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프로그래머스-코딩 기초 트레이닝

https://school.programmers.co.kr/learn/challenges/training?order=acceptance_desc 

 

코딩 기초 트레이닝 | 프로그래머스 스쿨

프로그래밍 언어를 처음 접하거나 프로그래밍 언어에 익숙해지고 싶다면? 코딩 역량을 탄탄하게 길러주는 코딩 기초 트레이닝으로 차근차근 코딩에 대한 기초 감각을 길러 보세요.

school.programmers.co.kr

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코딩 기초 트레이닝 캘린더

 

 

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DAY1.출력

Q.문자열 출력하기

str = input()
print(str)

Q.a와 b 출력하기

a, b = map(int, input().strip().split(' '))
print('a =', a)
print('b =', b)

 

Q.문자열 반복해서 출력하기

a, b = input().strip().split(' ')
b = int(b)

for i in range(b) :
    print(a, end='')

Q.대소문자 바꿔서 출력하기

str = input()
str2='' 
for x in str : 
    if x.isupper() :
        str2+=x.lower()
    else:
        str2+=x.upper()
print(str2)

'''
오답(에러)
주어진 코드에서 출력이 원하는 대로 나오지 않는 이유는 
반복문 안에서 x 변수에 대한 변환 작업은 원본 문자열인 str에 적용되지 않기 때문입니다. 
Python에서 문자열은 불변(immutable)하며, 문자열을 수정하려면 새로운 문자열을 생성해야 합니다.
'''
for i in range(len(str)) : 
    if str[i].isupper() : 
        str[i].lower()
        print(str[i])
    else :
        str[i]=str[i].upper()
print(str)

Q.특수문자 출력하기

출력예시 : !@#$%^&*(\'"<>?:;

print('!@#$%^&*(\\\'"<>?:;')

 

Day 2.출력, 연산
Q.덧셈식 출력하기

a, b = map(int, input().strip().split(' '))
print(a,'+',b,'=',a+b)

Q. 문자열 붙여서 출력하기

str1, str2 = input().strip().split(' ')
print(str1+str2)
#print(str1, str2, sep='')

Q. 문자열 돌리기

str = input()
for x in str :
    print(x)

Q. 홀짝 구분하기

a = int(input())
if a%2==0 :
    print(a,'is even')
else :
    print(a, 'is odd')

Q. 문자열 겹쳐쓰기

def solution(my_string, overwrite_string, s):
    answer = ''
    answer+=my_string[0:s]
    answer+=overwrite_string
    if len(answer)<len(my_string):
        answer+=my_string[len(answer):]
    # answer = my_string[:s]+overwrite_string+my_string[s+len(overwrite_string):]
    return answer

 

Day 3.연산
Q.문자열 섞기

def solution(str1, str2):
    answer = ''
    for i in range(len(str1)):
        answer+=str1[i]+str2[i]
    return answer

Q.문자 리스트를 문자열로 변환하기

def solution(arr):
    answer = ''
    for x in arr:
        answer+=x
    return answer

Q.문자열 곱하기

def solution(my_string, k):
    answer = ''
    for i in range(k):
        answer+=my_string
    return answer

Q.더 크게 합치기

def solution(a, b):
    answer = 0
    x=int(str(a)+str(b))
    y=int(str(b)+str(a))
    if x>y or x==y:
        answer=x
    else :
        answer=y

    return answer

Q.두 수의 연산값 비교하기

def solution(a, b):
    answer = 0
    x=int(str(a)+str(b))
    y=2*int(a)*(b)
    if x>y or x==y :
        answer=x
    else : 
        answer=y
    return answer
    #return max(int(str(a) + str(b)), 2 * a * b)
    #return int(str(a) + str(b)) if int(str(a) + str(b)) > 2*a*b else 2*a*b

Day 4.연산, 조건문
Q.n의 배수

def solution(num, n):
    answer = 1 if num%n==0 else 0
    return answer

Q.공배수

def solution(number, n, m):
    #오답
    #answer = 1 if number%(n*m)==0 else 0
    answer = 1 if (number%n==0 and number%m==0) else 0 
    return answer

Q.홀짝에 따라 다른 값 반환하기

def solution(n):
    answer = 0
    if n%2==0 :
        for i in range(2,n+1,2) :
            answer+=i*i
    else : 
        for i in range(1,n+1,2) :
            answer+=i
    return answer

Q.조건 문자열

def solution(ineq, eq, n, m):
    answer = 0
    #ㅎㅎ?
    if ineq=='<':
        if eq=='=' and n<=m :
            answer = 1 
        if eq=='!' and n<m :
            answer = 1
    elif ineq=='>':
        if eq=='=' and n>=m :
            answer = 1 
        if eq=='!' and n>m :
            answer = 1
    return answer

Q.flag에 따라 다른 값 반환하기

def solution(a, b, flag):
    answer = 0
    answer = a+b if flag else a-b
    return answer

 

DAY 5.조건문
Q.코드 처리하기

def solution(code):
    ret = ''
    mode=0
    for i in range(len(code)) : 
        x=str(code[i])
        if mode==0 :
            if x!='1' and i%2==0  :
                ret+=x
            elif x=='1': 
                mode=1
        else :
            if x!='1' and i%2!=0 :
                ret+=x
            elif x=='1': 
                mode=0
        #print(i, x, mode, ret, sep=' ')
    ret= 'EMPTY' if ret=='' else ret 
    # return answer if answer else 'EMPTY'
    return ret

