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Count of ‘GFG’ Subsequences in the given string
• Difficulty Level : Medium
• Last Updated : 05 May, 2021

Given a string of length n of capital letters. The task is to find the count of ‘GFG’ subsequence in the given string.
Examples:

```Input : str[] = "GFGFG"
Output : 4
GFGFG, GFGFG, GFGFG, GFGFG

Input : str[] = "ABCFGFPG"
Output : 1```

To find the number of “GFG” subsequences in the given string, observe for each ‘F’ if we know number of ‘G’ before and after it. Then the number of “GFG” subsequence for that ‘F’ is equal to product of number of ‘G’ before and after that ‘F’.
So, the idea is to maintain an array, arr[], where arr[i] store number of ‘G’ before index i, if ith character of the string is ‘F’ and number of ‘F’ before index i, if the ith character is ‘G’.
Also, we will calculate and store the number of “GFG” subsequence in result whenever we encounter ‘G’.
Below is the implementation of this approach:

## C++

 `// CPP Program to find the "GFG" subsequence in``// the given string``#include ``using` `namespace` `std;``#define MAX 100` `// Print the count of "GFG" subsequence in the string``void` `countSubsequence(``char` `s[], ``int` `n)``{``    ``int` `cntG = 0, cntF = 0, result = 0, C=0;` `    ``// Traversing the given string``    ``for` `(``int` `i = 0; i < n; i++) {``        ``switch` `(s[i]) {` `        ``// If the character is 'G', increment``        ``// the count of 'G', increase the result``        ``// and update the array.``        ``case` `'G'``:``            ``cntG++;``            ``result+=C;``            ``break``;` `        ``// If the character is 'F', increment``        ``// the count of 'F' and update the array.``        ``case` `'F'``:``            ``cntF++;``            ``C+=cntG;``            ``break``;` `        ``// Ignore other character.``        ``default``:``            ``continue``;``        ``}``    ``}` `    ``cout << result << endl;``}` `// Driven Program``int` `main()``{``    ``char` `s[] = ``"GFGFG"``;``    ``int` `n = ``strlen``(s);``    ``countSubsequence(s, n);``    ``return` `0;``}`

## Java

 `// Java Program to find the "GFG" subsequence``// in the given string` `public` `class` `GFG {` `    ``static` `int` `max = ``100``;``        ` `    ``// Print the count of "GFG" subsequence``    ``// in the string``    ``static` `void` `countSubsequence(String s, ``int` `n)``    ``{``        ``int` `cntG = ``0``, cntF = ``0``, result = ``0``, C=``0``;``    ` `        ``// Traversing the given string``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``switch` `(s.charAt(i)) {``    ` `            ``// If the character is 'G',``            ``// increment the count of 'G',``            ``// increase the result and``            ``// update the array.``            ``case` `'G'``:``                ``cntG++;``                ``result+=C;``                ``break``;``    ` `            ``// If the character is 'F',``            ``// increment the count of 'F'``            ``// and update the array.``            ``case` `'F'``:``                ``cntF++;``                ``C+=cntG;``                ``break``;``    ` `            ``// Ignore other character.``            ``default``:``                ``continue``;``            ``}``        ``}``    ` `        ``System.out.println(result);``    ``}``    ` `    ``// Driver code   ``    ``public` `static` `void` `main(String args[]) {``        ``String s= ``"GFGFG"``;``        ``int` `n = s.length();``        ``countSubsequence(s, n);``    ``}``}` `// This code is contributed by Sam007`

## Python3

 `# Python 3 Program to find the "GFG"``# subsequence in the given string``MAX` `=` `100` `# Print the count of "GFG" subsequence``# in the string``def` `countSubsequence(s, n):``    ``cntG ``=` `0``    ``cntF ``=` `0``    ``result ``=` `0``    ``C``=``0`  `    ``# Traversing the given string``    ``for` `i ``in` `range``(n):``        ``if` `(s[i] ``=``=` `'G'``):``            ` `            ``# If the character is 'G', increment``            ``# the count of 'G', increase the result``            ``# and update the array.``            ``cntG ``+``=` `1``            ``result ``+``=` `C``            ``continue` `        ``# If the character is 'F', increment``        ``# the count of 'F' and update the array.``        ``if` `(s[i] ``=``=` `'F'``):``            ``cntF ``+``=` `1``            ``C ``+``=` `cntG``            ``continue` `        ``# Ignore other character.``        ``else``:``            ``continue``    ` `    ``print``(result)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``s ``=` `"GFGFG"``    ``n ``=` `len``(s)``    ``countSubsequence(s, n)``    ` `# This code is contributed by``# Sanjit_Prasad`

## C#

 `// C# Program to find the "GFG" subsequence``// in the given string``using` `System;` `class` `GFG {``        ` `    ``// Print the count of "GFG" subsequence``    ``// in the string``    ``static` `void` `countSubsequence(``string` `s,``                                     ``int` `n)``    ``{``        ``int` `cntG = 0, cntF = 0, result = 0, C=0;``    ` `        ``// Traversing the given string``        ``for` `(``int` `i = 0; i < n; i++) {``            ``switch` `(s[i]) {``    ` `            ``// If the character is 'G',``            ``// increment the count of 'G',``            ``// increase the result and``            ``// update the array.``            ``case` `'G'``:``                ``cntG++;``                ``result += C;``                ``break``;``    ` `            ``// If the character is 'F',``            ``// increment the count of 'F'``            ``// and update the array.``            ``case` `'F'``:``                ``cntF++;``                ``C+=cntG;``                ``break``;``    ` `            ``// Ignore other character.``            ``default``:``                ``continue``;``            ``}``        ``}``    ` `        ``Console.WriteLine(result);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``string` `s= ``"GFGFG"``;``        ``int` `n = s.Length;``        ``countSubsequence(s, n);``    ``}``}` `// This code is contributed by Sam007.`

## Javascript

 ``
Output:
`4`

Time Complexity : O(n)

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