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# 14. 문자열 다루기 기본
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def solution(s):
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answer = True
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if len(s) == 4 or len(s) == 6:
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for i in s:
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if i.isdigit():
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answer = True
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else:
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answer = False
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break
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else:
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answer = False
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return answer
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s_1 = 'a234'
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s_2 = '1234'
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# print(solution(s_1))
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# print(solution(s_2))
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def solution_best(s):
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answer = s.isdigit() and len(s) in (4, 6)
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return answer
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# print(solution_best(s_1))
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# print(solution_best(s_2))
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import re
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def solution_re(s):
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answer = bool(re.match('^(\d{4}|\d{6}$)', s))
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return answer
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# print(solution_re(s_1))
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# print(solution_re(s_2))

Day8/SortOnMyOwnString.py

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# 16. 문자열 내 마음대로 정렬하기
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# 내가 풀지 못했음... 왜 어렵게 생각했을까 ㅠㅠㅠ
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def solution(strings, n):
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answer = []
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for i in range(len(strings)):
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strings[i] = strings[i][n] + strings[i]
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# print(strings)
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strings.sort()
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print(strings)
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for i in range(len(strings)):
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answer.append(strings[i][1:])
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return answer
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strings_1 = ['sun', 'bed', 'car']
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strings_2 = ['abce', 'abcd', 'cdx']
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n_1 = 1
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n_2 = 2
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# print(solution(strings_1, n_1))
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# print(solution(strings_2, n_2))
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def solution_best(strings, n):
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answer = sorted(sorted(strings), key = lambda x : x[n])
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return answer
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print(solution_best(strings_1, n_1))
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print(solution_best(strings_2, n_2))

Day8/StringInP_And_Y_Count.py

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# 8. 문자열 내 p와 y의 개수
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def solution(s):
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answer = True
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# p와 y 세는 변수 선언
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p_cnt = 0
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y_cnt = 0
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# 반복문 돌면서 개수 세기
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for i in s:
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if i == 'p' or i == 'p'.upper():
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p_cnt += 1
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elif i == 'y' or i == 'y'.upper():
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y_cnt += 1
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else:
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continue
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if p_cnt == y_cnt:
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answer = True
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else:
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answer = False
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return answer
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s_1 = 'pPoooyY'
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s_2 = 'Pyy'
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print(solution(s_1))
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print(solution(s_2))
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def solution_best(s):
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return s.lower().count('p') == s.lower().count('y')
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print(solution_best(s_1))
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print(solution_best(s_2))

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