package com.fishercoder.solutions;

/**
 * 922. Sort Array By Parity II
 *
 * Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.
 * Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.
 * You may return any answer array that satisfies this condition.
 *
 * Example 1:
 *
 * Input: [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.
 *
 *
 * Note:
 * 2 <= A.length <= 20000
 * A.length % 2 == 0
 * 0 <= A[i] <= 1000
 * */
public class _922 {
    public static class Solution1 {
        public int[] sortArrayByParityII(int[] A) {
            for (int i = 0, j = 1; i < A.length - 1 && j < A.length;) {
                if (A[i] % 2 != 0 && A[j] % 2 == 0) {
                    int tmp = A[i];
                    A[i] = A[j];
                    A[j] = tmp;
                    i += 2;
                    j += 2;
                }
                while (i < A.length - 1 && A[i] % 2 == 0) {
                    i += 2;
                }
                while (j < A.length && A[j] % 2 != 0) {
                    j += 2;
                }
            }
            return A;
        }
    }
}

package com.fishercoder.solutions;

public class _922 {
    public static class Solution1 {
        public int[] sortArrayByParityII(int[] A) {
            for (int i = 0, j = 1; i < A.length - 1 && j < A.length; ) {
                if (A[i] % 2 != 0 && A[j] % 2 == 0) {
                    int tmp = A[i];
                    A[i] = A[j];
                    A[j] = tmp;
                    i += 2;
                    j += 2;
                }
                while (i < A.length - 1 && A[i] % 2 == 0) {
                    i += 2;
                }
                while (j < A.length && A[j] % 2 != 0) {
                    j += 2;
                }
            }
            return A;
        }
    }
}

refactor 922