LeetCode 3701 - Compute Alternating Sum

Difficulty: 🟢 Easy
Topics: Array, Simulation

Solution

Problem Understanding

This problem asks us to compute the alternating sum of an integer array nums. The alternating sum is calculated by taking the sum of elements at even indices (0, 2, 4, ...) and subtracting the sum of elements at odd indices (1, 3, 5, ...). For example, given nums = [1, 3, 5, 7], we compute 1 - 3 + 5 - 7 = -4.

The input nums is an array of integers with length between 1 and 100, and each element is between 1 and 100. This guarantees that we will never have an empty array or values outside the positive integer range, which simplifies handling edge cases like empty input or negative numbers. Edge cases to consider include arrays of length 1 (only one element at index 0, which is even), arrays of length 2, and arrays where all elements are the same.

The expected output is a single integer representing the alternating sum.

Approaches

The most straightforward way to solve this problem is to iterate through the array while keeping a running sum. For each element, we check its index: if it is even, we add it to the sum; if it is odd, we subtract it. This is correct because it directly follows the definition of alternating sum. Given the constraints, this method is efficient, but we will also discuss why it is optimal.

A brute-force variation could separately sum the even-indexed elements and the odd-indexed elements and then subtract the odd sum from the even sum. While logically identical, it requires extra storage for two sums, which is unnecessary.

The key insight is that we can compute the alternating sum in a single pass without additional data structures. We can simply iterate once, adding or subtracting each element depending on its index parity.

Approach	Time Complexity	Space Complexity	Notes
Brute Force (two sums)	O(n)	O(1)	Compute even and odd sums separately, then subtract
Optimal (single pass)	O(n)	O(1)	Compute alternating sum directly while iterating

Algorithm Walkthrough

Initialize a variable total to zero. This will hold the running alternating sum.
Iterate through the array using a loop with index i from 0 to len(nums) - 1.
For each element nums[i], check if i is even. If i % 2 == 0, add nums[i] to total.
If i is odd, subtract nums[i] from total.
After the loop ends, return total.

Why it works: At every iteration, the algorithm maintains the invariant that total is equal to the alternating sum of all elements seen so far. By processing indices in order and applying the add/subtract rule, the algorithm guarantees correctness.

Python Solution

from typing import List

class Solution:
    def alternatingSum(self, nums: List[int]) -> int:
        total = 0
        for i, value in enumerate(nums):
            if i % 2 == 0:
                total += value
            else:
                total -= value
        return total

In this Python implementation, we initialize total to zero. We use enumerate to access both the index and the value of each element. For even indices, we add the value; for odd indices, we subtract it. Finally, we return the accumulated total.

Go Solution

func alternatingSum(nums []int) int {
    total := 0
    for i, value := range nums {
        if i%2 == 0 {
            total += value
        } else {
            total -= value
        }
    }
    return total
}

In Go, we declare total as an integer and iterate over the slice nums with range, obtaining both the index and the value. The logic mirrors the Python solution. Go handles integer arithmetic safely within the given constraints, so no additional handling for overflow is needed.

Worked Examples

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

i	nums[i]	Action	total
0	1	+1	1
1	3	-3	-2
2	5	+5	3
3	7	-7	-4

Output: -4

Example 2: nums = [100]

i	nums[i]	Action	total
0	100	+100	100

Output: 100

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	We iterate through the array once, performing constant-time operations per element
Space	O(1)	Only a single integer variable `total` is used, independent of input size

The algorithm is linear in time and constant in space, which is optimal for the problem size and constraints.

Test Cases

# Provided examples
assert Solution().alternatingSum([1,3,5,7]) == -4  # alternating sum with multiple elements
assert Solution().alternatingSum([100]) == 100    # single element array

# Boundary values
assert Solution().alternatingSum([1, 1]) == 0     # sum cancels out
assert Solution().alternatingSum([1, 2, 3]) == 2  # small odd-length array
assert Solution().alternatingSum([100]*100) == 0  # maximum length with same elements

# Stress test
assert Solution().alternatingSum(list(range(1, 101))) == -50  # sum of first 100 numbers with alternating signs

Test	Why
[1,3,5,7]	Standard alternating sum check
[100]	Single-element array edge case
[1, 1]	Even-length array where sum cancels
[1,2,3]	Odd-length small array
[100]*100	Maximum length array with uniform values
list(range(1, 101))	Large input to verify correctness and performance

Edge Cases

One important edge case is an array of length one. Since the only element is at index 0, the alternating sum equals the element itself. Our solution handles this because the loop still processes index 0 correctly.

A second edge case is an array of length two, where the sum might cancel out. For instance, [1, 1] should return 0. The solution correctly subtracts the second element from the first.

A third edge case is an array with all identical elements, especially at maximum allowed length. This can test integer arithmetic consistency and performance. Since the values are within the allowed range, the algorithm handles this without overflow, correctly alternating addition and subtraction across all indices.