[프로그래머스] Python3 알고리즘

인형뽑기

https://programmers.co.kr/learn/courses/30/lessons/64061

import numpy as np

def solution(board, moves):
    answer = 0
    a = np.array(board).T.tolist()
    
    # 뽑은 값을 저장해 놓을 리스트
    lst = list()
    
    for i in moves:
        n=0
        try:
            while n ==0:
                n = a[i-1].pop(0)
            lst.append(n)
            if (len(lst)>1) & (n == lst[-2]):
                answer += 1
                del lst[-1]
                del lst[-1]
        except:
            pass
        
    # 사라진 인형들 개수
    return answer*2

모의고사

https://programmers.co.kr/learn/courses/30/lessons/42840

def solution(answers):
    p = [[1, 2, 3, 4, 5],
         [2, 1, 2, 3, 2, 4, 2, 5],
         [3, 3, 1, 1, 2, 2, 4, 4, 5, 5]]
    s = [0] * len(p)

    for q, a in enumerate(answers):
        for i, v in enumerate(p):
            if a == v[q % len(v)]:
                s[i] += 1
    return [i + 1 for i, v in enumerate(s) if v == max(s)] 

두 정수 사이의 합

https://programmers.co.kr/learn/courses/30/lessons/12912

def solution(a, b):
    return sum(range(min(a,b), max(a,b) + 1))

2016년

https://programmers.co.kr/learn/courses/30/lessons/12901

from datetime import datetime

def solution(a, b):
    week = ['MON','TUE','WED','THU','FRI','SAT','SUN']
    day = datetime(year = 2016, month = a, day =b)
    return week[day.weekday()]

같은 숫자는 싫어

https://programmers.co.kr/learn/courses/30/lessons/12906

def solution(arr):
    answer = [arr[0]]
    
    for i in range(1,len(arr)):
        if arr[i] != arr[i-1]:
            answer.append(arr[i])
    
    return answer

def no_continuous(s):
    a = []
    for i in s:
        if a[-1:] == [i]: continue
        a.append(i)
    return a

문자열 다루기 기본

https://programmers.co.kr/learn/courses/30/lessons/12918

def solution(s):
    if len(s) not in (4,6):
        return False
    
    for ch in s:
        if ord(ch) not in range(48, 58):
            return False

    return True

문자열 내 맘대로 정렬하기

https://programmers.co.kr/learn/courses/30/lessons/12915

def solution(strings, n):
    return sorted(strings, key = lambda x : (x[n], x))

나누어 떨어지는 숫자 배열

https://programmers.co.kr/learn/courses/30/lessons/12910

def solution(arr, divisor):
    return sorted([i for i in arr if i % divisor == 0]) or [-1]

문자열 내 p와 y의 개수

https://programmers.co.kr/learn/courses/30/lessons/12916

def solution(s):
    if s.lower().count('p') == s.lower().count('y'):
        return True
    else:
        return False

문자열 내림차순으로 배치하기

https://programmers.co.kr/learn/courses/30/lessons/12917

def solution(s):
    return ''.join(sorted(s, reverse=True))

문자열 다루기 기본

https://programmers.co.kr/learn/courses/30/lessons/12918

[정규식]

• \d: 숫자
• \D: 숫자가 아닌 것
• \s: whitespace (\t,\n,\r,\f,\v)
• \S: whitespace가 아닌 것
• \w: 문자 + 숫자와 매치
• \W: 문자 + 숫자가 아닌 것

def solution(s):
    if len(s) in (4, 6):
        try: 
            int(s)
            return True
        except:
            return False
    else:
        return False

서울에서 김서방 찾기

https://programmers.co.kr/learn/courses/30/lessons/12919

def solution(seoul):
    for i,j in enumerate(seoul):
        if j == 'Kim':
            return "김서방은 {}에 있다".format(i)

def findKim(seoul):
    return "김서방은 {}에 있다".format(seoul.index('Kim'))

소수찾기

https://programmers.co.kr/learn/courses/30/lessons/12921

• 에라토스테네스의 체

def solution(n):
    a = [False, False] + [True]*(n-1)
    answer = []
    for i in range(2,n+1):
        if a[i]:
            answer.append(i)
            for j in range(2*i, n+1, i):
                a[j] = False
    return len(answer)

def solution(n):
    num=set(range(2,n+1))

    for i in range(2,n+1):
        if i in num:
            num-=set(range(2*i,n+1,i))
    return len(num)

수박수박수

https://programmers.co.kr/learn/courses/30/lessons/12922

def solution(n):
    b = '수박'*n
    answer = ''
    for i in range(n):
        answer = answer + b[i]
    return answer

def water_melon(n):
    s = "수박" * n
    return s[:n]

