LeetCode The Hard Way
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2164 - Sort Even and Odd Indices Independently (Easy)

https://leetcode.com/problems/sort-even-and-odd-indices-independently/

Problem Statement

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

Approach 1: Iteration

The simplest approach is to store odd-indexed numbers and even-indexed numbers separately, then sort them and merge them. One approach to store them separately is to check the parity (odd or even) of the iterating variable ( i in this example solution), and store them to their corresponding lists.

We then sort the two lists. We note that odd-indexed numbers need to be non-increasing, so we need to call reverse. Another approach to sort in reverse order is odds.sort(reverse = True).

We then put the numbers back to nums using a similar method of retrieving the parity of the iterating variable.

Written by@heiheihang
def sortEvenOdd(self, nums: List[int]) -> List[int]:

        #initialize the two lists for odd-indexed numbers and even-indexed numbers
        odds = []
        evens = []

        #place the numbers into the two lists accordingly
        for i in range(len(nums)):
            if(i % 2 == 1):
                odds.append(nums[i])
            else:
                evens.append(nums[i])

        #sort the two lists
        odds.sort()
        evens.sort()

        #as we want odd-indexed numbers to be non-increasing, we need to reverse it
        odds.reverse()

        #initialize result
        result = []

        #iterate all numbers
        for i in range(len(nums)):

            #put the numbers back
            #we have i//2 because we take turns visiting odds and evens
            if(i % 2 == 1):
                result.append(odds[i//2])
            else:
                result.append(evens[i//2])

        #return result
        return result

2163 - Minimum Difference in Sums After Removal of Elements (Hard)

Author: @wkw | https://leetcode.com/problems/minimum-difference-in-sums-after-removal-of-elements/

2165 - Smallest Value of the Rearranged Number (Medium)

Authors: @heiheihang, @wkw | https://leetcode.com/problems/smallest-value-of-the-rearranged-number/

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2164 - Sort Even and Odd Indices Independently (Easy)Problem LinkProblem StatementApproach 1: Iteration