프로그래머스 풀이 (Level 1)

SELECT.sql

## 모든 레코드 조회하기
# https://programmers.co.kr/learn/courses/30/lessons/59034
SELECT * FROM ANIMAL_INS ORDER BY ANIMAL_ID;

## 역순 정렬하기
# https://programmers.co.kr/learn/courses/30/lessons/59035
SELECT NAME, DATETIME FROM ANIMAL_INS ORDER BY ANIMAL_ID DESC;

## 아픈 동물 찾기
# https://programmers.co.kr/learn/courses/30/lessons/59036
SELECT ANIMAL_ID, NAME  
FROM ANIMAL_INS
WHERE INTAKE_CONDITION = 'Sick'
ORDER BY ANIMAL_ID;
  
## 어린 동물 찾기
# https://programmers.co.kr/learn/courses/30/lessons/59037
SELECT ANIMAL_ID, NAME
FROM ANIMAL_INS
WHERE INTAKE_CONDITION != 'Aged'
ORDER BY ANIMAL_ID;

## 동물의 아이디와 이름
# https://programmers.co.kr/learn/courses/30/lessons/59403
SELECT ANIMAL_ID, NAME
FROM ANIMAL_INS
ORDER BY ANIMAL_ID;

## 여러 기준으로 정렬하기
# https://programmers.co.kr/learn/courses/30/lessons/59404
SELECT ANIMAL_ID, NAME, DATETIME
FROM ANIMAL_INS
ORDER BY NAME, DATETIME desc

## 상위 n개 레코드
# https://programmers.co.kr/learn/courses/30/lessons/59405
SELECT NAME 
FROM ANIMAL_INS
ORDER BY DATETIME
LIMIT 1

SUM, MAX, MIN.sql

## 최댓값 구하기
# https://programmers.co.kr/learn/courses/30/lessons/59415

SELECT DATETIME 
FROM ANIMAL_INS
ORDER BY DATETIME desc
LIMIT 1;

## 최솟값 구하기
# https://programmers.co.kr/learn/courses/30/lessons/59038
SELECT DATETIME 
FROM ANIMAL_INS
ORDER BY DATETIME 
LIMIT 1;

## 동물 수 구하기
# https://programmers.co.kr/learn/courses/30/lessons/59406
SELECT COUNT(*)
FROM ANIMAL_INS

내적.py

## 내적
# https://programmers.co.kr/learn/courses/30/lessons/70128
def solution(a, b):
    prod = 0
    for n, m in zip(a, b):
        prod += n*m
    return prod

## 다른 사람풀이
def solution(a, b):
    return sum([x*y for x, y in zip(a, b)])

로또의 최고 순위와 최저 순위.py

def solution(lottos, win_nums):
    rank = [6,6,5,4,3,2,1]
    count_zero =  lottos.count(0)
    common_count = 0
    for num in lottos:
        if num in win_nums:
            common_count += 1
    
    return [rank[common_count + count_zero], rank[common_count]]

모의고사 .py

# https://programmers.co.kr/learn/courses/30/lessons/42840
def solution(answers):
    scores = [0]*4
    w1 = [1,2,3,4,5]
    w2 = [2,1,2,3,2,4,2,5]
    w3 = [3,3,1,1,2,2,4,4,5,5]
    for idx, ans in enumerate(answers):
        r1, r2, r3 = w1[idx%5], w2[idx%8], w3[idx%10]
        if r1 == ans: scores[1] += 1
        if r2 == ans: scores[2] += 1
        if r3 == ans: scores[3] += 1

    return [idx for idx, val in enumerate(scores) if val == max(scores) and idx != 0]

문자열 압축.py

## 문자열 압축 (다시 풀어야함)
# https://programmers.co.kr/learn/courses/30/lessons/60057
def solution(s):
    length = len(s)
    cand = [length] # 1-len 까지 압축했을 때 길이 값
    
    for size in range(1, length):
        compressed = ''
        splited = [s[i:i+size] for i in range(0, length, size)]
        count = 1
        
        for j in range(1, len(splited)):
            prev, curr = splited[j-1], splited[j]
            if prev == curr: 
                count += 1
            else: 
                compressed += (str(count) + prev) if count > 1 else prev
                count = 1
        
        compressed += (str(count) + splited[-1]) if count > 1 else splited[-1]
        cand.append(len(compressed))
    
    return min(cand)

