내일은 알고리즘

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

def solution(n):
    num = [0 for i in range(10)]
    
    for i in str(n):
        num[int(i)] += 1
    
    answer = ""
    for i in range(10):
        for j in range(num[9-i]):
            answer += str(9-i)
    
    return int(answer)

https://school.programmers.co.kr/learn/courses/30/lessons/12947

def solution(x):
    div = 0
    for i in str(x):
        div += int(i)
    
    return x % div == 0

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

def solution(s):
    p, y  = 0, 0
    s = s.lower()
    
    for i in s:
        if i == 'p':
            p += 1
        elif i == 'y':
            y += 1
    return p == y

https://school.programmers.co.kr/learn/courses/30/lessons/12934?language=python3 

import math


def solution(n):
    x = math.sqrt(n)
    xd = x // 1

    if x == xd:
        return (x + 1) * (x + 1)
    else:
        return -1

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

분노의 프로그래머스 1레벨 퍼먹기에 도전합니다.

def solution(n):
    answer = list(str(n))
    answer.reverse()
    
    for i in range(len(answer)):
        answer[i] = int(answer[i])
        
    return answer

https://school.programmers.co.kr/learn/courses/30/lessons/59415

SELECT datetime
FROM animal_ins
ORDER BY datetime DESC
LIMIT 1

이로서 프로그래머스 SQL 1레벨 문제를 전부 풀었습니다.

2레벨 3레벨은 주말에 풀어볼게요!

https://school.programmers.co.kr/learn/courses/30/lessons/59034

SELECT *
FROM animal_ins
ORDER BY animal_id ASC

모든 정보를 조회하는 문제라 정말 간단히 끝납니다.

https://school.programmers.co.kr/learn/courses/30/lessons/59039

SELECT animal_id
FROM animal_ins
WHERE NAME is NULL
ORDER BY animal_id ASC