시저암호

https://programmers.co.kr/learn/courses/30/lessons/12926

def solution(s,n):
    abc = 'abcdefghijklmnopqrstuvwxyz'
    ABC = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
    answer = ''
    for i in s:
        if i in abc:
            index = abc.find(i)
            index += n
            answer = answer + abc[index%26]
        elif i in ABC:
            index = ABC.find(i)
            index += n
            answer = answer + ABC[index%26]
        else:
            answer = answer + ' '
    return answer

def caesar(s, n):
    s = list(s)
    for i in range(len(s)):
        if s[i].isupper():
            s[i]=chr((ord(s[i])-ord('A')+ n)%26+ord('A'))
        elif s[i].islower():
            s[i]=chr((ord(s[i])-ord('a')+ n)%26+ord('a'))

    return "".join(s)

약수의 합

https://programmers.co.kr/learn/courses/30/lessons/12928

def solution(n):
    answer = 0
    for i in range(1,n+1):
        if n % i == 0:
            answer += i
    return answer

def sumDivisor(num):
    # num / 2 의 수들만 검사하면 성능 약 2배 향상된다.
    return num + sum([i for i in range(1, (num // 2) + 1) if num % i == 0])

이상한 문자 만들기

https://programmers.co.kr/learn/courses/30/lessons/12930

def solution(a):
    lst = a.split(' ')
    answer = []
    for i in lst:
        v = list(i)
        for z,x in enumerate(v):
            if z % 2 == 0:
                v[z] = x.upper()
            else:
                v[z] = x.lower()
        answer.append(''.join(v))
    return ' '.join(answer)

def toWeird(s):
    return " ".join(map(lambda x: "".join([a.lower() if i % 2 else a.upper() for i, a in enumerate(x)]), s.split(" ")))

자릿수 더하기

https://programmers.co.kr/learn/courses/30/lessons/12931

import timeit

def solution(n):
    start = timeit.default_timer()
    answer = 0
    for i in str(n):
        answer += int(i)
    stop = timeit.default_timer()
    return answer,  stop-start
    
solution(12)
'''
(3, 3.200000264769187e-06)
'''

def solution(n):
    start = timeit.default_timer()
    answer = 0
    
    while n > 0:
        answer += n % 10
        n  = n // 10
    stop = timeit.default_timer()
    return answer, stop-start
    
solution(12)
'''
(3, 1.600000359758269e-06)
'''

# 재귀구조
def sum_digit(number):
    start = timeit.default_timer()
    if number < 10:
        return number;
    a = (number % 10) + sum_digit(number // 10) 
    stop = timeit.default_timer()
    return a , stop-start
    
sum_digit(12)
'''
(3, 2.100000074278796e-06)
'''

자연수 뒤집어 배열로 만들기

https://programmers.co.kr/learn/courses/30/lessons/12932

def solution(n):
    start = timeit.default_timer()
    a = list(map(lambda x: int(x), str(n)))[::-1]
    stop = timeit.default_timer()
    return stop-start
    
solution(123)
'''
6.800000846851617e-06
'''

# reversed를 쓰는게 lambda보다 더 빠른 걸로.
def digit_reverse(n):
    start = timeit.default_timer()
    a = list(map(int, reversed(str(n))))
    stop = timeit.default_timer()
    return stop-start
    
digit_reverse(123)
'''
5.399999281507917e-06
'''

체육복

https://programmers.co.kr/learn/courses/30/lessons/42862

def solution(n,lost,reserve):
    
    set_reserve = set(reserve) - set(lost)
    set_lost = set(lost) - set(reserve)
    
    for i in set_reserve:
        if i-1 in set_lost:
            set_lost.remove(i-1)
        elif i+1 in set_lost:
            set_lost.remove(i+1)
    return n-len(set_lost)

def solution(n, lost, reserve):
    
    # 없는 학생 제외
    students = [1 if i+1 not in lost else 0 for i in range(n)]
    # 여벌있는 학생 포함
    students = [num+1 if index+1 in reserve else num for index, num in enumerate(students)]    
    
    for i in range(0, len(students) - 1):
        if students[i] == 0 and students[i+1] == 2:
            students[i] = 1
            students[i+1] -= 1
        elif students[i] == 2 and students[i+1] == 0:
            students[i] = 1
            students[i+1] = 1
    return sum([1 for i in students if i != 0])

정수 내림차순으로

https://programmers.co.kr/learn/courses/30/lessons/12933

def solution(n):
    n = sorted(list(str(n)))[::-1]
    return int(''.join(n))

def solution(n):
    answer = ''
    lst = []

    while n > 0:
        r = n % 10
        n = n // 10
        lst.append(r)

    lst.sort()
    lst.reverse()

    for j in range(0, len(lst)):
        answer += str(lst[j])

    return int(answer)

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