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Formatted question description: https://leetcode.ca/all/2164.html

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

Similar Questions:

Solution 1.

// OJ: https://leetcode.com/problems/sort-even-and-odd-indices-independently/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    vector<int> sortEvenOdd(vector<int>& A) {
        vector<int> odd, even, ans;
        for (int i = 0; i < A.size(); ++i) {
            if (i % 2) odd.push_back(A[i]);
            else even.push_back(A[i]);
        }
        sort(begin(odd), end(odd), greater<>());
        sort(begin(even), end(even));
        for (int i = 0, j = 0, k = 0; i < A.size(); ++i) {
            if (i % 2) ans.push_back(odd[j++]);
            else ans.push_back(even[k++]);
        }
        return ans;
    }
};

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