[프로그래머스] / [Level 1] / [Python] 옹알이 (2)

https://school.programmers.co.kr/learn/courses/30/lessons/133499

프로그래머스

---

 

def solution(babbling):
    answer = 0
    
    can = ["aya", "ye", "woo", "ma"]
    
    for idx in range(len(babbling)):
        can = ["aya", "ye", "woo", "ma"]
        previous = ""
        while True:
            for c in can:
                if babbling[idx][:len(c)] == c and previous != c:
                    babbling[idx] = babbling[idx][len(c):]
                    previous = c
                    break
            else:
                break
            
            if babbling[idx] == "":
                answer += 1
                break
    
    return answer

 

replace를 사용하여 조금 더 깔끔하게

def solution(babbling):
    can = ["aya", "ye", "woo", "ma"]
    for k, i in enumerate(babbling):
        for j in can:
            if j*2 in i:
                break
            babbling[k]=babbling[k].replace(j, " ")
        babbling[k]=babbling[k].replace(" ", "")
    return babbling.count("")