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Data Science LAB

[Python] ν”„λ‘œκ·Έλž˜λ¨ΈμŠ€ μ½”λ”©ν…ŒμŠ€νŠΈ μ—°μŠ΅ level 2 (μ˜€ν”ˆ μ±„νŒ…λ°©) λ³Έλ¬Έ

πŸ“ Coding Test/Programmers

[Python] ν”„λ‘œκ·Έλž˜λ¨ΈμŠ€ μ½”λ”©ν…ŒμŠ€νŠΈ μ—°μŠ΅ level 2 (μ˜€ν”ˆ μ±„νŒ…λ°©)

γ…… γ…œ γ…” γ…‡ 2022. 10. 27. 19:26
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1. 문제 μ„€λͺ…

μΉ΄μΉ΄μ˜€ν†‘ μ˜€ν”ˆμ±„νŒ…λ°©μ—μ„œλŠ” μΉœκ΅¬κ°€ μ•„λ‹Œ μ‚¬λžŒλ“€κ³Ό λŒ€ν™”λ₯Ό ν•  수 μžˆλŠ”λ°, 본래 λ‹‰λ„€μž„μ΄ μ•„λ‹Œ κ°€μƒμ˜ λ‹‰λ„€μž„μ„ μ‚¬μš©ν•˜μ—¬ μ±„νŒ…λ°©μ— λ“€μ–΄κ°ˆ 수 μžˆλ‹€.

μ‹ μž…μ‚¬μ›μΈ κΉ€ν¬λ£¨λŠ” μΉ΄μΉ΄μ˜€ν†‘ μ˜€ν”ˆ μ±„νŒ…λ°©μ„ κ°œμ„€ν•œ μ‚¬λžŒμ„ μœ„ν•΄, λ‹€μ–‘ν•œ μ‚¬λžŒλ“€μ΄ λ“€μ–΄μ˜€κ³ , λ‚˜κ°€λŠ” 것을 μ§€μΌœλ³Ό 수 μžˆλŠ” κ΄€λ¦¬μžμ°½μ„ λ§Œλ“€κΈ°λ‘œ ν–ˆλ‹€. μ±„νŒ…λ°©μ— λˆ„κ΅°κ°€ λ“€μ–΄μ˜€λ©΄ λ‹€μŒ λ©”μ‹œμ§€κ°€ 좜λ ₯λœλ‹€.

"[λ‹‰λ„€μž„]λ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€."

μ±„νŒ…λ°©μ—μ„œ λˆ„κ΅°κ°€ λ‚˜κ°€λ©΄ λ‹€μŒ λ©”μ‹œμ§€κ°€ 좜λ ₯λœλ‹€.

"[λ‹‰λ„€μž„]λ‹˜μ΄ λ‚˜κ°”μŠ΅λ‹ˆλ‹€."

μ±„νŒ…λ°©μ—μ„œ λ‹‰λ„€μž„μ„ λ³€κ²½ν•˜λŠ” 방법은 λ‹€μŒκ³Ό 같이 두 가지이닀.

  • μ±„νŒ…λ°©μ„ λ‚˜κ°„ ν›„, μƒˆλ‘œμš΄ λ‹‰λ„€μž„μœΌλ‘œ λ‹€μ‹œ λ“€μ–΄κ°„λ‹€.
  • μ±„νŒ…λ°©μ—μ„œ λ‹‰λ„€μž„μ„ λ³€κ²½ν•œλ‹€.

λ‹‰λ„€μž„μ„ λ³€κ²½ν•  λ•ŒλŠ” 기쑴에 μ±„νŒ…λ°©μ— 좜λ ₯λ˜μ–΄ 있던 λ©”μ‹œμ§€μ˜ λ‹‰λ„€μž„λ„ μ „λΆ€ λ³€κ²½λœλ‹€.

예λ₯Ό λ“€μ–΄, μ±„νŒ…λ°©μ— "Muzi"와 "Prodo"λΌλŠ” λ‹‰λ„€μž„μ„ μ‚¬μš©ν•˜λŠ” μ‚¬λžŒμ΄ μˆœμ„œλŒ€λ‘œ λ“€μ–΄μ˜€λ©΄ μ±„νŒ…λ°©μ—λŠ” λ‹€μŒκ³Ό 같이 λ©”μ‹œμ§€κ°€ 좜λ ₯λœλ‹€.

"Muziλ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€."
"Prodoλ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€."

