922. Sort Array By Parity II

🟩 Easy

Question

Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.

Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.

You may return any answer array that satisfies this condition.

Example 1:

Input: [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.

Note:

1. 2 <= A.length <= 20000

2. A.length % 2 == 0

3. 0 <= A[i] <= 1000

Two Path

• Time complexity: O(n)

• Space complexity: O(n)

def sortArrayByParityII(self, A):
    ans = [None] * len(A)
    ans[::2] = (x for x in A if x % 2 == 0)
    ans[1::2] = (x for x in A if x % 2 == 1)
    return ans

Read / Write Heads

• Time complexity: O(n)

• Space complexity: O(1)

def sortArrayByParityII(self, A):
    j = 1
    for i in range(0, len(A), 2):
        if A[i] % 2: # for each even index, if is odd
            while A[j] % 2: # draft an element from odd partition
                j += 2
            A[i], A[j] = A[j], A[i]
    return A