Q.등차수열의 특정한 항만 더하기

def solution(a, d, included):
    answer = 0
    for i in range(len(included)) : 
        if included[i]==True : 
            answer+=a+(d*i)    
    return answer

Q.주사위 게임 2

def solution(a, b, c):
    answer = 0
    if a==b==c : 
        answer+=(a+b+c)*(a*a+b*b+c*c)*(a*a*a+b*b*b+c*c*c)
    elif a==b or a==c or b==c :
        answer+=(a+b+c)*(a*a+b*b+c*c)
    else :
        answer+=(a+b+c)
    return answer
    #answer *= a**(i+1)+b**(i+1)+c**(i+1)

Q.원소들의 곱과 합

def solution(num_list):
    answer = 0
    sum1=0
    sum2=1
    for x in num_list :
        sum1+=x
        sum2*=x
    if sum1**2>sum2 :
        answer = 1    
    elif sum1**2<sum2 : 
        answer = 0 
    return answer
    #sum(num_list)

Q.이어 붙인 수

def solution(num_list):
    answer = 0
    a=''
    b=''
    for x in num_list : 
        if x%2==0 : 
            a+=str(x)
        else : 
            b+=str(x)
    answer = int(a) + int(b)
    return answer

DAY 6.조건문, 반복문
Q.마지막 두 원소

def solution(num_list):
    answer = []
    size=len(num_list)
    if num_list[size-1] > num_list[size-2] :
        num_list.append(num_list[size-1]-num_list[size-2])
    else : 
        num_list.append(num_list[size-1]*2)
    answer=num_list
    return answer
    # a=num_list[-1]
    # b=num_list[-2]

Q.수 조작하기 1

def solution(n, control):
    for x in control :
        if x=='w' : 
            n=n+1 
        elif x=='s':
            n=n-1
        elif x=='d':
            n=n+10
        elif x=='a':
            n=n-10
    return n

Q.수 조작하기 2

def solution(numLog):
    answer = ''
    for i in range(1,len(numLog)) :
        tmp = numLog[i]-numLog[i-1]
        if tmp==1 :
            answer+='w'
        elif tmp==-1:
            answer+='s'
        elif tmp==10:
            answer+='d'
        elif tmp==-10:
            answer+='a'
    return answer

Q.수열과 구간 쿼리 3

def solution(arr, queries):
    for x in queries : 
        i,j=x[0],x[1]
        arr[i],arr[j] = arr[j],arr[i]
    return arr
    #for a,b in queries:
    #    arr[a],arr[b]=arr[b],arr[a]

Q.수열과 구간 쿼리 2

def solution(arr, queries):
    answer = []
    for s,e,k in queries : 
        tmp=arr[s:e+1]
        minValue = 2147000000 
        for x in tmp :
            if x>k and minValue>x : 
                minValue = x
        if minValue!=2147000000 : 
            answer.append(minValue)
        else :
            answer.append(-1)
    return answer
    #m=float('inf')
    #min(tmp)

DAY 7.반복문
Q.수열과 구간 쿼리 4

def solution(arr, queries):
    for s,e,k in queries : 
        #tmp = arr[s:e+1]
        for i in range(s,e+1) : 
            if i%k==0 : 
                arr[i]=arr[i]+1
    return arr

Q.배열 만들기 2

def solution(l, r):
    answer = []
    #*
    for num in range(l,r+1):
        if not set(str(num)) - set(['0','5']) :
            answer.append(num)
'''
set(str(num))은 num을 문자열로 변환하고, 문자열을 중복되지 않는 집합(set)으로 변환합니다. 집합은 중복된 요소를 포함하지 않으므로, 각 문자가 한 번만 나타납니다.
set(['0', '5'])는 문자 '0'과 '5'로 이루어진 문자열 리스트를 집합으로 변환합니다.
그런 다음 set(str(num)) - set(['0', '5'])는 num을 문자열로 변환한 집합에서 '0'과 '5'로 이루어진 집합을 뺀 결과를 나타냅니다. 만약 num의 문자열 표현에 '0' 또는 '5'가 없다면, 이 연산 결과는 빈 집합이 됩니다.
'''
    return answer if answer else [-1]

Q.카운트 업

def solution(start_num, end_num):
    return list(range(start_num, end_num+1))
#return [i for i in range(start,end+1)]

Q.콜라츠 수열 만들기

def solution(n):
    answer=[n]    
    while n>1 : 
        if n%2==0 : 
            n=n//2
        else :
            n=3*n+1
        answer.append(n)
    return answer

Q.배열 만들기 4

def solution(arr):
    stk = []
    i=0
    while i < len(arr) :
        if len(stk)==0 : 
            stk.append(arr[i])
            i+=1
        else : 
            if stk[-1] < arr[i] :
                stk.append(arr[i])
                i+=1
            else : 
                stk.pop() #*
    return stk

DAY 8.조건문,문자열
Q.간단한 논리 연산

def solution(x1, x2, x3, x4):
    #논리합 OR ∨ , 논리곱 AND ∧ (True 1, False 0)
    answer = True
    a=(True if x1==True or x2==True else False)
    b=(True if x3==True or x4==True else False)
    answer = False if a==False or b==False else True 
    return answer
#return (x1 | x2) & (x3 | x4)
#return (x1 or x2) and (x3 or x4)

