I am trying to answer the following question: given a sorted array with some sequenced numbers and some non-sequenced numbers, write an algorithm that obtains a pair {start, end} for each group of consecutive numbers. Consecutive numbers have difference of 1 only.

So far, I can think of the brute force method only:

public static void main(String[] args) {
    int[] array = { 4, 5, 6, 7, 8, 9, 12, 15, 16, 17, 18, 20, 22, 23, 24, 27 };
    Map<Integer, Integer> list = new HashMap<Integer, Integer>();

    list = findIndex(array);
}

// Bruteforce
private static Map<Integer, Integer> findIndex(int[] array) {
    Map<Integer, Integer> list = new HashMap<Integer, Integer>();

    int x = -1, y = -1;

    int end = array.length;
    for (int i = 0; i < end; i++) {
        x = i;
        while (i < end - 1) {

            if (array[i] + 1 == array[i + 1]) {
                i++;
                y = i;
            } else {
                if (x != y && x >= 0) {
                    list.put(x, y);
                    System.out.println("i = " + x + " to j = " + y);
                    i = i + 1;
                    break;
                }
            }
        }

    }
    return list;
}

Output :

i = 0 to j = 5
i = 7 to j = 10
i = 12 to j = 14

It works fine, but how do I improve time complexity?

You don't need to nest loops for this:

int end = array.length;
if (end > 0) {
    int start = 0;
    for (int i = 1; i < end; i++) {
        if (array[i] != array[i - 1] + 1) {
            if (i - start > 1) {
                list.put(start, i - 1);
                System.out.println("i = " + start + " to j = " + (i - 1));
            }
            start = i;
        }
    }
    if (end - start > 1) {
        list.put(start, end - 1);
        System.out.println("i = " + start + " to j = " + (end - 1));
    }
}

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