Sort Array By Parity II

1. Define the problem

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. Use two pointers, one scanning even indices and one scanning odd indices , swapping values whenever the even-index pointer finds an odd value and the odd-index pointer finds an even value.

Constraints

2 ≤ nums.length ≤ 2 × 10⁴
nums.length is even.
half of the integers in nums are even.
0 ≤ nums_i ≤ 1000

Example

Inputnums = [4, 2, 5, 7]

Output[4, 5, 2, 7]

Explanation Indices 0 and 2 (even indices) hold even values 4 and 2; indices 1 and 3 (odd indices) hold odd values 5 and 7 — swapping the misplaced pair at indices 1 and 2 produces this arrangement.

2. Visualize the solution

Two pointers scan even and odd indices separately

Statusinit

evenIdx=0 (value 4) is already even — advance evenIdx by two.

What happens in this step

evenIdx = 0, oddIdx = 1
  nums[evenIdx] = nums[0] = 4

4 % 2 === 0 → already correct for an even index.
evenIdx += 2 → evenIdx becomes 2. oddIdx doesn't move this step.

Step 1 of 3

Steps to visualize

Point evenIdx at index 0 and oddIdx at index 1.
If nums_evenIdx is already even, move evenIdx forward by two.
If nums_oddIdx is already odd, move oddIdx forward by two.
Otherwise nums_evenIdx is odd and nums_oddIdx is even, so swap them and advance both pointers by two.
Stop once evenIdx passes the end of the array.

3. Walk through the code

Walk through the code

Same walkthrough, now with the code. Press Next to move one step and watch which lines run.

Two pointers scan even and odd indices separately

Statusinit

evenIdx=0 (value 4) is already even — advance evenIdx by two.

What happens in this step

evenIdx = 0, oddIdx = 1
  nums[evenIdx] = nums[0] = 4

4 % 2 === 0 → already correct for an even index.
evenIdx += 2 → evenIdx becomes 2. oddIdx doesn't move this step.

Step 1 of 3

4. Solution

Solution

solution.tsTypeScript

function sortArrayByParityII(nums) {
  let even = 0;
  let odd = 1;
  const n = nums.length;

  while (even < n && odd < n) {
    if (nums[even] % 2 === 0) {
      even += 2;
    } else if (nums[odd] % 2 === 1) {
      odd += 2;
    } else {
      const temp = nums[even];
      nums[even] = nums[odd];
      nums[odd] = temp;
      even += 2;
      odd += 2;
    }
  }

  return nums;
}

Time: O(n)
Space: O(1) (in place, excluding the returned array)

5. Test cases

Test cases

Input	Expected	Covers
`nums = [4, 2, 5, 7]`	`[4, 5, 2, 7]`	example from the docstring
`nums = [1, 2]`	`[2, 1]`	smallest valid input: only two elements
`nums = [0, 1, 2, 3]`	`[0, 1, 2, 3]`	already correctly arranged input needs no swaps
`nums = [2, 4, 1, 3]`	`[2, 1, 4, 3]`	input that needs exactly one swap
`nums = [1, 3, 5, 2, 4, 6]`	`[2, 3, 6, 1, 4, 5]`	a larger array needing multiple swaps
`nums = [-1, -4, -3, -6]`	`[-4, -1, -6, -3]`	negative values mixed among even and odd

Sort Array By Parity II