### 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

1
2
3
4
5
6
7
8
9
10
11
12
13

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