 # Repeated String Frequency of letter a in repeated string
There is a string ,s, of lowercase English letters that is repeated infinitely many times. Given an integer ,n, find and print the number of letter a's in the first n letters of the infinite string.

Example
s="abcac"
n=10
The substring we consider is abcacabcac, the first 10 characters of the infinite string. There are 4 occurrences of a in the substring.

Function Description
Complete the repeatedString function in the editor below.
repeatedString has the following parameter(s):
• s: a string to repeat
• n: the number of characters to consider
Returns
• int: the frequency of a in the substring
Input Format
The first line contains a single string,s.
The second line contains an integer,n.

Constraints
1≤∣s∣≤10
1≤n≤10^12
For 25 % of the test cases,n≤10^6 .

Sample Input
Sample Input 0
``````aba
10
``````
Sample Output 0
``````7
``````
Explanation 0 The first letters of the infinite string are `abaabaabaa`. Because there are 7 a's, we return 7 .

Sample Input 1
``````a
1000000000000
``````
Sample Output 1
``````1000000000000
``````
Explanation 1
Because all of the first n=`1000000000000`. letters of the infinite string are a, we return `1000000000000`.
Output :
Using Python 3 -
```				```
#!/bin/python3

import math
import os
import random
import re
import sys

#
# Complete the 'repeatedString' function below.
#
# The function is expected to return a LONG_INTEGER.
# The function accepts following parameters:
#  1. STRING s
#  2. LONG_INTEGER n
#

def repeatedString(s, n):
# below code is what you need
r = 0
l = len(s)
for i in range(0, l):
if s[i] == 'a':
r+= 1
r*= int(n / l)
for i in range(0, n % l):
if s[i] == 'a':
r+= 1
return r
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')

s = input()

n = int(input().strip())

result = repeatedString(s, n)

fptr.write(str(result) + '\n')

fptr.close()

```
```
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