μ±„νŒ…λ°©μ— 있던 μ‚¬λžŒμ΄ λ‚˜κ°€λ©΄ μ±„νŒ…λ°©μ—λŠ” λ‹€μŒκ³Ό 같이 λ©”μ‹œμ§€κ°€ λ‚¨λŠ”λ‹€.

"Muziλ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€."
"Prodoλ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€."
"Muziλ‹˜μ΄ λ‚˜κ°”μŠ΅λ‹ˆλ‹€."

Muziκ°€ λ‚˜κ°„ν›„ λ‹€μ‹œ λ“€μ–΄μ˜¬ λ•Œ, Prodo λΌλŠ” λ‹‰λ„€μž„μœΌλ‘œ λ“€μ–΄μ˜¬ 경우 기쑴에 μ±„νŒ…λ°©μ— λ‚¨μ•„μžˆλ˜ Muzi도 Prodo둜 λ‹€μŒκ³Ό 같이 λ³€κ²½λœλ‹€.

"Prodoλ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€."
"Prodoλ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€."
"Prodoλ‹˜μ΄ λ‚˜κ°”μŠ΅λ‹ˆλ‹€."
"Prodoλ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€."

μ±„νŒ…λ°©μ€ 쀑볡 λ‹‰λ„€μž„μ„ ν—ˆμš©ν•˜κΈ° λ•Œλ¬Έμ—, ν˜„μž¬ μ±„νŒ…λ°©μ—λŠ” ProdoλΌλŠ” λ‹‰λ„€μž„μ„ μ‚¬μš©ν•˜λŠ” μ‚¬λžŒμ΄ 두 λͺ…이 μžˆλ‹€. 이제, μ±„νŒ…λ°©μ— 두 번째둜 λ“€μ–΄μ™”λ˜ Prodoκ°€ Ryan으둜 λ‹‰λ„€μž„μ„ λ³€κ²½ν•˜λ©΄ μ±„νŒ…λ°© λ©”μ‹œμ§€λŠ” λ‹€μŒκ³Ό 같이 λ³€κ²½λœλ‹€.

"Prodoλ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€."
"Ryanλ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€."
"Prodoλ‹˜μ΄ λ‚˜κ°”μŠ΅λ‹ˆλ‹€."
"Prodoλ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€."

μ±„νŒ…λ°©μ— λ“€μ–΄μ˜€κ³  λ‚˜κ°€κ±°λ‚˜, λ‹‰λ„€μž„μ„ λ³€κ²½ν•œ 기둝이 λ‹΄κΈ΄ λ¬Έμžμ—΄ λ°°μ—΄ recordκ°€ λ§€κ°œλ³€μˆ˜λ‘œ μ£Όμ–΄μ§ˆ λ•Œ, λͺ¨λ“  기둝이 처리된 ν›„, μ΅œμ’…μ μœΌλ‘œ 방을 κ°œμ„€ν•œ μ‚¬λžŒμ΄ 보게 λ˜λŠ” λ©”μ‹œμ§€λ₯Ό λ¬Έμžμ—΄ λ°°μ—΄ ν˜•νƒœλ‘œ return ν•˜λ„λ‘ solution ν•¨μˆ˜λ₯Ό μ™„μ„±ν•˜λΌ.

 

 

 

2. μ œν•œ 사항

  • recordλŠ” λ‹€μŒκ³Ό 같은 λ¬Έμžμ—΄μ΄ λ‹΄κΈ΄ 배열이며, κΈΈμ΄λŠ” 1 μ΄μƒ 100,000 μ΄ν•˜μ΄λ‹€.
  • λ‹€μŒμ€ record에 λ‹΄κΈ΄ λ¬Έμžμ—΄μ— λŒ€ν•œ μ„€λͺ…이닀.
    • λͺ¨λ“  μœ μ €λŠ” [μœ μ € 아이디]둜 κ΅¬λΆ„ν•œλ‹€.
    • [μœ μ € 아이디] μ‚¬μš©μžκ°€ [λ‹‰λ„€μž„]으둜 μ±„νŒ…λ°©μ— μž…μž₯ - "Enter [μœ μ € 아이디] [λ‹‰λ„€μž„]" (ex. "Enter uid1234 Muzi")
    • [μœ μ € 아이디] μ‚¬μš©μžκ°€ μ±„νŒ…λ°©μ—μ„œ 퇴μž₯ - "Leave [μœ μ € 아이디]" (ex. "Leave uid1234")
    • [μœ μ € 아이디] μ‚¬μš©μžκ°€ λ‹‰λ„€μž„μ„ [λ‹‰λ„€μž„]으둜 λ³€κ²½ - "Change [μœ μ € 아이디] [λ‹‰λ„€μž„]" (ex. "Change uid1234 Muzi")
    • 첫 λ‹¨μ–΄λŠ” Enter, Leave, Change 쀑 ν•˜λ‚˜μ΄λ‹€.
    • 각 λ‹¨μ–΄λŠ” 곡백으둜 κ΅¬λΆ„λ˜μ–΄ 있으며, μ•ŒνŒŒλ²³ λŒ€λ¬Έμž, μ†Œλ¬Έμž, 숫자둜만 μ΄λ£¨μ–΄μ Έμžˆλ‹€.
    • μœ μ € 아이디와 λ‹‰λ„€μž„μ€ μ•ŒνŒŒλ²³ λŒ€λ¬Έμž, μ†Œλ¬Έμžλ₯Ό κ΅¬λ³„ν•œλ‹€.
    • μœ μ € 아이디와 λ‹‰λ„€μž„μ˜ κΈΈμ΄λŠ” 1 μ΄μƒ 10 μ΄ν•˜μ΄λ‹€.
    • μ±„νŒ…λ°©μ—μ„œ λ‚˜κ°„ μœ μ €κ°€ λ‹‰λ„€μž„μ„ λ³€κ²½ν•˜λŠ” λ“± 잘λͺ» 된 μž…λ ₯은 주어지지 μ•ŠλŠ”λ‹€.

