2164. Sort Even and Odd Indices Independently

Problem

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 (Java)

class Solution {
    public int[] sortEvenOdd(int[] nums) {
        int[] odd = new int[nums.length / 2];
        int[] even = new int[(nums.length + 1) / 2];
        int o = 0;
        int e = 0;
        for (int i = 0; i < nums.length; i++) {
            if (i % 2 == 0) {
                even[e] = nums[i];
                ++e;
            } else {
                odd[o] = nums[i];
                ++o;
            }
        }
        Arrays.sort(odd);
        Arrays.sort(even);
        e = 0;
        o = odd.length - 1;
        for (int i = 0; i < nums.length; i++) {
            if (i % 2 == 0) {
                nums[i] = even[e];
                ++e;
            } else {
                nums[i] = odd[o];
                --o;
            }
        }
        return nums;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).