LeetCode 922. Sort Array By Parity II

Description

https://leetcode.com/problems/sort-array-by-parity-ii/

Given an array of integers nums, half of the integers in nums are odd, and the other half are even.

Sort the array so that whenever nums[i] is odd, i is odd, and whenever nums[i] is even, i is even.

Return any answer array that satisfies this condition.

Example 1:

Input: nums = [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.

Example 2:

Input: nums = [2,3]
Output: [2,3]

Constraints:

• 2 <= nums.length <= 2 * 104
• nums.length is even.
• Half of the integers in nums are even.
• 0 <= nums[i] <= 1000

Follow Up: Could you solve it in-place?

Explanation

Iterate the list and store the odd, even numbers separately. And then created a new empty list and based on indices to add odd, even numbers.

Python Solution

class Solution:
    def sortArrayByParityII(self, nums: List[int]) -> List[int]:
        results = []
        
        odds = []
        evens = []
        for num in nums:
            if num % 2 == 0:
                evens.append(num)
            else:
                odds.append(num)
        
        
        i = 0
        while i < len(nums):
            if i % 2 == 0:
                value = evens.pop(0)
            else:
                value = odds.pop(0)
            results.append(value)
            
            i += 1
        
        return results