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Number of substrings that start with “geeks” and end with “for”

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  • Difficulty Level : Easy
  • Last Updated : 31 May, 2021

Given a string str consisting of lowercase English alphabets, the task is to find the count of substrings that start with “geeks” and end with “for”.
Examples: 
 

Input: str = “geeksforgeeksisforgeeks” 
Output:
“geeksfor”, “geeksforgeeksisfor” and “geeksisfor” 
are the only valid substrings.
Input: str = “geeksforgeeks” 
Output:
 

 

Naive approach: First set the counter to 0 then iterate over the string and whenever the substring “geeks” is encountered, from the very next indices again iterate over the string and try to find the substring “for”. If “for” is present then increment the counter and finally print it.
Efficient approach: Set two counter for the substrings “geeks” and “for”, say c1 and c2. On iterating, whenever the substring “geeks” is encountered, increment c1 and whenever “for” is encountered, set c2 = c2 + c1. This is because every occurrence of “geeks” will make a valid substring with the current found “for”. Finally, print c2.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function to return the count
// of required substrings
int countSubStr(string s, int n)
{
    int c1 = 0, c2 = 0;
 
    // For every index of the string
    for (int i = 0; i < n; i++) {
 
        // If the substring starting at
        // the current index is "geeks"
        if (s.substr(i, 5) == "geeks")
            c1++;
 
        // If the substring is "for"
        if (s.substr(i, 3) == "for")
            c2 = c2 + c1;
    }
 
    return c2;
}
 
// Driver code
int main()
{
    string s = "geeksforgeeksisforgeeks";
    int n = s.size();
 
    cout << countSubStr(s, n);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
    // Function to return the count
    // of required substrings
    static int countSubStr(String s, int n)
    {
        int c1 = 0, c2 = 0;
 
        // For every index of the string
        for (int i = 0; i < n; i++)
        {
 
            // If the substring starting at
            // the current index is "geeks"
            if (i < n - 5 &&
                "geeks".equals(s.substring(i, i + 5)))
            {
                c1++;
            }
 
            // If the substring is "for"
            if (i < n - 3 &&
                "for".equals(s.substring(i, i + 3)))
            {
                c2 = c2 + c1;
            }
        }
        return c2;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s = "geeksforgeeksisforgeeks";
        int n = s.length();
        System.out.println(countSubStr(s, n));
    }
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation of the approach
 
# Function to return the count
# of required substrings
def countSubStr(s, n) :
 
    c1 = 0; c2 = 0;
 
    # For every index of the string
    for i in range(n) :
 
        # If the substring starting at
        # the current index is "geeks"
        if (s[i : i + 5] == "geeks") :
            c1 += 1;
 
        # If the substring is "for"
        if (s[i :i+ 3] == "for") :
            c2 = c2 + c1;
 
    return c2;
 
# Driver code
if __name__ == "__main__" :
 
    s = "geeksforgeeksisforgeeks";
    n = len(s);
 
    print(countSubStr(s, n));
     
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
     
public class GFG
{
  
    // Function to return the count
    // of required substrings
    static int countSubStr(String s, int n)
    {
        int c1 = 0, c2 = 0;
  
        // For every index of the string
        for (int i = 0; i < n; i++)
        {
  
            // If the substring starting at
            // the current index is "geeks"
            if (i < n - 5 &&
                "geeks".Equals(s.Substring(i, 5)))
            {
                c1++;
            }
  
            // If the substring is "for"
            if (i < n - 3 &&
                "for".Equals(s.Substring(i, 3)))
            {
                c2 = c2 + c1;
            }
        }
        return c2;
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        String s = "geeksforgeeksisforgeeks";
        int n = s.Length;
        Console.WriteLine(countSubStr(s, n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function to return the count
// of required substrings
function countSubStr(s, n)
{
    var c1 = 0, c2 = 0;
 
    // For every index of the string
    for (var i = 0; i < n; i++) {
 
        // If the substring starting at
        // the current index is "geeks"
        if (s.substring(i, i+5) == "geeks")
            c1++;
 
        // If the substring is "for"
        if (s.substring(i,i+ 3) == "for")
            c2 = c2 + c1;
    }
 
    return c2;
}
 
// Driver code
var s = "geeksforgeeksisforgeeks";
var n = s.length;
document.write( countSubStr(s, n));
 
</script>

Output: 

3

 


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