LeetCode 905 - Sort Array By Parity

https://leetcode.com/articles/sort-array-by-parity/

Given an array A of non-negative integers, return an array consisting of all the even elements of A, followed by all the odd elements of A.

You may return any answer array that satisfies this condition.

Example 1:

Input: [3,1,2,4]
Output: [2,4,3,1]
The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Note:

1. 1 <= A.length <= 5000
2. 0 <= A[i] <= 5000

If we want to do the sort in-place, we can use quicksort, a standard textbook algorithm.

Algorithm

We'll maintain two pointers i and j. The loop invariant is everything below i has parity 0 (ie. A[k] % 2 == 0 when k < i), and everything above j has parity 1.

Then, there are 4 cases for (A[i] % 2, A[j] % 2):

• If it is (0, 1), then everything is correct: i++ and j--.
• If it is (1, 0), we swap them so they are correct, then continue.
• If it is (0, 0), only the i place is correct, so we i++ and continue.
• If it is (1, 1), only the j place is correct, so we j-- and continue.

Throughout all 4 cases, the loop invariant is maintained, and j-i is getting smaller. So eventually we will be done with the array sorted as desired.

• Time Complexity: $O(N)$, where $N$ is the length of A. Each step of the while loop makes j-idecrease by at least one. (Note that while quicksort is $O(N \log N)$ normally, this is $O(N)$ because we only need one pass to sort the elements.)
• Space Complexity: $O(1)$ in additional space complexity. 

  public int[] sortArrayByParity(int[] A) {

    int i = 0, j = A.length - 1;

    while (i < j) {

      if (A[i] % 2 > A[j] % 2) {

        int tmp = A[i];

        A[i] = A[j];

        A[j] = tmp;

      }

      if (A[i] % 2 == 0)

        i++;

      if (A[j] % 2 == 1)

        j--;

    }

    return A;

  }

  public int[] sortArrayByParity(int[] A) {

    int[] ans = new int[A.length];

    int t = 0;

    for (int i = 0; i < A.length; ++i)

      if (A[i] % 2 == 0)

        ans[t++] = A[i];

    for (int i = 0; i < A.length; ++i)

      if (A[i] % 2 == 1)

        ans[t++] = A[i];

    return ans;

  }

  public int[] sortArrayByParity(int[] A) {

    Integer[] B = new Integer[A.length];

    for (int t = 0; t < A.length; ++t)

      B[t] = A[t];

    Arrays.sort(B, (a, b) -> Integer.compare(a % 2, b % 2));

    for (int t = 0; t < A.length; ++t)

      A[t] = B[t];

    return A;

    /*

     * Alternative:

     * 

     * return Arrays.stream(A)

     * 

     * .boxed()

     * 

     * .sorted((a, b) -> Integer.compare(a%2, b%2))

     * 

     * .mapToInt(i -> i)

     * 

     * .toArray();

     * 

     */

  }