 

 

 

 

 

 

 

3. λ‚΄ 풀이

from collections import defaultdict
def solution(record):
    id_dict = defaultdict(list)
    new_record = []
    
    for id in record:
        ids = id.split(' ')
        new_record.append(ids)
        try:
            id_dict[ids[1]]=(ids[2])
        except:
            pass
    
    answer = []
    words = {'Enter':'λ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€.', 'Leave': 'λ‹˜μ΄ λ‚˜κ°”μŠ΅λ‹ˆλ‹€.'}
    for i in new_record:
        if i[0] == 'Change':
            continue
        else:
            answer.append(id_dict[i[1]]+words[i[0]])
    return answer

 

1. defaultdict(list)λ₯Ό μ΄μš©ν•˜μ—¬ default값이 list인 λ”•μ…”λ„ˆλ¦¬ μ„ μ–Έ

2. new_record 리슀트λ₯Ό μ„ μ–Έν•˜μ—¬ 곡백(' ')을 κΈ°μ€€μœΌλ‘œ record의 λ¬Έμžμ—΄ μš”μ†Œμ—μ„œ split된 리슀트λ₯Ό μΆ”κ°€

3. id별 κ°€μž₯ λ§ˆμ§€λ§‰μœΌλ‘œ μ‚¬μš©ν•œ λ‹‰λ„€μž„μ„ λ°›μ•„μ€Œ (forλ¬Έ μ‚¬μš©)

4. Leave의 경우, 곡백을 κΈ°μ€€μœΌλ‘œ λ¬Έμžμ—΄μ„ λΆ„ν• ν•˜λ©΄ index error λ°œμƒ -> try~except λ¬Έ μ‚¬μš©ν•˜μ—¬ ν•΄κ²°

5. for 문을 μ‚¬μš©ν•˜μ—¬ Change의 κ²½μš°λŠ” pass, Enter와 LeaveλŠ” 각각 상황에 맞좰 answer list에 μΆ”κ°€

 

 

 

 

4. λ‹€λ₯Έ μ‚¬λžŒμ˜ 풀이

def solution(record):
    answer = []
    namespace = {}
    printer = {'Enter':'λ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€.', 'Leave':'λ‹˜μ΄ λ‚˜κ°”μŠ΅λ‹ˆλ‹€.'}
    for r in record:
        rr = r.split(' ')
        if rr[0] in ['Enter', 'Change']:
            namespace[rr[1]] = rr[2]

    for r in record:
        if r.split(' ')[0] != 'Change':
            answer.append(namespace[r.split(' ')[1]] + printer[r.split(' ')[0]])

    return answer

 

 

 

 

def solution(record):
    user_id = {EC.split()[1]:EC.split()[-1] for EC in record if EC.startswith('Enter') or EC.startswith('Change')}
    return [f'{user_id[E_L.split()[1]]}λ‹˜μ΄ λ“€μ–΄μ™”μŠ΅λ‹ˆλ‹€.' if E_L.startswith('Enter') else f'{user_id[E_L.split()[1]]}λ‹˜μ΄ λ‚˜κ°”μŠ΅λ‹ˆλ‹€.' for E_L in record if not E_L.startswith('Change')]
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