Discussion on Sherlock and Pairs Challenge

2 years ago+ 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