Question

Below code works fine but it is in the complexity of O(n^2). Is it possible to do it O(n) or O(log n) time.

public class TwoRepeatingElements {
    public static void main(String[] args) {
        Integer array[] = {4, 2, 4, 5, 2, 3, 1, 2};
        findTwoRepeatingElements(array);


    }

    private static void findTwoRepeatingElements(Integer[] array) {
        int i, j;
        for(i = 0; i < array.length-1; i++) {
            for(j = i+1; j < array.length-1; j++) {
                if(array[i] == array[j]) {
                    System.out.println(array[i]);
                }
            }
        }

    }

}

Answer 1

This solution runs in O(n) time, notice that if you only need to find one repeated element, you can exit the loop as soon as you find it:

private static void findTwoRepeatingElements(Integer[] array) {
    Set<Integer> seen = new HashSet<Integer>();
    for (Integer i : array) {
        if (seen.contains(i))
            System.out.println(i);
        else
            seen.add(i);
    }
}

EDIT

If you need to print the repeated elements only once, this solution uses a little more memory since two sets are needed, but it’s still O(n). Take a look at this:

private static void findTwoRepeatingElements(Integer[] array) {

    Set<Integer> seen = new HashSet<Integer>();
    Set<Integer> repeated = new HashSet<Integer>();

    for (Integer i : array)
        if (!seen.add(i) && repeated.add(i))
            System.out.println(i);

}