Count of ‘GFG’ Subsequences in the given string

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 <bits/stdc++.h>
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

<script>


// Javascript Program to find the "GFG" subsequence in
// the given string
var MAX = 100;

// Print the count of "GFG" subsequence in the string
function countSubsequence( s, n)
{
    var cntG = 0, cntF = 0, result = 0, C=0;

    // Traversing the given string
    for (var 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;
        }
    }

    document.write( result );
}

// Driven Program
var s = "GFGFG";
var n = (s.length);
countSubsequence(s, n);

// This code is contributed by itsok.
</script>

Output

4

Complexity Analysis:

• Time Complexity : O(n) 
• Auxiliary Space: O(1).