Related Articles

# Count occurrences of a character in a repeated string

• Difficulty Level : Medium
• Last Updated : 23 Aug, 2021

Given an integer N and a lowercase string. The string is repeated infinitely. The task is to find the No. of occurrences of a given character x in first N letters.
Examples:

```Input : N = 10 str = "abcac"
Output : 4
Explanation: "abcacabcac" is the substring from the infinitely repeated string. In first 10 letters 'a' occurs 4  times.

Input: N = 10, str = "aba"
Output : 7```

Approach:
1. Find the occurrences of character ‘a’ in the given string.
2. Find the No. of repetitions which are required to find the ‘a’ occurrences.
3. Multiply the single string occurrences to the No. of repetitions.
4. If given n is not the multiple of given string size then we will find the ‘a’ occurrences in the remaining substring.
Below is the implementation of above approach:

## C++

 `// CPP program to find the occurrences of``// character x in the infinite repeated string``// upto length n``#include ``using` `namespace` `std;` `// Function to count the character 'a'``int` `countChar(string str, ``char` `x)``{``    ``int` `count = 0, n = 10;``    ``for` `(``int` `i = 0; i < str.size(); i++)``        ``if` `(str[i] == x)``            ``count++;` `    ``// atleast k repetition are required``    ``int` `repetitions = n / str.size();``    ``count = count * repetitions;` `    ``// if n is not the multiple of the string size``    ``// check for the remaining repeating character.``    ``for` `(``int` `i = 0; i < n % str.size(); i++) {``        ``if` `(str[i] == x)``            ``count++;``    ``}` `    ``return` `count;``}` `// Driver code``int` `main()``{``    ``string str = ``"abcac"``;``    ``cout << countChar(str, ``'a'``);``    ``return` `0;``}` `// This code is contributed by Surendra_Gangwar`

## Java

 `// Java program to find the occurrences``// of character x in the infinite``// repeated string upto length n``import` `java.util.*;``import` `java.lang.*;` `class` `GFG``{``// Function to count the character 'a'``static` `int` `countChar(String str, ``char` `x)``{``    ``int` `count = ``0``;``    ``int` `n = ``10``;``    ``for` `(``int` `i = ``0``; i < str.length(); i++)``        ``if` `(str.charAt(i) == x)``            ``count++;` `    ``// atleast k repetition are required``    ``int` `repetitions = n / str.length();``    ``count = count * repetitions;` `    ``// if n is not the multiple of the``    ``// string size check for the remaining``    ``// repeating character.``    ``for` `(``int` `i = ``0``;``            ``i < n % str.length(); i++)``    ``{``        ``if` `(str.charAt(i) == x)``            ``count++;``    ``}` `    ``return` `count;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``String str = ``"abcac"``;``    ``System.out.println(countChar(str, ``'a'``));``}``}` `// This code is contributed``// by Akanksha Rai`

## Python3

 `# Python3 program to find the occurrences of``# character x in the infinite repeated string``# upto length n` `# Function to count the character 'a'``def` `countChar(``str``, x):``    ``count ``=` `0``    ``for` `i ``in` `range``(``len``(``str``)):``        ``if` `(``str``[i] ``=``=` `x) :``            ``count ``+``=` `1``    ``n ``=` `10``    ` `    ``# atleast k repetition are required``    ``repetitions ``=` `n ``/``/` `len``(``str``)``    ``count ``=` `count ``*` `repetitions` `    ``# if n is not the multiple of the``    ``# string size check for the remaining``    ``# repeating character.``    ``l ``=` `n ``%` `len``(``str``)``    ``for` `i ``in` `range``(l):``        ``if` `(``str``[i] ``=``=` `x):``            ``count ``+``=` `1``    ``return` `count` `# Driver code``str` `=` `"abcac"``print``(countChar(``str``, ``'a'``))` `# This code is contributed``# by sahishelangia`

## C#

 `// C# program to find the occurrences``// of character x in the infinite``// repeated string upto length n``using` `System;` `class` `GFG``{``// Function to count the character 'a'``static` `int` `countChar(``string` `str, ``char` `x)``{``    ``int` `count = 0;``    ``int` `n = 10;``    ``for` `(``int` `i = 0; i < str.Length; i++)``        ``if` `(str[i] == x)``            ``count++;` `    ``// atleast k repetition are required``    ``int` `repetitions = n / str.Length;``    ``count = count * repetitions;` `    ``// if n is not the multiple of the``    ``// string size check for the remaining``    ``// repeating character.``    ``for` `(``int` `i = 0;``             ``i < n % str.Length; i++)``    ``{``        ``if` `(str[i] == x)``            ``count++;``    ``}` `    ``return` `count;``}` `// Driver code``public` `static` `void` `Main()``{``    ``string` `str = ``"abcac"``;``    ``Console.WriteLine(countChar(str, ``'a'``));``}``}` `// This code is contributed``// by Akanksha Rai`

## PHP

 ``

## Javascript

 ``

Output:

`4`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up