Even Odd: NLogN-Time, N-Space

Easy-LC problem today: Sort Even and Odd Indices Independently - LeetCode

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

Given the small constraints, we can afford to do a NLogN-Time and N-Space solution:
1/ Copy the even and odd elements in their own arrays
2/ Sort them independently
3/ Merge them back

Code is down below, cheers, ACC.


public int[] SortEvenOdd(int[] nums)
{
 int[] even = new int[nums.Length % 2 == 0 ? nums.Length / 2 : (nums.Length / 2 + 1)];
 int[] odd = new int[nums.Length / 2];
 int indexEven = 0;
 int indexOdd = 0;
 for (int i = 0; i < nums.Length; i++) { if (i % 2 == 0) even[indexEven++] = nums[i]; else odd[indexOdd++] = nums[i]; } Array.Sort(even); Array.Sort(odd); int[] retVal = new int[nums.Length]; indexEven = 0; indexOdd = odd.Length - 1; for (int i = 0; i < retVal.Length; i++) { if (i % 2 == 0) retVal[i] = even[indexEven++]; else retVal[i] = odd[indexOdd--]; } return retVal; }

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