class Solution:
    def sortArrayByParityII(self, nums: List[int]) -> List[int]:
        arr = [None]*len(nums)
        even,odd = 0,1
        for i in(nums):
            if i % 2 == 0:
                arr[even] = i
                even +=2
        for i in (nums):
            if i % 2 != 0:
                arr[odd] = i
                odd+=2
        return arr
class Solution {
    public int[] sortArrayByParityII(int[] nums) {
        
        int[] ans = new int[nums.length];
        
        int even_pointer = 0;
        int odd_pointer = 1;
        
        for(int i = 0; i < nums.length; i++){
            
            if(nums[i] % 2 == 0){
                ans[even_pointer] = nums[i];
                even_pointer += 2;
            } else{
                ans[odd_pointer] = nums[i];
                odd_pointer += 2;
            }
            
        }
        
        return ans;
    }
}
Sort Array By Parity II
Given an array of integers nums, half of the integers in nums are odd, and the other half are even.

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

Return any answer array that satisfies this condition.

&nbsp;
Example 1:

Input: nums = [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.


Example 2:

Input: nums = [2,3]
Output: [2,3]


&nbsp;
Constraints:


	2 &lt;= nums.length &lt;= 2 * 104
	nums.length is even.
	Half of the integers in nums are even.
	0 &lt;= nums[i] &lt;= 1000


&nbsp;
Follow Up: Could you solve it in-place?

var sortArrayByParityII = function(nums) {
    let arrEven = []
    let arrOdd = []
    let result = []
    for(let i in nums){
    nums[i]%2==0 ? arrEven.push(nums[i]) : arrOdd.push(nums[i])
    }
    for(let i in  arrEven){
        result.push(arrEven[i])
        result.push(arrOdd[i])
    }
   return result
};
class Solution {
public:
    vector<int> sortArrayByParityII(vector<int>& nums) {

        vector<int>ans(nums.size());

        int even_idx=0;
        int odd_idx=1;

        for(int i=0;i<nums.size();i++)
        {
            if((nums[i]%2)==0) //the num is even
            {
                ans[even_idx]=nums[i];
                even_idx+=2;
            }
            else //the num is odd
            {
                ans[odd_idx]=nums[i];
                odd_idx+=2;
            }
        }
        return ans;
    }
};