Q.주사위 게임 3

#런타임에러 ㅠ ㅠ 
def solution(a, b, c, d):
    answer = 0
    x=set([a,b,c,d])
    y=list([a,b,c,d])

    if len(x)==1 : 
        answer=1111*a
    elif len(x)==4 :
        answer=min(x)
    elif len(x)==3 :
        ch=[0]*7
        p=0
        for i in y:
            ch.insert(i,(ch[i]+1))
        for i in ch:
            if ch[i]>1 :
                p=i
                break
        tmp=y
        tmp.remove(p) #list.remove(value) 하나만 제거함
        tmp.remove(p)
        q=tmp[0]
        r=tmp[1]
        answer=q*r
    elif len(x)==2 :
        ch=list([0]*7)
        for i in y:
            ch[i]=ch[i]+1
        #answer=int(str(ch[0])+str(ch[1])+str(ch[2])+str(ch[3])+str(ch[4])+str(ch[5])+str(ch[6]))
        for i in range(len(ch)):
            if ch[i]==3:
                answer=333
                tmp=y
                p=i
                tmp.remove(p)
                tmp.remove(p)
                tmp.remove(p)
                q=tmp[0]
                answer=(10*p+q)**2
                break
                
        if answer==0 :
            tmp=list(x)
            p=tmp[0]
            q=tmp[1]
            answer=(p+q)*abs(p-q)
    return answer

#정답 코드 https://school.programmers.co.kr/questions/47795 
def solution(a, b, c, d):
    answer = 0
    arr=[a,b,c,d]
    counts=[arr.count(i) for i in arr] #각 숫자의 출현 횟수를 저장
    
    if max(counts)==4 : #모두 같은 경우
        answer=1111*a
    elif max(counts)==3 : #pppq
        p=arr[counts.index(3)]
        q=arr[counts.index(1)]
        answer=(10*p+q)**2
    elif max(counts)==2: #ppqr or ppqq 
        if min(counts) == 2:
            return (a + c) * abs(a - c) if a == b else (a + b) * abs(a - b) #?
        else:
            p = arr[counts.index(2)]
            return (a * b * c * d) / p**2 #?
    else : #모두 다른 경우
        answer=min(arr)
        
    return answer

Q.글자 이어 붙여 문자열 만들기

def solution(my_string, index_list):
    answer = ''
    for x in index_list : 
        answer+=my_string[x]
    return answer
Q.9로 나눈 나머지
def solution(number):
    return int(number)%9

Q.문자열 여러 번 뒤집기

def solution(my_string, queries):
    for s,e in queries : 
        '''
        for i in range(e-s+1//2) : 
            #my_string[s+i], my_string[e-i] = my_string[e-i],my_string[s+i] #오류남
        '''  
        my_string = my_string[:s] + my_string[s:e+1][::-1] + my_string[e+1:]
        #my_string[start:end+1][::-1]는 파이썬에서 문자열을 일부분 선택한 후 그 부분을 뒤집는 연산을 수행하는 코드
    return my_string

DAY 9.문자열
Q.배열 만들기 5

def solution(intStrs, k, s, l):
    answer = []
    for x in (intStrs) :
        x=int(x[s:s+l])
        if x>k :
            answer.append(x)
    return answer

Q.부분 문자열 이어 붙여 문자열 만들기

def solution(my_strings, parts):
    answer = ''
    for i in range(len(my_strings)) :
        str=my_strings[i]
        s,e = parts[i]
        answer+=str[s:e+1]
    return answer

Q.문자열의 뒤의 n글자

def solution(my_string, n):
    return my_string[len(my_string)-n:]
    #return my_string[-n:]

Q.접미사 배열

def solution(my_string):
    answer = []
    size=len(my_string)
    for i in range (size):
        answer.append(my_string[i:])
    answer.sort()
    return answer

Q.접미사인지 확인하기

def solution(my_string, is_suffix):
    answer = 0
    for i in range(len(my_string)) :
        if(my_string[i:] == is_suffix) :
            answer=1
        if answer==1:
            break 
    else : 
        answer=0
        
    return answer
    #return int(my_string.endswith(is_suffix)) #endsWith()는 특정문자열로 어떤 string이 특정문자열로 끝나는지를 검사해서 그 값을 Boolean값으로 반환
    #return int(my_string[-len(is_suffix):] == is_suffix)

DAY 10.문자열
Q.문자열의 앞의 n글자

def solution(my_string, n):
    return my_string [:n]

Q.접두사인지 확인하기

def solution(my_string, is_prefix):
    #answer = 1 if my_string.startswith(is_prefix) else 0 
    answer = 1 if my_string[:len(is_prefix)] == is_prefix else 0
    return answer

Q.문자열 뒤집기

def solution(my_string, s, e):
    return my_string[:s]+my_string[s:e+1:][::-1]+my_string[e+1:]

Q.세로 읽기

def solution(my_string, m, c):
    return my_string[c-1::m]

Q.qr code

def solution(q, r, code):
    answer = ''
    for i in range(len(code)) :
        if i%q == r :
            answer+=code[i]
    return answer

DAY 11.리스트(배열)
Q.문자 개수 세기

def solution(my_string):
    answer = []
    #아스키코드 변환 ord():문자>숫자/chr():숫자>문자
    answer=[my_string.count(chr(i)) for i in range(ord('A'), ord('Z') + 1)] + [my_string.count(chr(i)) for i in range(ord('a'), ord('z') + 1)]
    return answer

Q.배열 만들기 1

def solution(n, k):
    answer = []
    for i in range(1,n+1) :
        if i%k==0 :
            answer.append(i)
    return answer
    #return [i for i in range(k,n+1,k)]
    #return [i for i in range(1, n+1) if i % k == 0]

