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. Sales by Match
  2. Discussions

Sales by Match

  • nadiyakhurshid98
     + 0 comments

    Solution in python with O(n):

    def sockMerchant(n, ar):
    # Write your code here
    count = {}
    no_of_pair = 0
    for i in ar:
        if i not in count:
            count[i] = 1
            continue
        count[i] += 1
    for key in count:
        if count[key] >= 2:
            no_of_pair += int(count[key]/2)

    print(count)
print(no_of_pair)
return int(no_of_pair)