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Sort Even and Odd Indices Independently

Difficulty: Easy

Problem 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.
  2. Sort the values at even indices of nums in non-decreasing order.

Return the array formed after rearranging the values of nums.

Key Insights

  • Values at odd indices must be sorted in non-increasing order.
  • Values at even indices must be sorted in non-decreasing order.
  • The rearrangement of values occurs independently for even and odd indices.
  • The final output array will contain the rearranged values based on the above sorting rules.

Space and Time Complexity

Time Complexity: O(n log n), where n is the length of the input array (due to the sorting operations). Space Complexity: O(n), for storing the sorted values.

Solution

To solve this problem, we will:

  1. Extract the values at odd indices and sort them in non-increasing order.
  2. Extract the values at even indices and sort them in non-decreasing order.
  3. Reconstruct the original array by placing the sorted values back into their respective indices.

We will use two lists to hold the sorted values for even and odd indices, and then combine them to form the final result.

Code Solutions

def sortEvenOdd(nums):
    # Extract values at even indices
    even_indices = [nums[i] for i in range(0, len(nums), 2)]
    # Extract values at odd indices
    odd_indices = [nums[i] for i in range(1, len(nums), 2)]
    
    # Sort even indices in non-decreasing order
    even_indices.sort()
    # Sort odd indices in non-increasing order
    odd_indices.sort(reverse=True)
    
    # Reconstruct the final array
    result = []
    even_pointer, odd_pointer = 0, 0
    
    for i in range(len(nums)):
        if i % 2 == 0:
            result.append(even_indices[even_pointer])
            even_pointer += 1
        else:
            result.append(odd_indices[odd_pointer])
            odd_pointer += 1
            
    return result
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