g2101_2200.s2164_sort_even_and_odd_indices_independently

Class Solution

java.lang.Object

g2101_2200.s2164_sort_even_and_odd_indices_independently.Solution

public class Solution
extends Object

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

Constructor Summary

Constructors

Constructor and Description
Solution()

Method Summary

Modifier and Type	Method and Description
int[]	sortEvenOdd(int[] nums)

Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Detail

Solution

public Solution()

Method Detail

sortEvenOdd

public int[] sortEvenOdd(int[] nums)