Nsryjdtyk

I've spent the best part of a day on this question, it is marked as Expert which is way above my level but I gave it my best attempt. There are about fifteen submission test cases and my solution manages to satisfy the first four. However, from there on in it hits timeouts due to performance issues.

Hackerrank question here

There doesn't seem to be much information or solutions for this question around, would appreciate if anyone had any suggestion on how I can improve the performance. Maybe I need to go back to the drawing board with my approach.

• N = number of elements in the array


• K = The height difference allowed


• H = Array of heights


• queries = 2D array of ranges within which you must find the number of matching pairs (i.e. how many fighters in this range meet the height requirements to fight eachother)

My approach is to copy the subset (I know this is slow and would like to get around it somehow)and then run a quicksort over this new subset.

With this we can now quickly figure out what heights are within range of eachother (i.e. if we get height = 1 and we know k = 2, then the maximum height he can fight against is going to be 3).

I hope this is clear and would appreciate any suggestions or advice, Thank you.

    static int solve(int n, int k, int h, int queries) {

    // N = no. heights

    // K = Difference

    // L & R = First and Last fighters

    int results = new int[queries.length];

    int count = 1;

    int pairs = 0;

    int l, r =0;



    for(int j=0; j<queries.length; j++) {

        l = queries[j][0];

        r = queries[j][1];

        int range = Arrays.copyOfRange(h, l, r+1);



        int sortedSubset = sort(range);



        for(int i=0; i<sortedSubset.length-1; i++) {

            count = i+1;

            while((count != sortedSubset.length) && sortedSubset[count] <= (sortedSubset[i]+k)) { 



                // While the current number is still in range of the number we are checking i.e. (number + k)

                pairs++;

                count++;

           }

        }

        results[j] = pairs;

        pairs = 0;

    }



    return results;

}

I suggest you do the following:

1. Sort the entire heights array once.


2. Finds all the matching pairs for the full array (i.e. l=0; r=h.length-1;).


3. For each matching pair you find, loop over the queries array, and add it to the counts of all the subsets that contain both fighters. For example, if the pair h[2],h[5] is matching, increment results[j] for all j such that queries[j][0] <=2 and 5 <= queries[j][1].

I believe running a single sort instead of queries.length sorts will improve the performance.

The only downside of this algorithm is that if all the queries cover a small subset of the full heights array, the algorithm will do a lot of unnecessary work (finding matching pairs for indices that we'll never need to count).