New Year Chaos Solution - HackerRank

Problem

It is New Year’s Day and people are in line for the Wonderland rollercoaster ride. Each person wears a sticker indicating their initial position in the queue from 1 to n. Any person can bribe the person directly in front of them to swap positions, but they still wear their original sticker. One person can bribe at most two others.

Determine the minimum number of bribes that took place to get to a given queue order. Print the number of bribes, or, if anyone has bribed more than two people, print Too chaotic.

See full description and example input & output in the HackerRank page.

Solution

A person, wearing the number q[i], was initally at position q[i] - 1 (e.g. a person with number 3 was at the index 2 at first). The person can bribe others at most twice, so the person can move left at most 2 positions. When inspect the list, we compare each person’s original position, i.e. q[i] - 1, with the current position i, if (q[i] - 1) - i is greater than 2, then this person bribes too many people. Our program prints Too chaotic in this case.

Next, we calculate how many bribes happen. Say, the initial list is:

  1 2 3 4 5

And, its current state is:

  1 2 5 3 4

For every person X, we check people in the left of this person, if any of them’s number is greater than X’s, then that person must bribe X.

Below is a Java version solution.

public static void minimumBribes(List<Integer> q) {
    int bribes = 0;
    for (int i = 0; i < q.size(); i++) {
        if (q.get(i) - 1 - i > 2) { // this person has done too many bribes
            System.out.println("Too chaotic");
            return;
        } else {
            for (int j = 0; j < i; j++) {
                // for every person in the left of "person q[i]", 
                // if that person's number is greater, 
                // than "person q[i]" got bribed by that person.
                if (q.get(j) > q.get(i)) {
                    bribes++;
                }
            }
        }
    }
    System.out.println(bribes);
}

Time complexity is O(n^2).

An Optimization

When calculate bribes, the above solution has done many unnecessary checks. A person X, wearing the number q[i], was initally at position q[i] - 1. People, who were in the right of X, can do bribe to be positioned at q[i] - 1 or q[i] - 2. The bribe limit is 2, it’s impossible for people who were in the right of X to be at index less than q[i] - 2. So we can improve the previous solution to let j start with q[i] - 2. The value q[i] - 2 may be negative, to avoid the out-of-bound exception, let j start with 0 when q[i] - 2 is negative.

public static void minimumBribes(List<Integer> q) {
    int bribes = 0;
    for (int i = 0; i < q.size(); i++) {
        if (q.get(i) - 1 - i > 2) {
            System.out.println("Too chaotic");
            return;
        } else {
            int j = Math.max(0, q.get(i) - 2);
            while (j < i) {
                if (q.get(j) > q.get(i)) {
                    bribes++;
                }
                j++;
            }
        }
    }
    System.out.println(bribes);
}

Note that, you can add another check, q.get(i) - 1 == i. In this case, the person hasn’t change the position, so no bribe happens on this person. We can skip bribe calculation in this case.