LeetCode in Kotlin

3701. Compute Alternating Sum

Easy

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

Solution

class Solution {
    fun alternatingSum(nums: IntArray): Int {
        var sum = 0
        for (i in nums.indices) {
            val num = nums[i]
            if (i % 2 == 0) {
                sum += num
            } else {
                sum -= num
            }
        }
        return sum
    }
}

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