# 2164. Sort Even and Odd Indices Independently

## Description

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

## Solutions

• class Solution {
public int[] sortEvenOdd(int[] nums) {
int n = nums.length;
int[] a = new int[(n + 1) >> 1];
int[] b = new int[n >> 1];
for (int i = 0, j = 0; j < n >> 1; i += 2, ++j) {
a[j] = nums[i];
b[j] = nums[i + 1];
}
if (n % 2 == 1) {
a[a.length - 1] = nums[n - 1];
}
Arrays.sort(a);
Arrays.sort(b);
int[] ans = new int[n];
for (int i = 0, j = 0; j < a.length; i += 2, ++j) {
ans[i] = a[j];
}
for (int i = 1, j = b.length - 1; j >= 0; i += 2, --j) {
ans[i] = b[j];
}
return ans;
}
}

• class Solution {
public:
vector<int> sortEvenOdd(vector<int>& nums) {
int n = nums.size();
vector<int> a;
vector<int> b;
for (int i = 0; i < n; ++i) {
if (i % 2 == 0)
a.push_back(nums[i]);
else
b.push_back(nums[i]);
}
sort(a.begin(), a.end());
sort(b.begin(), b.end(), greater<int>());
vector<int> ans(n);
for (int i = 0, j = 0; j < a.size(); i += 2, ++j) ans[i] = a[j];
for (int i = 1, j = 0; j < b.size(); i += 2, ++j) ans[i] = b[j];
return ans;
}
};

• class Solution:
def sortEvenOdd(self, nums: List[int]) -> List[int]:
a = sorted(nums[::2])
b = sorted(nums[1::2], reverse=True)
nums[::2] = a
nums[1::2] = b
return nums


• func sortEvenOdd(nums []int) []int {
n := len(nums)
var a []int
var b []int
for i, v := range nums {
if i%2 == 0 {
a = append(a, v)
} else {
b = append(b, v)
}
}
ans := make([]int, n)
sort.Ints(a)
sort.Slice(b, func(i, j int) bool {
return b[i] > b[j]
})
for i, j := 0, 0; j < len(a); i, j = i+2, j+1 {
ans[i] = a[j]
}
for i, j := 1, 0; j < len(b); i, j = i+2, j+1 {
ans[i] = b[j]
}
return ans
}