We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Mathematics
  3. Combinatorics
  4. Sherlock and Pairs
  5. Discussions

Sherlock and Pairs

Sort 128 Discussions, By:

  • rbpimenta
     + 0 comments

    For those who are trying in Java 8, you will need to change the given method to work with long and not int.

    You also need to change that in the main, that is automatically generated by Hacker Rank.

  • devcana15
     + 0 comments

    Can anyone help what is wrong with the following code. Only three out of 17 cases are working. https://www.hackerrank.com/challenges/sherlock-and-pairs/submissions/code/288605771

  • murshid_islam
     + 0 comments

    Here's my Python code that passed all t he test cases: 

    import collections

def solve(a):
    # if every int appears only once in the list, then return 0
    if len(a) == len(set(a)):
        return 0
    else:
        # create a dict where the keys are elements of list 'a' 
        # and the values are frequencies of the elements of 'a'
        ctr_dict = collections.Counter(a)
        total_pairs = 0
        for x in ctr_dict.values():
            # if an integer appears 'n' times, 
            # there are n*(n-1) distinct pairs of indices (i,j)
            total_pairs += x*(x-1)
        return total_pairs

  • emmanuelpr015
     + 1 comment

    In Python 3, I recommend to use a dictionary to counts the number of repetitions for each integer. As many people have told, the number of possible different pairs if a number is repeated N times is N*(N-1), not over 2 as the combinatory coefficient nCr is, because one must count twice, i.e. (0,1) and (1,0).

    def solve(a):
    # Write your code here
    a_dic = {}
    pairs = 0
    #loop to count the number of elements
    for i in range(len(a)):
        if a[i] not in a_dic:
            a_dic[a[i]] = 0 #{a[i]:0}
        a_dic[a[i]] += 1 #{a[i]:+1} add one
    #loop for check whats integers taht are more than once
    for k in a_dic:
        val = a_dic[k]
        if val > 1: #condition more than once
            pairs += val * (val-1)  #possible pairs n*(n-1)
    return pairs

  • mdpopescu
     + 0 comments

    Ok this was stupid. Why use this line in the pre-existing code

    int result = Result.solve(a);

    when the result won't fit in an int in most cases?

Load more conversations