## Algorithm

Problem Name: 922. 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.

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]
```

Constraints:

• `2 <= nums.length <= 2 * 104`
• `nums.length` is even.
• Half of the integers in `nums` are even.
• `0 <= nums[i] <= 1000`

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int[] sortArrayByParityII(int[] A) {
int start = 1;
for (int i = 0; i  <  A.length; i += 2) {
if (A[i] % 2 == 1) {
while (A[start] % 2 != 0) {
start += 2;
}
int temp = A[i];
A[i] = A[start];
A[start] = temp;
}
}
return A;
}
}
``````
Copy The Code &

Input

cmd
nums = [4,2,5,7]

Output

cmd
[4,5,2,7]

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const sortArrayByParityII = function(A) {
const res = []
const odd = []
const even = []
for(let i = 0, len = A.length; i  <  len; i++) {
if(A[i] % 2 === 0) even.push(A[i])
else odd.push(A[i])
}
for(let i = 0, len = odd.length; i < len; i++) {
res.push(even[i], odd[i]>
}
return res
};
``````
Copy The Code &

Input

cmd
nums = [4,2,5,7]

Output

cmd
[4,5,2,7]

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def sortArrayByParityII(self, A):
even, odd = [a for a in A if not a % 2], [a for a in A if a % 2]
return [even.pop() if not i % 2 else odd.pop() for i in range(len(A))]
``````
Copy The Code &

Input

cmd
nums = [2,3]

Output

cmd
[2,3]

### #4 Code Example with C# Programming

```Code - C# Programming```

``````
namespace LeetCode
{
public class _0922_SortArrayByParityII
{
public int[] SortArrayByParityII(int[] A)
{
var oddIndex = 1;
for (int evenIndex = 0; evenIndex  <  A.Length; evenIndex += 2)
{
if (A[evenIndex] % 2 == 1)
{
while (A[oddIndex] % 2 == 1)
oddIndex += 2;

var temp = A[evenIndex];
A[evenIndex] = A[oddIndex];
A[oddIndex] = temp;
}
}

return A;
}
}
}
``````
Copy The Code &

Input

cmd
nums = [2,3]

Output

cmd
[2,3]