Prepare > Python > Collections > Company Logo

Company Logo | HackerRank

 

문제

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A newly opened multinational brand has decided to base their company logo on the three most common characters in the company name. They are now trying out various combinations of company names and logos based on this condition. Given a string , which is the company name in lowercase letters, your task is to find the top three most common characters in the string.

• Print the three most common characters along with their occurrence count.
• Sort in descending order of occurrence count.
• If the occurrence count is the same, sort the characters in alphabetical order.

Input Format

A single line of input containing the string S

Output Format

Print the three most common characters along with their occurrence count each on a separate line.
Sort output in descending order of occurrence count.
If the occurrence count is the same, sort the characters in alphabetical order.

 

=> S의 알파벳중 가장 반복이 많은 3개를 그 횟수와 함께 출력하라. 순서는 반복많은순, 알파벳 순이다.

 

 

 

 

코드

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import math
import os
import random
import re
import sys


if __name__ == '__main__':
    s = input()
    
---------------------------------------------------------------------

from collections import Counter


s_count = Counter(s)
S = sorted(s_count.items(), key=lambda x: (-x[1], x[0]))

for i in range(3):
    print(*S[i])

 

 

 

 

노트

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sorted()

sorted(iterable, key= , reverse= )

• 원본유지가 특징임
• iterabel: 정렬할 대상
• key: 정렬기준(함수식 필요)
• reverse: 역순 여부. Default는 오름차순(False)

 

 

딕셔너리에서 활용

d = {'e': 1, 'b': 2, 'd': 3, 'a': 2, 'c': 4}

 

 

key로 정렬, key리스트 리턴

sorted(d)

# ['a', 'b', 'c', 'd', 'e']

 

 

key 역순 정렬, key리스트 리턴

sorted(d, reverse=True)

# ['e', 'd', 'c', 'b', 'a']

• reverse 파라미터 이용

 

 

key로 정렬, key-value쌍 리턴

sorted(d.items())

# [('a', 2), ('b', 2), ('c', 4), ('d', 3), ('e', 1)]

• d.items() 이용

 

 

value로 정렬, value리스트 리턴

sorted(d.values())

# [1, 2, 2, 3, 4]

• reverse 파라미터 사용 가능

 

 

value로 정렬, key-value쌍 리턴

sorted(d.items(), key = lambda x: x[1])

# [('e', 1), ('b', 2), ('a', 2), ('d', 3), ('c', 4)]

 

 

value내림차순 정렬, key-value쌍 리턴

sorted(d.items(), key = lambda x: -x[1])

# [('c', 4), ('d', 3), ('b', 2), ('a', 2), ('e', 1)]

 

 

 

value오름차순, key오름차순, key-value쌍 리턴

sorted(d.items(), key = lambda x: (x[1], x[0]))

# [('e', 1), ('a', 2), ('b', 2), ('d', 3), ('c', 4)]

 

 

value내림차순, key오름차순, key-value쌍 리턴

sorted(d.items(), key = lambda x: (-x[1], x[0]))

# [('c', 4), ('d', 3), ('a', 2), ('b', 2), ('e', 1)]

 

???

sorted(d.items(), key = lambda x: x[1], reverse=True)

# [('c', 4), ('d', 3), ('b', 2), ('a', 2), ('e', 1)]

 

 

 

 

참조

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[Python] 딕셔너리(Dictionary) 정렬 방법

[파이썬] sorted() 정렬 함수 파헤치기