LeetCode in Kotlin

2164. Sort Even and Odd Indices Independently

Easy

You are given a 0-indexed integer array nums. Rearrange the values of nums according to the following rules:

1. Sort the values at odd indices of nums in non-increasing order.
   • For example, if nums = [4,**1**,2,**3**] before this step, it becomes [4,**3**,2,**1**] after. The values at odd indices 1 and 3 are sorted in non-increasing order.
2. Sort the values at even indices of nums in non-decreasing order.
   • For example, if nums = [**4**,1,**2**,3] before this step, it becomes [**2**,1,**4**,3] after. The values at even indices 0 and 2 are sorted in non-decreasing order.

Return the array formed after rearranging the values of nums.

Example 1:

Input: nums = [4,1,2,3]

Output: [2,3,4,1]

Explanation:

First, we sort the values present at odd indices (1 and 3) in non-increasing order.

So, nums changes from [4,1,2,3] to [4,3,2,1].

Next, we sort the values present at even indices (0 and 2) in non-decreasing order. So, nums changes from [4,1,2,3] to [2,3,4,1].

Thus, the array formed after rearranging the values is [2,3,4,1].

Example 2:

Input: nums = [2,1]

Output: [2,1]

Explanation: Since there is exactly one odd index and one even index, no rearrangement of values takes place.

The resultant array formed is [2,1], which is the same as the initial array.

Constraints:

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

Solution

class Solution {
    fun sortEvenOdd(nums: IntArray): IntArray {
        val odd = IntArray(nums.size / 2)
        val even = IntArray((nums.size + 1) / 2)
        var o = 0
        var e = 0
        for (i in nums.indices) {
            if (i % 2 == 0) {
                even[e] = nums[i]
                ++e
            } else {
                odd[o] = nums[i]
                ++o
            }
        }
        odd.sort()
        even.sort()
        e = 0
        o = odd.size - 1
        for (i in nums.indices) {
            if (i % 2 == 0) {
                nums[i] = even[e]
                ++e
            } else {
                nums[i] = odd[o]
                --o
            }
        }
        return nums
    }
}

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