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  • + 0 comments
    def repeatedString(s: str, n: int) -> int:
        return s.count('a') * (n // len(s)) + s[:n % len(s)].count('a')
    
  • + 0 comments

    Python solution

    def repeatedString(s, n):
        a_count = 0
        for c in s:
            if c=="a":
                a_count+=1
        
        repeated = a_count*(n//len(s))
        mod = n%len(s)
        for c in s[:mod]:
            if c=="a":
                repeated+=1
        return repeated
    
  • + 0 comments

    my python solution (still learning python)

    def repeatedString(s, n):
        # Write your code here
        s_len = len(s)
        fits = n // s_len
        missing = n - s_len * fits
        missing_part = s[:missing]
        
        result = s.count("a") * fits + missing_part.count("a")
        return result
    
  • + 0 comments

    python! solution

    def repeatedString(s, n):
        return n//len(s) * (s.count('a')) + s.count('a', 0, n%len(s))
    
  • + 0 comments

    Here is my python solution

    def check(arr,index):
        sub_arr=''
        sac=0
        for i in range(index):
            sub_arr=sub_arr+arr[i]
        for j in sub_arr:
            if j=="a":
                sac+=1
        return sac
    ac=0
    arr=input()
    n=int(input())
    for i in arr:
        if i=='a':
            ac+=1
    if(len(arr)-ac==0):
        print(n)
    else:
        if(len(arr)==n):
            print(ac)
        elif(len(arr)<n):
            q,r=n//len(arr),n%len(arr)
            if r==0:
                print(q*ac)
            else:
                sac=check(arr,r)
                print((q*ac)+sac)
        elif(len(arr)>n):
            sac=check(arr,n)
            print(sac)