You are given an integer array, nums, where exactly half of the elements are even, and the other half are odd.

Rearrange nums such that:

• All even numbers are placed at even indexes $[0, 2, 4, ...]$.

• All odd numbers are placed at odd indexes $[1, 3, 5, ...]$.

You may return any valid arrangement that satisfies these conditions.

Constraints:

• $2 \leq$ nums.length $\leq 10^3$

• nums.length is even.

• Half of the integers in nums are even.

• $0\leq$ nums[i] $\leq1000$

The essence of this solution is to use a modified cyclic sort approach combined with the flavor of two pointers to rearrange the array such that even numbers are placed at even indexes and odd numbers at odd indexes. The approach uses two pointers: an even pointer for even indexes and an odd pointer for odd indexes. The even pointer starts at index 000 and moves across even indexes, while the odd pointer starts at index 111 and moves across odd indexes. The algorithm iterates through the array and checks whether the numbers at these indices are correctly placed. If the number at the even pointer is even, the pointer moves forward to the next even index. Similarly, if the number at the odd pointer is odd, it moves forward to the next odd index. However, if the number at the even pointer is odd and the number at the odd pointer is even, they are misplaced and need to be swapped. After swapping, both pointers continue to their respective next positions. This process repeats until all elements are correctly placed. The algorithm restores order with minimal operations by ensuring that misplaced numbers are swapped into their correct positions. Although inspired by the principles of cyclic sort, where elements are placed in their correct positions, this approach uses a modified cyclic sort approach that organizes elements based on parity rather than numerical value. It is an in-place approach, making it both simple and optimal.

Now, let’s look at the steps of the solution:

1. We initialize two pointers, i = 0 and j = 1. i moves through even indexes $[0, 2, 4, ...]$ and j moves through odd indexes $[1, 3, 5, ...]$.

2. We iterate through the array until both i and j are within the bounds of nums.

   1. If nums[i] is even (correct placement), we move i forward by $2$ to check the next even index because it is already at its correct index.

   2. If nums[j] is odd (correct placement), we move j forward by $2$ to check the next odd index because it is already at its correct index.

   3. Otherwise, if nums[i] is odd and nums[j] is even, we swap them to fix their positions. This follows a modified cyclic sort approach where elements are swapped directly to their rightful places without repeated scanning. After swapping, both pointers (i and j) are moved forward by $2$.

3. After iterating through the array, we return the modified nums, where all even numbers appear at even indexes, and all odd numbers at odd indexes.

Let’s look at the following illustration to get a better understanding of the solution: