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

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: 3
“geeksfor”, “geeksforgeeksisfor” and “geeksisfor”
are the only valid substrings.

Input: str = “geeksforgeeks”
Output: 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 `

Output:

```3
```

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : 29AjayKumar, AnkitRai01