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Count occurrences of a substring recursively

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  • Difficulty Level : Easy
  • Last Updated : 26 Aug, 2022
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Given two strings str1 and str2, the task is to count the number of times str2 occurs in str1 using recursion.

Examples: 

Input: str1 = “geeksforgeeks”, str2 = “geek”
Output: 2
Explanation: The occurrences of str2 in str1 are starting at index {0, 8}

Input: str1 = “aaaaa”, str2 = “aaa”
Output: 3
Explanation: The occurrences of str2 in str1 are starting at index {0,1,2

Approach: We have already discussed other approaches in our previous article but here we are going to solve this problem using recursion.

Algorithm: 

  • If size of string str2 is greater then string str1 or size of string str1 is 0 then, return 0.
  • Otherwise, Check if string str2 is present in str1 as substring or not.
    • if present then, increment the count of occurrence and recursively call for other substring.
    • else, recursively call for other substring.
    • return count from every recursively call as answer.

Below is the implementation of the above approach:

C++




// Recursive C++ program to count the number of
// times string str2 occurs in string str1
 
#include <iostream>
#include <string>
using namespace std;
 
// Function to count the number of
// times string str2 occurs in string str1
int countSubstring(string str1, string str2)
{
    int n1 = str1.length();
    int n2 = str2.length();
 
    // Base Case
    if (n1 == 0 || n1 < n2)
        return 0;
 
    // Recursive Case
    // Checking if the first substring matches
    if (str1.substr(0, n2).compare(str2) == 0)
        return countSubstring(str1.substr(1), str2) + 1;
 
    // Otherwise, return the count from
    // the remaining index
    return countSubstring(str1.substr(1), str2);
}
 
 
// Driver function
int main()
{
    string str1 = "geeksforgeeks", str2 = "geeks";
    cout << countSubstring(str1, str2) << endl;
   
      str1 = "aaaaa", str2 = "aaa";
    cout << countSubstring(str1, str2) << endl;
 
    return 0;
}

Java




// Recursive Java program for
// counting number of substrings
class GFG
{
 
// Recursive function to
// count the number of
// occurrences of "hi" in str.
static int countSubstring(String str1,
                         String str2)
{
    int n1 = str1.length();
    int n2 = str2.length();
 
    // Base Case
    if (n1 == 0 || n1 < n2)
        return 0;
 
    // Recursive Case
    // Checking if the first
    // substring matches
    if (str1.substring(0, n2).equals(str2))
        return countSubstring(str1.substring(1),
                                            str2) + 1;
 
    // Otherwise, return the count
    // from the remaining index
    return countSubstring(str1.substring(1),
                                        str2);
}
 
// Driver Code
public static void main(String args[])
{
    String str1 = "geeksforgeeks",
           str2 = "geeks";
    System.out.println(countSubstring(str1,
                                     str2));
 
    str1 = "aaaaa";
    str2 = "aaa";
    System.out.println(countSubstring(str1,
                                     str2));
 
}
}
 
// This code is contributed
// by Arnab Kundu

Python3




# Recursive Python3 program for
# counting number of substrings
 
# Recursive function to
# count the number of
# occurrences of "hi" in str.
def countSubstring(str1, str2):
     
    n1 = len(str1);
    n2 = len(str2);
     
    # Base Case
    if (n1 == 0 or n1 < n2):
        return 0;
 
    # Recursive Case
    # Checking if the first
    # substring matches
    if (str1[0 : n2] == str2):
        return countSubstring(str1[1:],
                             str2) + 1;
 
    # Otherwise, return the count
    # from the remaining index
    return countSubstring(str1[1:],
                         str2);
 
# Driver Code
if __name__ == '__main__':
     
    str1 = "geeksforgeeks";
    str2 = "geeks";
    print(countSubstring(str1, str2));
 
    str1 = "aaaaa";
    str2 = "aaa";
    print(countSubstring(str1, str2));
 
# This code is contributed by Princi Singh

C#




// Recursive C# program for
// counting number of substrings
using System;
class GFG
{
 
// Recursive function to
// count the number of
// occurrences of "hi" in str.
static int countSubstring(String str1,
                         String str2)
{
    int n1 = str1.Length;
    int n2 = str2.Length;
 
    // Base Case
    if (n1 == 0 || n1 < n2)
        return 0;
 
    // Recursive Case
    // Checking if the first
    // substring matches
    if (str1.Substring(0, n2).Equals(str2))
        return countSubstring(str1.Substring(1),
                                            str2) + 1;
 
    // Otherwise, return the
    // count from the remaining
    // index
    return countSubstring(str1.Substring(1),
                                        str2);
}
 
// Driver Code
public static void Main()
{
    string str1 = "geeksforgeeks",
           str2 = "geeks";
    Console.Write(countSubstring(str1,
                                str2));
    Console.Write("\n");
     
    str1 = "aaaaa";
    str2 = "aaa";
    Console.Write(countSubstring(str1,
                                str2));
 
}
}
 
// This code is contributed
// by Smita

Javascript




<script>
 
// Recursive js program for counting number of substrings
 
// Recursive function to count
// the number of occurrences of "hi" in str.
function countSubstring( str1, str2){
    let n1 = str1.length;
    let n2 = str2.length;
 
    // Base Case
    if (n1 == 0 || n1 < n2)
        return 0;
 
    // Recursive Case
    // Checking if the first substring matches
    if (str1.substr(0, n2) == (str2))
        return countSubstring(str1.substr(1), str2) + 1;
 
    // Otherwise, return the count from
    // the remaining index
    return countSubstring(str1.substr(1), str2);
}
 
// Driver function
let str1 = "geeksforgeeks", str2 = "geeks";
document.write( countSubstring(str1, str2),'<br>');
str1 = "aaaaa", str2 = "aaa";
document.write( countSubstring(str1, str2),'<br>');
 
 
</script>

Output

2
3

Complexity Analysis:

  • Time Complexity: O(N2) , (i.e. N is maximum length between str1 and str2)
  • Auxiliary Space: O(1)

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