Q.글자 지우기

def solution(my_string, indices):
    answer=''
    for i in range(len(my_string)) : 
        if i not in indices :
            answer +=my_string[i]
    return answer

Q.카운트 다운

def solution(start, end_num):
    answer=[]
    ret=start
    while ret >= end_num :
        answer.append(ret)
        ret-=1
    return answer
    #return list(range(start,end-1,-1))

Q.가까운 1 찾기

def solution(arr, idx):
    answer = 0
    for i in range(len(arr)):
        if i >= idx and arr[i] == 1 :
            answer=i
            break
    else :
        answer=-1 
    
    return answer

DAY 12.리스트(배열)
Q.리스트 자르기

def solution(n, slicer, num_list):
    answer = []
    a,b,c=slicer 
    if n==1 :
        num_list=num_list[0:b+1]
    elif n==2:
        num_list=num_list[a:]
    elif n==3:
        num_list=num_list[a:b+1]
    elif n==4:
        num_list=num_list[a:b+1:c]
    answer=num_list
    return answer

Q.첫 번째로 나오는 음수

def solution(num_list):
    answer = 0
    for x in num_list :
        if x<0 : 
            answer=num_list.index(x) #index(): x를 찾은 가장 첫 번째 인덱스
            break
    else : 
        answer=-1
    return answer

Q.배열 만들기 3

def solution(arr, intervals):
    answer = []
    for s,e in intervals :
        for i in arr[s:e+1] :
            answer.append(i)
    return answer
    #for a,b in intervals: answer+=arr[a:b+1]

Q.2의 영역

def solution(arr):
    answer = []
    s=0
    e=0
    for i in range(len(arr)):
        if arr[i] == 2 : 
            s=i
            break 
    for j in range(len(arr)-1,-1,-1):
        if arr[j] == 2:
            e=j
            break
    if arr.count(2)==0 :
        answer.append(-1)
    else :
        answer+=arr[s:e+1]
    return answer
    
    #if 2 not in arr:
    #    return [-1]
    #return arr[arr.index(2) : len(arr) - arr[::-1].index(2)]

Q.배열 조각하기

def solution(arr, query):
    answer = []
    for i in range(len(query)) :
        if i%2==0 :
            arr=arr[:query[i]+1]
        else :
            arr=arr[query[i]:]
    return arr

DAY 13.리스트(배열)
Q.n 번째 원소부터

def solution(num_list, n):
    return num_list[n-1:]

Q.순서 바꾸기

def solution(num_list, n):
    return num_list[n:]+num_list[:n]

Q.왼쪽 오른쪽

def solution(str_list):
    answer = []
    for i in str_list : 
        if i=="l" :
            answer=str_list[:str_list.index("l")]
            break
        elif i=="r":
            answer=str_list[str_list.index("r")+1:]
            break
    return answer

Q.n번째 원소까지

def solution(num_list, n):
    return num_list[:n]

Q.n개 간격의 원소들

def solution(num_list, n):
    return num_list[::n]

DAY 14.리스트(배열)
Q.홀수vs짝수

def solution(num_list):
    answer = 0
    sum1=0
    sum2=0
    for i in range(len(num_list)):
        if i%2==0:
            sum1+=num_list[i]
        else :
            sum2+=num_list[i]
    answer=sum1 if sum1>=sum2 else sum2 
    return answer
    #return max(sum(num_list[::2]), sum(num_list[1::2]))

Q.5명씩

def solution(names):
    answer = []
    for i in range(0,len(names),5):
        answer.append(names[i])
    return answer
    #return names[::5]

Q.할 일 목록

def solution(todo_list, finished):
    answer = []
    for i in range(len(todo_list)) :
        if not finished[i] :
            answer.append(todo_list[i])
    return answer
    #return [work for idx, work in enumerate(todo_list) if not finished[idx]]

Q.n보다 커질 때까지 더하기

def solution(numbers, n):
    answer = 0
    i=0
    while answer<=n:
        answer+=numbers[i]
        i+=1
    return answer

Q.수열과 구간 쿼리1

def solution(arr, queries):
    answer = []
    '''
    #테스트 케이스 하나만 정답 .. 나머지 케이스 모두 틀림
    for i in range(len(arr)):
        if i<len(queries):
            s,e = queries[i]
            if s<=i<=e :
                for j in range (s,e+1) : 
                   arr[j] +=1
    '''
    
    for i,q in enumerate(queries) :
        s=q[0]
        e=q[1]
        #if s<=i<=e : 
        for k in range(s,e+1) :
            arr[k]+=1
    return arr

DAY 15.리스트(배열),문자열
Q.조건에 맞게 수열 변환하기 1

def solution(arr):
    answer = []
    for idx,x in enumerate(arr) : 
        if x>=50 and x%2==0 :
            arr[idx]=x//2
        elif x<50 and x%2!=0 :
            arr[idx]=x*2
    return arr

