Binary Search O(log n) algorithm to find duplicate in sequential list?

Question

Does anyone know a faster-than-linear algorithm for finding a duplicate in a sequential list of numbers? I'm working in Java now but any language or psuedo-code is fine.

For example, given this int[] input:

0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 7 | 8 | 9

Output would be either index or value '7'.

I know the obvious traversal at O(n) linear time, but I'm trying to see if this is possible via binary search at O(log n) time.

Answer 1

If you assume the numbers must start at 0 and be increasing by 1 you can compare the middle to the index. If the middle is the same go higher, if the middle is not, go lower.

This will give you binary search time O(log2 N). The only difference is that you are comparing with the index, rather than a fixed value.

---

public static void main(String... args) {
    int[] array = {0, 1, 2, 3, 4, 5, 6, 7, 7, 8, 9};
    int duplicate = findDuplicate(array);
    System.out.println(duplicate);
}

private static int findDuplicate(int[] array) {
    int low = 0;
    int high = array.length - 1;

    while (low <= high) {
        int mid = (low + high) >>> 1;
        int midVal = array[mid];

        if (midVal == mid)
            low = mid + 1;
        else
            high = mid - 1;
    }
    return high;
}