Number of subsequences of the form a^i b^j c^k

Given a string, count number of subsequences of the form aibjck, i.e., it consists of i ’a’ characters, followed by j ’b’ characters, followed by k ’c’ characters where i >= 1, j >=1 and k >= 1.

Note: Two subsequences are considered different if the set of array indexes picked for the 2 subsequences are different.

Expected Time Complexity : O(n)

Examples:

Input  : abbc
Output : 3
Subsequences are abc, abc and abbc

Input  : abcabc
Output : 7
Subsequences are abc, abc, abbc, aabc
abcc, abc and abc

Asked in : Amazon

We traverse given string. For every character encounter, we do following:

1) Initialize counts of different subsequences caused by different combination of ‘a’. Let this count be aCount.

2) Initialize counts of different subsequences caused by different combination of ‘b’. Let this count be bCount.

3) Initialize counts of different subsequences caused by different combination of ‘c’. Let this count be cCount.

4) Traverse all characters of given string. Do following for current character s[i]
    If current character is ‘a’, then there are following possibilities :
    a) Current character begins a new subsequence.
    b) Current character is part of aCount subsequences.
    c) Current character is not part of aCount subsequences.
    Therefore we do aCount = (1 + 2 * aCount);

    If current character is ‘b’, then there are following possibilities :
    a) Current character begins a new subsequence of b’s with aCount subsequences.
    b) Current character is part of bCount subsequences.
    c) Current character is not part of bCount subsequences.
    Therefore we do bCount = (aCount + 2 * bCount);

    If current character is ‘c’, then there are following possibilities :
    a) Current character begins a new subsequence of c’s with bCount subsequences.
    b) Current character is part of cCount subsequences.
    c) Current character is not part of cCount subsequences.
    Therefore we do cCount = (bCount + 2 * cCount);

5) Finally we return cCount;

Below is C++ implementation of the idea.

// C++ program to count subsequences of the
// form a^i b^j c^k
#include <bits/stdc++.h>
using namespace std;

// Returns count of subsequences of the form
// a^i b^j c^k
int countSubsequences(string s)
{
    // Initialize counts of different subsequences
    // caused by different combination of 'a'
    int aCount = 0;

    // Initialize counts of different subsequences
    // caused by different combination of 'a' and
    // different combination of 'b'
    int bCount = 0;

    // Initialize counts of different subsequences
    // caused by different combination of 'a', 'b'
    // and 'c'.
    int cCount = 0;

    // Traverse all characters of given string
    for (unsigned int i=0; i<s.size(); i++)
    {
        /* If current character is 'a', then
           there are following possibilities :
             a) Current character begins a new
                subsequence.
             b) Current character is part of aCount
                subsequences.
             c) Current character is not part of
                aCount subsequences. */
        if (s[i] == 'a')
            aCount = (1 + 2 * aCount);

        /* If current character is 'b', then
           there are following possibilities :
             a) Current character begins a new
                subsequence of b's with aCount
                subsequences.
             b) Current character is part of bCount
                subsequences.
             c) Current character is not part of
                bCount subsequences. */
        else if (s[i] == 'b')
            bCount = (aCount + 2 * bCount);

        /* If current character is 'c', then
           there are following possibilities :
             a) Current character begins a new
                subsequence of c's with bCount
                subsequences.
             b) Current character is part of cCount
                subsequences.
             c) Current character is not part of
                cCount subsequences. */
        else if (s[i] == 'c')
            cCount = (bCount + 2 * cCount);
    }

    return cCount;
}

// Driver code
int main()
{
    string s = "abbc";
    cout << countSubsequences(s) << endl;
    return 0;
}

Output:

  3

Time Complexity : O(n)

This article is contributed by Mr. Somesh Awasthi. 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.

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