Q.조건에 맞게 수열 변환하기 2 

def solution(arr):
    '''
    #* ㅠ 
    answer = 0
    arr2=[]
    arr2.append(arr.copy()) #arr2.append(arr) 안됨. 
    for x in range(len(arr)) : 
        tmp=[]
        for i in range(len(arr)) : 
            
            x = arr[i]
            if x>=50 and x%2==0 :
                arr[i]=x//2
            elif x<50 and x%2!=0 :
                arr[i]=x*2+1
            tmp.append(arr[i])
        arr2.append(tmp)
        
    for i in range(len(arr2)) :
        if arr2[i] == arr2[i+1] : 
            answer = i
            break
    '''
    
    idx = 0
    pre = arr 
    while True : 
        tmp = []
        for x in pre :
            if x>=50 and x%2==0 :
                tmp.append(x//2)
            elif x<50 and x%2!=0 :
                tmp.append(x*2+1)
            else : 
                tmp.append(x)
        if pre == tmp : 
            break
        
        idx+=1
        pre=tmp
    answer = idx     
    return answer

Q.1로 만들기

def solution(num_list):
    answer = 0
    for x in num_list : 
        while x>1 :
            if x%2==0 :
                answer+=1
                x=x//2
            elif x%2!=0 :
                answer+=1
                x=(x-1)//2
    return answer

Q.길이에 따른 연산

def solution(num_list):
    answer = 0
    if len(num_list)>=11 :
        for i in num_list : 
            answer+=i
    elif len(num_list)<=10 :
        answer+=1
        for i in num_list : 
            answer*=i
    return answer
    #return sum(num_list) if len(num_list)>=11 else prod(num_list)

Q.원하는 문자열 찾기

def solution(myString,pat):
            
    answer = 0
    myString=myString.upper()
    pat=pat.upper()
    #if(myString.startswith(pat)) : answer=1 ##?? 안되는건가 ?
    
    '''
    #정확성: 90.9 ㅠ 
    for i in range(len(myString)-len(pat)) :
        if myString[i:i+len(pat)] == pat :
            answer = 1 
            break
    '''
    
    if pat in myString : 
        answer=1
        
    return answer
    #return int(pat.lower() in myString.lower())

DAY 16.문자열
Q.대문자로 바꾸기

def solution(myString):
    return myString.upper()

Q.소문자로 바꾸기

def solution(myString):
    return myString.lower()

Q.배열에서 문자열 대소문자 변환하기

def solution(strArr):
    for i in range(len(strArr)):
        if i%2==0:
            strArr[i]=strArr[i].lower()
        else : 
            strArr[i]=strArr[i].upper()
    return strArr

Q.A 강조하기

def solution(myString):
    answer = ''
    for x in myString : 
        if x=='a':
            answer+=x.upper()
        elif x!="A" and x.isupper() :
            answer+=x.lower()
        else :
            answer+=x
    return answer
    #return myString.lower().replace('a', 'A')

Q.특정한 문자를 대문자로 바꾸기

def solution(my_string, alp):
    answer = ''
    '''
    for x in my_string :
        if x==alp :
            answer+=x.upper()
        else :
            answer+=x
    '''
    answer = my_string.replace(alp,alp.upper())
    return answer

DAY 17.문자열
Q.특정 문자열로 끝나는 가장 긴 부분 문자열 찾기

def solution(myString, pat):
    answer = ''
    ''' # 틀림 
    for i in range (len(myString)-len(pat),0,-1):
        if myString[i:i+len(pat)] == pat :
            answer = myString[:i+1] #ㅠ 틀림 .
            break
    '''
    
    #index(), find() <--> rindex(), rfind() : 뒤에서 부터
    #find(찾을 문자열), find(찾을 문자열, 시작 위치), find(찾을 문자열, 시작 위치, 종료 위치) 
    #만약 없을 경우 find()에서는 -1을 리턴하는 반면, index()에서는 ValueError가 발생
    
    answer = myString[:myString.rfind(pat)+len(pat)]
    return answer

Q.문자열이 몇 번 등장하는지

def solution(myString, pat):
    answer = 0
    for i in range(len(myString)-len(pat)+1) :
        if myString[i:i+len(pat)] == pat : 
            answer+=1
    return answer

Q.ad 제거하기

def solution(strArr):
    answer = []
    ''' 틀림 ㅠ 
    for x in strArr : 
        if x.find("ad")>0 :
            strArr.remove(x) 
    '''
    for x in strArr : 
        if x.find("ad")<0 :
            answer.append(x)

    return strArr
    #return [word for word in strArr if 'ad' not in word]

Q.공백으로 구분하기 1

def solution(my_string):
    answer=list(my_string.split())
    return answer
    #return my_string.split()

Q.공백으로 구분하기 2

def solution(my_string):
    return my_string.split()

DAY 18.문자열
Q.x 사이의 개수 

def solution(myString):
    answer = []
    for x in myString.split('x') :
        answer.append(len(x))
    return answer
    #return [len(w) for w in myString.split('x')]

Q.문자열 잘라서 정렬하기

def solution(myString):
    answer = []
    tmp=myString.split('x')
    for i in tmp :
        if i!="" :
            answer.append(i)
    answer.sort()
    return answer
    #return sorted(ch for ch in myString.split('x') if ch)
    '''
    myString.split('x'): 문자열 myString을 'x'를 구분자로 사용하여 분할한 후 각 부분을 리스트로 반환합니다. 예를 들어, "ABCxDEFxGHI"를 "x"로 분할하면 ['ABC', 'DEF', 'GHI']가 됩니다.
    for ch in myString.split('x') if ch: 리스트 컴프리헨션을 사용하여 분할된 문자열 중 빈 문자열("")을 필터링하여 제거합니다. 이 부분은 if ch 조건을 사용하여 빈 문자열이 아닌 경우에만 포함합니다.
    sorted(): 리스트를 정렬합니다. 위 코드에서는 문자열이 사전순으로 정렬됩니다.
    '''

Q.간단한 식 계산하기

def solution(binomial):
    answer = 0
    a,op,b = binomial.split()
    if op=='+' : 
        answer=int(a)+int(b)
    elif op=='-':
        answer=int(a)-int(b)
    elif op=='*':
        answer=int(a)*int(b)
    return answer
    #return eval(binomial)
    #expression (=식)을 문자열로 받아서, 실행하는 함수

Q.문자열 바꿔서 찾기

def solution(myString, pat):
    answer = 0
    tmp = ""
    for ch in myString :
        if ch=="B" : 
            tmp+="A" 
        else :
            tmp+="B"
    answer = 1 if tmp.find(pat)>=0 else 0 
    return answer

Q.rny_string

def solution(rny_string):
    answer = ''
    for i in range(len(rny_string)) :
        if rny_string[i]=='m' :
            answer+='rn'
        else : 
            answer+=rny_string[i]
    return answer
    #return rny_string.replace('m', 'rn')

DAY 19.문자열,리스트(배열)
Q.세 개의 구분자

def solution(myStr):
    answer = []
    x1=[]
    for x in myStr.split('a') : 
        tmp = list(ch for ch in x.split('b') if ch)
        x1+=tmp
    x2=[]
    for x in x1 : 
        tmp = list(ch for ch in x.split('c') if ch)
        x2+=tmp
        
    answer = x2 if len(x2)>0 else ["EMPTY"]
    return answer

Q.배열의 원소만큼 추가하기

def solution(arr):
    answer = []
    for i in arr : 
        for x in range (i) :
            answer.append(i)
    return answer

Q.반 배열에 추가, 삭제하기

def solution(arr, flag):
    answer = []
    for i in range(len(arr)) :
        if flag[i] :
            for j in range (arr[i]*2) :
                answer.append(arr[i])
        else : 
            for j in range (arr[i]) :
                answer.pop()
    return answer

Q.배열 만들기 6

def solution(arr):
    stk = []
    for i in range(len(arr)) : 
        x=arr[i]
        if len(stk)==0 : 
            stk.append(x)
        elif stk[len(stk)-1]==x :
            stk.pop()
        elif stk[len(stk)-1]!=x :
            stk.append(x)
    if len(stk)==0 : 
        stk.append(-1)
    return stk
    #elif stk[-1] == arr[i]:

Q.무작위로 K개의 수 뽑기

def solution(arr, k):
    answer = []
    idx=0
    while idx<len(arr) :
        if arr[idx] not in answer : 
            answer.append(arr[idx])
        if len(answer)==k : 
            break 
        idx+=1
    if len(answer)<k :
        while len(answer)<k:
            answer.append(-1)
    return answer

DAY 20.함수(메서드)
Q.배열의 길이를 2의 거듭제곱으로 만들기

def solution(arr): #ㅠ
    answer = []
    i=1
    while i<len(arr):
        i=i*2
    for j in range(i-len(arr)) :
        arr.append(0)
    return arr 
    '''
    a = 1
    b = len(arr)
    while a < b :
        a *= 2
    return arr + [0] * (a-b)
    '''

Q.배열 비교하기

def solution(arr1, arr2):
    answer = 0
    if len(arr1)==len(arr2) :
        sum1=0
        sum2=0
        for i in range (len(arr1)) : 
            sum1+=arr1[i]
            sum2+=arr2[i]
        if sum1==sum2 :
            answer = 0
        else : 
            answer = 1 if sum1>sum2 else -1
    else : 
        answer = 1 if len(arr1)>len(arr2) else -1
    return answer
    #sum(arr1), sum(arr2)

Q.문자열 묶기

def solution(strArr):
    arr=[0]*(len(strArr)+1)
    for x in strArr : 
        arr[len(x)]+=1
    return max(arr)
    '''
    a=[0]*31
    for x in strArr: a[len(x)]+=1
    return max(a)
    '''

Q.배열의 길이에 따라 다른 연산하기

def solution(arr, n):
    if len(arr)%2==0 : 
        for i in range (1,len(arr),2) :
            arr[i]+=n
    else : 
        for i in range(0,len(arr),2) : 
            arr[i]+=n
    return arr

    
Q.뒤에서 5등까지

def solution(num_list):
    answer = []
    for i in range(5):
        x=min(num_list)
        answer.append(x)
        num_list.remove(x)
    return answer
    #return sorted(num_list)[:5]

DAY 21.함수(메서드)
Q.뒤에서 5등 위로

def solution(num_list):
    answer = []
    #list.sort() 본 리스트를 정렬해서 반환
    #sorted(리스트) 는 본 리스트를 기반하여 정렬한 새로운 리스트를 반환
    return sorted(num_list)[5:]

Q.전국 대회 선발 고사

def solution(rank, attendance):
    answer = 0
    rank2=[]
    for i in range(len(rank)) : 
        if attendance[i]==True : 
            rank2.append(rank[i])
    a,b,c = sorted(rank2)[:3]
    answer = 10000*rank.index(a) + 100*rank.index(b) + rank.index(c)
    return answer

Q.정수 부분

def solution(flo):
    return int(flo)
    ##flo//1

Q.문자열 정수의 합

def solution(num_str):
    answer = 0
    for x in num_str : 
        answer +=int(x)
    return answer

Q.문자열을 정수로 변환하기

def solution(n_str):
    return int(n_str)

