overview

Problem Overview: You are given an array of integers and need to compute its alternating sum. The pattern is simple: add the first element, subtract the second, add the third, subtract the fourth, and continue this pattern until the end of the array.

Approach 1: Index Parity Simulation (O(n) time, O(1) space)

The most direct way is to iterate through the array once and decide whether to add or subtract each value based on its index. If the index is even, add the value to the running total; if the index is odd, subtract it. This works because the alternating pattern directly maps to index parity. The algorithm performs a single pass over the array and uses only one accumulator variable, making it both time and memory efficient. This approach fits naturally with problems involving arrays and straightforward simulation.

Approach 2: Sign Toggle Simulation (O(n) time, O(1) space)

Another clean implementation uses a sign variable that flips after each element. Start with sign = 1 and multiply the current element by the sign before adding it to the result. After processing an element, flip the sign using sign *= -1. This avoids checking index parity and keeps the loop logic minimal. The method still processes each element exactly once, so the time complexity remains O(n) with constant O(1) space.

Recommended for interviews: Interviewers expect the linear simulation solution. Showing the index-parity version demonstrates that you recognize the alternating pattern quickly. The sign-toggle variation shows cleaner reasoning and often results in simpler code. Both run in O(n) time with O(1) space, which is optimal for this problem.

Solution

We can directly traverse the array nums. For each index i, if i is even, we add nums[i] to the answer; otherwise, we subtract nums[i] from the answer.

Finally, we return the answer.

The time complexity is O(n), where n is the length of array nums. The space complexity is O(1).

Code

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Python

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Go

TypeScript

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Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Index Parity SimulationO(n)O(1)Best when implementing directly from the definition of alternating addition and subtraction using index parity.
Sign Toggle SimulationO(n)O(1)Cleaner implementation when you want minimal conditional checks inside the loop.

Video Solution

3701. Compute Alternating Sum (Leetcode Easy) • Programming Live with Larry • 359 views views

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Frequently Asked Questions

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).

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