Find The Parity Outlier

You are given an array (which will have a length of at least 3, but could be very large) containing integers. The array is either entirely comprised of odd integers or entirely comprised of even integers except for a single integer N. Write a method that takes the array as an argument and returns this "outlier" N.
Examples

[2, 4, 0, 100, 4, 11, 2602, 36]
Should return: 11 (the only odd number)

[160, 3, 1719, 19, 11, 13, -21]
Should return: 160 (the only even number)

Good Solution1:

import java.util.Arrays;

public class FindOutlier{
    public static int find(int[] integers) {
        // Since we are warned the array may be very large, we should avoid counting values any more than we need to.

        // We only need the first 3 integers to determine whether we are chasing odds or evens.
        // So, take the first 3 integers and compute the value of Math.abs(i) % 2 on each of them.
        // It will be 0 for even numbers and 1 for odd numbers.
        // Now, add them. If sum is 0 or 1, then we are chasing odds. If sum is 2 or 3, then we are chasing evens.
        int sum = Arrays.stream(integers).limit(3).map(i -> Math.abs(i) % 2).sum();
        int mod = (sum == 0 || sum == 1) ? 1 : 0;

        return Arrays.stream(integers).parallel() // call parallel to get as much bang for the buck on a "large" array
                .filter(n -> Math.abs(n) % 2 == mod).findFirst().getAsInt();
    }
}

Good Solution2:

public class FindOutlier {

    static int find(int[] integers) {
        int oddcount = 0, odd = 0, evencount = 0, even = 0;
        for (int i : integers) {
            if (i % 2 == 0) {
                even = i;
                evencount++;
            } else {
                odd = i;
                oddcount++;
            }
            if (evencount > 0 && oddcount > 0) {
                if (evencount > 1) {
                    return odd;
                } else if (oddcount > 1) {
                    return even;
                }
            }
        }
        return evencount > 1 ? odd : even;
    }
}