3701. Compute Alternating Sum (Leetcode Easy)
Compute Alternating Sum - Video Solution
Watch 10 video solutions for Compute Alternating Sum, a easy level problem involving Array, Simulation. This walkthrough by Programming Live with Larry has 359 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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] = 1andnums[2] = 5because 0 and 2 are even numbers. - Elements at odd indices are
nums[1] = 3andnums[3] = 7because 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] = 100because 0 is an even number. - There are no elements on odd indices.
- The alternating sum is
nums[0] = 100.
Constraints:
1 <= nums.length <= 1001 <= 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.
Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Index Parity Simulation | O(n) | O(1) | Best when implementing directly from the definition of alternating addition and subtraction using index parity. |
| Sign Toggle Simulation | O(n) | O(1) | Cleaner implementation when you want minimal conditional checks inside the loop. |