Level 4 무지의 먹방 라이브 <2019 KAKAO BLIND RECRUITMENT> Python 3

https://programmers.co.kr/learn/courses/30/lessons/42891?language=python3 

코딩테스트 연습 - 무지의 먹방 라이브

 

최종 코드

• GitHub -> https://github.com/hwayeon351/Programmers-Algorithms/blob/main/level4_무지의먹방라이브.py

GitHub - hwayeon351/Programmers-Algorithms: 프로그래머스 알고리즘 소스 코드 모음

from heapq import heappush, heappop
def solution(food_times, k):
    if sum(food_times) <= k: return -1
    hp = []
    for i in range(1, len(food_times)+1):
        heappush(hp,[food_times[i-1], i])
    eat = 0
    length = len(food_times)
    prev_eat = 0
    while True:
        if eat + (hp[0][0] - prev_eat)*length >= k: break
        now_eat = heappop(hp)[0]
        eat += (now_eat - prev_eat)*length
        length -= 1
        prev_eat = now_eat
    hp.sort(key = lambda x:x[1])
    k -= eat
    return hp[k%length][1]

 

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풀이 과정

from heapq import heappush, heappop
def solution(food_times, k):
    if sum(food_times) <= k: return -1
    hp = []
    for i in range(1, len(food_times)+1):
        heappush(hp,[food_times[i-1], i])
    eat = 0
    length = len(food_times)
    prev_eat = 0
    while True:
        if eat + (hp[0][0] - prev_eat)*length >= k: break
        now_eat = heappop(hp)[0]
        eat += (now_eat - prev_eat)*length
        length -= 1
        prev_eat = now_eat
    hp.sort(key = lambda x:x[1])
    k -= eat
    return hp[k%length][1]

---

오답 노트

• 정확성만 통과한 코드

#정확성 통과 코드
def solution(food_times, k):
    answer = 0
    if k >= sum(food_times): return -1
    while k > -1:
        for i in range(len(food_times)):
            if food_times[i] > 0:
                food_times[i] -= 1
                k -= 1
                if k == -1: return i+1
    return answer

정확성만 통과한 코드 결과

 

1시간 23분동안 작성한 코드

#무지의 먹방 라이브 실패코드
def solution(food_times, k):
    answer = 0
    if k >= sum(food_times): return -1
    remain = []
    done = 0
    while k > len(food_times)-done:
        remain.clear()
        eat1 = k//(len(food_times)-done)
        eat2 = k%(len(food_times)-done)
        for i in range(len(food_times)):
            if food_times[i] > eat1: 
                food_times[i] -= eat1
                remain.append(i)
            else: 
                eat2 += abs(food_times[i]-eat1)
                food_times[i] = 0
                done += 1
        k = eat2
    if k == 0:
        return remain[0]+1
    return remain[k%len(remain)]+1

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참고

• 1. 오픈채팅방 ->2021.09.02 - [코테 노트/프로그래머스] - Level 2 오픈채팅방 <2019 KAKAO BLIND RECRUITMENT> Python 3

• 2. 실패율 ->2021.09.02 - [코테 노트/프로그래머스] - Level 1 실패율 <2019 KAKAO BLIND RECRUITMENT> Python 3

• 3. 후보키 -> 2021.09.02 - [코테 노트/프로그래머스] - Level 2 후보키 <2019 KAKAO BLIND RECRUITMENT> Python 3

• 5. 길 찾기 게임 ->2021.09.02 - [코테 노트/프로그래머스] - Level 3 길 찾기 게임 <2019 KAKAO BLIND RECRUITMENT> Python 3

• 6. 매칭 점수 -> 2022.04.01 - [코테 노트/프로그래머스] - Level 3 매칭 점수 <2019 KAKAO BLIND RECRUITMENT> Python 3

Level 3 매칭 점수 <2019 KAKAO BLIND RECRUITMENT> Python 3

 

• 7. 블록 게임 ->