# Sorting a Java Array by Parity ## The challenge

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`

## The solution

```class Solution {
// return a sorted primitive int array
public int[] sortArrayByParity(int[] A) {

// keep track of place
int lastIndex = 0;

// loop through input array
for (int i=0; i<A.length; i++) {
// if we found an even number
if (A[i]%2==0) {
// keep track of the values
int lastValue = A[lastIndex];
int thisValue = A[i];

// swap the values
A[i] = lastValue;
A[lastIndex] = thisValue;

// increment our swap index
lastIndex++;
}
}

// return a sorted array
return A;
}
}
```
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