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

Given a string, count number of subsequences of the form aⁱb^jc^k, 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

Approach:

We traverse given string. For every character encounter, we do the 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 :O(1). 
     • Current character begins a new subsequence. 
     • Current character is part of aCount subsequences. 
     • 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 : 
     • Current character begins a new subsequence of b's with aCount subsequences. 
     • Current character is part of bCount subsequences. 
     • 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 : 
     • Current character begins a new subsequence of c's with bCount subsequences. 
     • Current character is part of cCount subsequences. 
     • Current character is not part of cCount subsequences. 
   •     Therefore we do cCount = (bCount + 2 * cCount);
5. Finally we return cCount;

Explanation of approach with help of example:

• aCount is the number of subsequences of the letter 'a'.
• Consider this example: aa.
• We can see that aCount for this is 3, because we can choose these possibilities: (xa, ax, aa) (x means we did not use that character). Note also that this is independent of characters in between, i.e. the aCount of aa and ccbabbbcac are the same because both have exactly 2 a's.
• Now, adding 1 a, we now have the following new subsequences: each of the old subsequences, each of the old subsequences + the new a, and the new letter a, alone. So a total of aCount + aCount + 1 subsequences.
• Now, let's consider bCount, the number of subsequences with some a's and then some b's. in 'aab', we see that bCount should be 3 (axb, xab, aab) because it is just the number of ways we can choose subsequences of the first two a's, and then b. So every time we add a b, the number of ways increases by aCount.
• Let's find bCount for 'aabb'. We have already determined that aab has 3 subsequences, so certainly we still have those. Additionally, we can add the new b onto any of these subsequences, to get 3 more. Finally, we have to count the subsequences that are made without using any other b's, and by the logic in the last paragraph, that is just aCount. So, bCount after this is just the old bCount*2 + aCount;
• cCount is similar. 

Below is the implementation of above idea: 

3

Complexity Analysis:

• Time Complexity:O(n). 
  One traversal of the string is needed.
• Auxiliary Space:O(1). 
  No extra space is needed.

DP Based Approach:

Base Case: Set dp[0][0][0] = 1 as an empty string is also a valid subsequence.

DP Recurrence and Transition: Iterate through the string character by character. For each character s[idx]:

Update dp[i][j][k] based on the recurrence relation considering the current character:

If s[idx] == 'a', update dp[i+1][j][k] += dp[i][j][k] to account for 'a'.

If s[idx] == 'b', update dp[i][j+1][k] += dp[i][j][k] to account for 'b'.

If s[idx] == 'c', update dp[i][j][k+1] += dp[i][j][k] to account for 'c'.

Final Sub-problem: Finally, the count of subsequences of the form a^i b^j c^k where i >= 1, j >= 1, and k >= 1 will be stored in dp[n][n][n].

Complexity Analysis: 

Time Complexity: O(n). 

Auxiliary Space: O(n). 

This article is contributed by Mr. Somesh Awasthi.