class Solution:
    def sortArrayByParityII(self, nums: List[int]) -> List[int]:
        even=[]
        odd=[]
        final=[]
        for i in nums:
            if i %2==0:
                even.append(i)
            else:
                odd.append(i)
        for i in range(len(nums)//2):
            final.append(even[i])
            final.append(odd[i])
        return final
        


'''

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]

'''