Sort Array By Parity

View Sort Array By Parity on LeetCode

Statistics

Time Spent Coding
2 minutes

Time Complexity
O(n) - Every element in the input list must be seen, resulting in the O(b) time complexity.

Space Complexity
O(n) - Every element in the input list is duplicated and placed in either evens or odds, resulting in the O(n) space complexity.

Runtime Beats
99.69% of other submissions

Memory Beats
51.77% of other sumbissions

Explanation

For every element n in nums, if the modulo of 2 is larger than 0, append that number to the odds list; else append n to evens.

Finally, return the combination of both lists.

Solution

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

        for n in nums:
            if n%2:
                odds.append(n)
            else:
                evens.append(n)

        
        return evens + odds