DAY 22.함수(메서드), 조건문 활용
Q.0 떼기

def solution(n_str):
    answer = ''
    while True : 
        if n_str[0]=='0' : 
            n_str=n_str[1:]
        else :
            break
    return n_str
    #return n_str.lstrip('0')
    #return str(int(n_str))
    '''
    'abcdefg'.strip('a') : 인자로 전달된 문자를 String의 왼쪽과 오른쪽에서 제거합니다.
    'abcdefg'.lstrip('a') : 인자로 전달된 문자를 String의 왼쪽에서 제거합니다.
    'abcdefg'.rstrip('a') : 인자로 전달된 문자를 String의 오른쪽에서 제거합니다.
    인자가 없을 경우 공백 제거
    '''

Q.두 수의 합

    def solution(a, b):
    return str(int(a)+int(b))

Q.문자열로 변환

def solution(n):
    return str(n)

Q.배열의 원소 삭제하기

def solution(arr, delete_list):
    for x in delete_list : 
        if x in arr : 
            arr.remove(x)
    return arr

Q.부분 문자열인지 확인하기

def solution(my_string, target):
    answer = 0 if my_string.find(target)<0 else 1
    return answer

DAY 23.조건문 활용
Q.부분 문자열

def solution(str1, str2):
    return 0 if str2.find(str1)<0 else 1 
    #return int(str1 in str2)

Q.꼬리 문자열

def solution(str_list, ex):
    answer = ''
    for x in str_list : 
        if ex not in x :
            answer +=x 
    return answer

Q.정수 찾기

def solution(num_list, n):
    return 1 if n in num_list else 0 
    #return int(n in num_list)

Q.주사위 게임 1

def solution(a, b):
    answer = 0
    if a%2!=0 and b%2!=0 :
        answer = a**2 + b**2 
    elif a%2==0 and b%2==0 : 
        answer = abs(a-b) #abs():절댓값 함수
    else : 
        answer = 2*(a+b)
    return answer

Q.날짜 비교하기

def solution(date1, date2):
    answer = -1
    for i in range (3) :
        if date1[i]<date2[i] : 
            answer=1 
            break 
        elif date1[i]>date2[i]:
            answer=0
            break
    if answer==-1 : 
        answer=0
    return answer

DAY 24.조건문 활용, 반복문 활용, 이차원 리스트(배열)
Q.커피 심부름

def solution(order):
    answer = 0
    for x in order : 
        if x=="iceamericano" or x=="americanoice"or x=="hotamericano" or x=="americanohot" or x=="americano" or x=="anything":
            answer+=4500
        elif x=="icecafelatte" or x=="cafelatteice" or x=="hotcafelatte" or x=="cafelattehot" or x=="cafelatte" :
            answer+=5000
    return answer
    #if 'latte' in want:

Q.그림 확대

def solution(picture, k):
    answer = []
    for x in picture :
        tmp=''
        for i in x :
            tmp+=i*k
        for j in range(k) :
            answer.append(tmp)
    return answer

Q.조건에 맞게 수열 변환하기 3

def solution(arr, k):
    if k%2==0:
        for i,x in enumerate(arr) :
            arr[i]=arr[i]+k
    else :
        for i,x in enumerate(arr) :
            arr[i]=arr[i]*k
    return arr
    #return [i*k if k%2!=0 else i+k for i in arr]

Q.l로 만들기

def solution(myString):
    answer = ''
    #아스키코드 변환 함수
    #ord():문자>숫자, chr():숫자>문자
    for x in myString :
        if ord(x)<ord('l'):
        #if i < "l":
            answer+='l'
        else :
            answer +=x
    return answer

Q.특별한 이차원 배열 1

def solution(n):
    answer=[]
    for i in range (n) : 
        tmp=[]
        for j in range(n) :
            if i==j :
                tmp.append(1)
            else :
                tmp.append(0)
        answer.append(tmp)
    return answer
    '''
    answer=[[0]*n for i in range(n)]
    for i in range(n): answer[i][i]=1
    '''

DAY 25.이차원 리스트(배열)
Q.정수를 나선형으로 배치하기 

#**
def solution(n):
    answer=[[0]*n for _ in range(n)]
    #시계방향 : 우,하,좌,상
    dx=[0,1,0,-1]
    dy=[1,0,-1,0]
    
    x,y=0,0
    answer[x][y]=1
    k=2
    while k<=n*n :
        for i in range(4):
            while True :
                nx=x+dx[i]
                ny=y+dy[i]
                if nx>=n or ny>=n or nx<0 or ny<0 or answer[nx][ny]!=0:
                    break
                answer[nx][ny]=k
                x=nx
                y=ny
                k+=1
    return answer

Q.특별한 이차원 배열 2

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

Q.정사각형으로 만들기

def solution(arr):
    #x=len(arr) #행의 수
    #y=len(arr[0]) #열의 수
    if len(arr) > len(arr[0]) : #행의 수가 더 많을 경우
        for i in range(len(arr)) :
            for j in range(len(arr)-(len(arr[i]))) :
                arr[i].append(0)
    elif len(arr) < len(arr[0]) : #열의 수가 더 많을 경우
        for i in range(len(arr[0])-len(arr)):
            arr.append([0]*len(arr[i]))
    return arr

Q.이차원 배열 다각선 순회하기

def solution(board, k):
    answer = 0
    for i in range(len(board)) : 
        for j in range(len(board[0])) :
            if i+j<=k :
                answer +=board[i][j]
    return answer

---

사용 문법 정리

-입출력
str = input()
a = int(input())
a, b = map(int, input().strip().split(' '))

-
길이 len(str)
min(counts)
max(counts)
sum(counts)

