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  1. Practice
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  3. Greedy
  4. Luck Balance
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Luck Balance

  • RodneyShag + 1 comment

    Java O(n) solution

    Using QuickSelect, you can get an O(n) average-case solution.

    The full solution is in my HackerRank solutions. Here is a snippet of the code.

    public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    int N = scan.nextInt();
    int K = scan.nextInt();
    ArrayList<Integer> contest = new ArrayList<>(N);
    int savedLuck = 0;
    for (int i = 0; i < N; i++) {
        int luck = scan.nextInt();
        int importance = scan.nextInt();
        if (importance == 0) {
            savedLuck += luck; // lose every non-important contest
        } else {
            contest.add(luck);
        }
    }
    scan.close();
        
    /* Compete in "important" contests */
    quickselect(contest, 0, contest.size() - 1, contest.size() - K);
    for (int i = 0; i < contest.size(); i++) {
        if (i < contest.size() - K) {
            savedLuck -= contest.get(i); // win contest
        } else {
            savedLuck += contest.get(i); // lose contest
        }
    }
    System.out.println(savedLuck);
}

    Let me know if you have any questions.

    • tat_lim + 2 comments

      Agreed! The average-case solution for QuickSelect is indeed O(n). Unfortunately, when the list is sorted (or "nearly" sorted), your algorithm will degenerate to the worst-case of O(n*n)... :(

      • RodneyShag + 1 comment

        Hi. I agree, O(n^2) can happen, but this worst-case O(n^2) happens extremely rarely. It's O(n) 99.99%+ of the time for randomly generated data. To get the degenerate case, the user would have to provide data in a specific way to break quickselect based on how the pivot is selected. Such cases are mostly a concern if a hacker is intentionally trying to break quickselect.

        Quickselect doesn't become O(n^2) just because the data is sorted. 3 things have to align for the runtime to reach O(n^2), for sorted data, here are the 3 things (from Wikipedia)
        1) Data is sorted
        2) User is searching for the maximum element in a set (which is usually not the case, since a sorting algorithm is not a good way to find the max element)
        3) The first element is used as a pivot

        If you look into how I implemented QuickSelect you will see that I do not choose the starting element as a pivot, so my code will still run in O(n) on a sorted list.

        As a personal preference, I'm happy getting O(n) 99% of the time with the rare case of O(n^2).

        Cheers!

        • tat_lim + 0 comments

          Agreed! For a high-volume website, however, we cannot bet on the worst-case of O(n*n) even if it has a mere chance of 0.01%, especially when millions of searches are conducted per day. FYI, 1 million multiplied by 0.01% = 100.

      • rhinodavid + 1 comment

        A good implementation of the algorithm will have randomization built in, such that the order of the list does not matter.

        • tat_lim + 0 comments

          Agreed! On the other hand, even with randomization builtin, QuickSelect still has the worst-case of O(n*n), however improbable it may be.