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Company Logo

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  • jaithalal680
     + 0 comments

    def count(s): s = sorted(s) count_letters = {} for char in s: count_letters[char] = count_letters.get(char, 0)+1 arr = sorted(count_letters.items(), key=lambda x: x[1] , reverse=True)

    for i in range(3):
    print(arr[i][0], arr[i][1])

    if name == 'main': count(input())

  • daturpinusn
     + 0 comments

    Did all the work in one line: s = Counter(sorted(list(input()))).most_common(3)

    Still needed two lines to print the output: for i in s: print(i[0], i[1])

    But basically the line takes the input(), puts the letters in a list, sorts them, then makes them a counter and takes the top 3 most common. Because the Counter takes the first from the list if the counts match, it is in alphabetical order as it counts.

  • yashdeora98294
     + 0 comments

    Here is HackerRank Company Logo in python solution - https://programmingoneonone.com/hackerrank-company-logo-solution-in-python.html

  • TheresaKenny09
     + 0 comments

    It really sharpens your skills in identifying character frequency and sorting with custom rules. Highly recommend it to anyone brushing up on Python for interviews or real-world tasks. Cricbet99 Login Registration Online

  • arohi_agrawal111
     + 0 comments
    from collections import Counter


if __name__ == '__main__':
  s = input()
  count = Counter(s)
  sort = sorted(count.items(), key = lambda x: (-x[1], x[0]))
  
  for a,b in sort[:3]:
    print(a,b)