How do you solve Sort Even and Odd Indices Independently on LeetCode?

Sort Even and Odd Indices Independently (LeetCode #2164) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Elements at even and odd indices can be treated independently.

What is the brute force solution for Sort Even and Odd Indices Independently?

The brute force approach for Sort Even and Odd Indices Independently is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves separating the elements at even and odd indices, sorting them independently, and then merging them back into the original array. This is straightforward but inefficient due to the sorting step. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Sort Even and Odd Indices Independently?

Sort Even and Odd Indices Independently uses the following concepts: Array, Sorting. The recommended patterns to study are: Array, Sorting.

What are the interview tips for Sort Even and Odd Indices Independently?

Clearly explain your thought process while coding. Consider edge cases and test your solution with them. Practice writing clean and efficient code.

What are common mistakes when solving Sort Even and Odd Indices Independently?

Confusing the order of sorting for even and odd indices. Not handling edge cases like arrays with only one element.

#2164

Sort Even and Odd Indices Independently

Easy

ArraySortingArraySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(n)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution directly sorts the even and odd indexed elements in a single pass through the array, which is more efficient. By using two separate lists and sorting them independently, we achieve the desired result without unnecessary complexity.

⚙️

Algorithm

4 steps

1Step 1: Initialize two empty lists for even and odd indexed elements.
2Step 2: Iterate through the original array and populate the even and odd lists based on the index.
3Step 3: Sort the even list in non-decreasing order and the odd list in non-increasing order.
4Step 4: Create a new result array and fill it by alternating between the sorted even and odd lists.

solution.py13 lines

1def sortEvenOdd(nums):
2    even = sorted(nums[i] for i in range(0, len(nums), 2))
3    odd = sorted((nums[i] for i in range(1, len(nums), 2)), reverse=True)
4    result = []
5    even_index, odd_index = 0, 0
6    for i in range(len(nums)):
7        if i % 2 == 0:
8            result.append(even[even_index])
9            even_index += 1
10        else:
11            result.append(odd[odd_index])
12            odd_index += 1
13    return result

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting of the even and odd lists. The space complexity is O(n) because we are using additional space to store the sorted lists.

1Elements at even and odd indices can be treated independently.
2Sorting can be performed separately for both groups.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.