-print
print(a,'+',b,'=',a+b)
print(str1, str2, sep='')
print(a, end='')

-대소문자 변환
x.isupper()
x.islower()
x.lower()
x.upper()

-공백 제거 
strip()

-반복문 for문 
for i in range(len(str)) : 
    if str[i].isupper() : 

for x in str :
    print(x)

for i in range(0,len(names),5):
        answer.append(names[i])

while answer<=n:
        answer+=numbers[i]
        i+=1

for i,q in enumerate(queries) :
        s=q[0]
        e=q[1]

for idx,x in enumerate(arr) : 
        if x>=50 and x%2==0 :
            arr[idx]=x//2


-조건문 if문
if ineq=='<':
    answer=1
elif ineq=='>':
    answer=1
return answer 

if 2 not in arr: #arr에 2가 없다면
    return -1 

if not finished[i] :
    answer.append(todo_list[i])

if 'latte' in want:

-return 
answer = a+b if flag else a-b

- 변환
x=int(str(a)+str(b))
x=str(code[i])

-연산
짝수even    a%2==0 
홀수odd     a%2==1
배수 num%n==0 
공배수 number%n==0 and number%m==0  #number가 n의 배수이면서 m의 배수 

-숫자
나머지 n%i==0
몫(소숫값) n=n/2
몫(정수) n=n//2
거듭제곱 n**2
정수화 int(a) 
정수화 a=a//1 

- 최대/최솟값
minValue = 2147000000
m=float('inf')

-리스트[]
answer=[]
answer=list()
answer.append(1)
answer.pop()
answer.pop(4)
answer.remove(x) #하나만 제거함

-set : 중복되지 않는 집합(set) 
answer=set()
차집합 set(str(num)) - set(['0','5']) 

-짧게
return [i for i in range(start,end+1)]
return [i for i in range(k,n+1,k)]
return [i for i in range(1, n+1) if i % k == 0]
return list(range(start,end-1,-1))
return [work for idx, work in enumerate(todo_list) if not finished[idx]]
return sum(num_list) if len(num_list)>=11 else prod(num_list)
return [word for word in strArr if 'ad' not in word]
return sorted(ch for ch in myString.split('x') if ch)
return sorted(num_list)[:5]
return [i*k if k%2!=0 else i+k for i in arr]

-논리연산
#논리합 OR ∨ , 논리곱 AND ∧ (True 1, False 0)
a=(True if x1==True or x2==True else False)
b=(True if x3==True or x4==True else False)
#return (x1 | x2) & (x3 | x4)
#return (x1 or x2) and (x3 or x4)

-정렬
-sort()/sorted()
sort() : answer.sort(), list.sort(reverse=True)
    리스트 원본값을 직접 수정함
sorted() : ret=sorted(list), ret=sorted(list,reverse=True)
    리스트 원본 값은 그대로이고 정렬 값을 반환함

-함수
count() : counts=[arr.count(i) for i in arr] #각 숫자의 출현 횟수를 저장
copy() : arr2.append(arr.copy()) #arr2.append(arr) 안됨. 
abs() : 절댓값

-find( )/index() <--> rindex(), rfind() : 뒤에서 부터 
find() : x.find("ad")
    #find(찾을 문자열), find(찾을 문자열, 시작 위치), find(찾을 문자열, 시작 위치, 종료 위치) 
    찾는 문자가 없는 경우에 -1을 출력
    문자열만 사용이 가능
index() : num_list.index(x) x를 찾은 가장 첫 번째 인덱스 
    찾는 문자가 없는 경우에 ValueError 에러가 발생
    문자열, 리스트, 튜플 자료형에서 사용 가능하고 딕셔너리 자료형에는 사용 불가 
myString.rfind(pat)

-문자열 관련
endsWith(),startswith():  my_string.endswith('abc')
    endsWith() 는 특정문자열로 어떤 string이 특정문자열로 끝나는지를 검사해서 그 값을 Boolean값으로 반환 
replace() : replace('a', 'A') , my_string.replace(alp,alp.upper()) 
split() : myString.split(), myString.split(''), myString.split('x') 
strip() 
    인자가 없을 경우 공백 제거 
    'abcdefg'.strip('a') : 인자로 전달된 문자를 String의 왼쪽과 오른쪽에서 제거합니다.
    'abcdefg'.lstrip('a') : 인자로 전달된 문자를 String의 왼쪽에서 제거합니다.
    'abcdefg'.rstrip('a') : 인자로 전달된 문자를 String의 오른쪽에서 제거합니다.

-아스키코드 변환
ord() : ord('A') 문자>숫자
chr() : chr(65) 숫자>문자

-슬라이싱
my_string[len(my_string)-n:]
my_string[-n:]
num_list[::n] #n의 간격 

- from math import prod 
prod() : num_list 원소값을 모두 곱한 값을 리턴
return sum(num_list) if len(num_list)>=11 else prod(num_list) 

-2차원 배열
answer=[[0]*n for i in range(n)]
answer=[[0]*n for _ in range(n)]
for s,e in intervals :
for a,b in intervals: answer+=arr[a:b+1]
return [work for idx, work in enumerate(todo_list) if not finished[idx]]
for i,q in enumerate(queries) :

-2차원 배열 방향 
answer=[[0]*n for _ in range(n)]
#ex 시계방향 : 우,하,좌,상
   dx=[0,1,0,-1]
   dy=[1,0,-1,0]

-변수 바꾸기 
a,b = b,a
arr[i],arr[j] = arr[j],arr[i]