Sort Array By Parity

Given an integer array nums, move all the even integers at the beginning of the array followed by all the odd integers.

Return any array that satisfies this condition.

 

Example 1:

Input: nums = [3,1,2,4]
Output: [2,4,3,1]
Explanation: The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Example 2:

Input: nums = [0]
Output: [0]

 

Constraints:

  • 1 <= nums.length <= 5000
  • 0 <= nums[i] <= 5000




 class Solution:
    def sortArrayByParity(self, nums: List[int]) -> List[int]:

        odd = []
        even = []

        for num in nums:
            if num % 2 == 0:
                even.append(num)
            else:
                odd.append(num)

        return even+odd

Random Note

From python 3.7 dict guarantees that order will be kept as they inserted, and popitem will use LIFO order but we need FIFO type system. so we need OrderedDict which have popIten(last = T/F) for this req. One thing, next(iter(dict)) will return the first key of the dict

Python