Sort Array By Parity

Question

Given an array A of non-negative integers, return an array consisting of all the even elements of A, followed by all the odd elements of A.

You may return any answer array that satisfies this condition.

Example 1:

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

Solution

Approach 1: Sort

Since the array is already sorted, we can keep two pointers and , where is the slow-runner while is the fast-runner. As long as , we increment to skip the duplicate.

When we encounter , the duplicate run has ended so we must copy its value to . is then incremented and we repeat the same process again reaches the end of array.

class Solution:
    def sortArrayByParity(self, A):
        even=[]
        odd=[]
        for i in range(0,len(A)):
            if A[i]%2==0:
                even.append(A[i])
            else:
                odd.append(A[i])
        return even+odd


if __name__ == "__main__":
    nums = [3,1,2,4]
    answer=Solution()
    results=answer.sortArrayByParity(nums)
    print(results)

Runtime: 68 ms, faster than 83.75% of Python3 online submissions for Sort Array By Parity.
Memory Usage: 13.8 MB, less than 5.69% of Python3 online submissions for Sort Array By Parity.

Use a custom comparator when sorting, to sort by parity.

class Solution(object):
    def sortArrayByParity(self, A):
        A.sort(key = lambda x: x % 2)
        return A

Time Complexity: O(NlogN), where N is the length of A.
Space Complexity: O(N) for the sort, depending on the built-in implementation of sort.

Approach 2: Two Pass

Write all the even elements first, then write all the odd elements.

class Solution(object):
    def sortArrayByParity(self, A):
        return ([x for x in A if x % 2 == 0] +
                [x for x in A if x % 2 == 1])

Time Complexity: O(N), where N is the length of A.
Space Complexity: O(N) for the sort, depending on the built-in implementation of sort.

Approach 3: In-Place

Intuition

If we want to do the sort in-place, we can use quicksort, a standard textbook algorithm.

Algorithm

We'll maintain two pointers i and j. The loop invariant is everything below i has parity 0 (ie. A[k] % 2 == 0 when k < i), and everything above j has parity 1.

Then, there are 4 cases for (A[i] % 2, A[j] % 2):

If it is (0, 1), then everything is correct: i++ and j--.
If it is (1, 0), we swap them so they are correct, then continue.
If it is (0, 0), only the i place is correct, so we i++ and continue.
If it is (1, 1), only the j place is correct, so we j-- and continue.

Throughout all 4 cases, the loop invariant is maintained, and j-i is getting smaller. So eventually we will be done with the array sorted as desired.

class Solution(object):
    def sortArrayByParity(self, A):
        i, j = 0, len(A) - 1
        while i < j:
            if A[i] % 2 > A[j] % 2:
                A[i], A[j] = A[j], A[i] # 直接交换，不用中间变量

            if A[i] % 2 == 0: i += 1
            if A[j] % 2 == 1: j -= 1

        return A

Time Complexity: O(N), where N is the length of A. Each step of the while loop makes j-idecrease by at least one. (Note that while quicksort is O(NlogN) normally, this is O(N)because we only need one pass to sort the elements.)
Space Complexity: O(1) in additional space complexity.

Txing

Q905 Sort Array By Parity