Package 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 Details

    • Solution

      public Solution()

  • Method Details

    • sortEvenOdd

      public int[] sortEvenOdd(int[] nums)