부족한 금액 계산하기.py

## 위클리 챌린지 1주차
# https://programmers.co.kr/learn/courses/30/lessons/82612

def solution(price, money, count):
    fee = 0
    for i in range(count):
        fee += price*(i + 1)
    
    return fee - money if fee - money > 0 else 0

숫자 문자열과 영단어.py

## 숫자 문자열과 영단어
# https://programmers.co.kr/learn/courses/30/lessons/81301
def solution(s):
    words = ['zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine']
    word2num = {num: str(idx) for idx, num in enumerate(words)}
    res = []
    i = 0
    while i < len(s):
        if s[i].isdigit():
            res.append(s[i])
            i += 1
        else:
            en = i + 1
            while en < len(s) and not s[en].isdigit():
                if s[i:en] in word2num: break
                en += 1
            res.append(word2num[s[i:en]])
            i = en
    
    return int(''.join(res))

## 다른 사람 풀이
num_dic = {"zero":"0", "one":"1", "two":"2", "three":"3", "four":"4", "five":"5", "six":"6", "seven":"7", "eight":"8", "nine":"9"}
def solution(s):
    answer = s
    for key, value in num_dic.items():
        answer = answer.replace(key, value)
    
    return int(answer)

신규 아이디 추천.py

import re
def solution(new_id):
    new_id = re.sub('[^a-z0-9\-_.]', '', new_id.lower())
    tmp = []
    for i in range(len(new_id)):
        if len(tmp) and new_id[i] == new_id[i - 1] and new_id[i] == '.':
            continue
        tmp.append(new_id[i])
    
    # remove head, tail '.'
    while len(tmp):
        if tmp[0] == '.': tmp.pop(0)
        elif tmp[-1] == '.': tmp.pop()
        else: break
    
    if len(tmp) == 0: tmp.append('a')
    if len(tmp) > 15: 
        tmp = tmp[:15]
        if tmp[-1] == '.': tmp.pop()
    while len(tmp) < 3:
        tmp.append(tmp[-1])
    
    return ''.join(tmp)

완주하지 못한 선수.py

## 완주하지 못한 선수
# https://programmers.co.kr/learn/courses/30/lessons/42576
# 아래처럼 짜면 시간 초과
def solution(participant, completion):
    res = participant.copy()
    for completed in completion:
        res.remove(completed)
    
    return res

# 그렇다면 해쉬를 이용해야 할 것 같다는 생각이 듬.
import collections
def solution(participant, completion):
    counter = collections.Counter(participant)
    for completed in completion:
        counter[completed] -= 1
    
    for k, v in counter.items():
        if v == 1: return k
        
## 다른 사람의 풀이
import collections

def solution(participant, completion):
    answer = collections.Counter(participant) - collections.Counter(completion)
    return list(answer.keys())[0]

## 또는 정렬의 성질을 이용해도 O(nlogn)이라는 걸..
def solution(participant, completion):
    participant.sort(); completion.sort()
    for p, c in zip(participant, completion):
        if p != c:
            return p
        
    return participant[-1]

직업군 추천하기.py

## 위클리 챌린지 4주차
# https://programmers.co.kr/learn/courses/30/lessons/84325?language=python3
from typing import List

def solution(table: List[str], languages: List[str], preference: List[int]):
    recommended = []
    table = [[s.split()[0], s.split()[1:]] for s in table]
    for field, langs in table:
        score = 0
        for idx, lang in enumerate(langs):
            # 언어 선호도 X 직업군 언어 점수
            if lang in languages:
                score += (preference[languages.index(lang)])*(5 - idx)            
                
        recommended.append([score, field])
    recommended.sort(key=lambda x: (-x[0], x[1]))
    
    return recommended[0][1]