2164. Sort Even and Odd Indices Independently

#### QUESTION:

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

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. 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 {
func sortEvenOdd(_ nums: [Int]) -> [Int] {
var result:[Int] = []
var flag:Bool = true
var evenArr:[Int] = []
var oddArr:[Int] = []
for num in nums {
if flag {
evenArr.append(num)
} else {
oddArr.append(num)
}
flag = !flag
}
evenArr = evenArr.sorted()
oddArr = oddArr.sorted(by: >)
flag = true
for index in 0...nums.count - 1 {
if flag {
result.append(evenArr[0])
evenArr.remove(at: 0)
} else {
result.append(oddArr[0])
oddArr.remove(at: 0)
}
flag = !flag
}
return result
}
}
``````
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