Given a string, count number of subsequences of the form a^{i}b^{j}c^{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 :abbcOutput :3 Subsequences are abc, abc and abbcInput :abcabcOutput :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 :

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;**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:

## C++

`// 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 the 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;` `}` |

## Java

`// Java program to count subsequences of the` `// form a^i b^j c^k` `public` `class` `No_of_subsequence {` ` ` `// Returns count of subsequences of the form` ` ` `// a^i b^j c^k` ` ` `static` `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` `(` `int` `i = ` `0` `; i < s.length(); 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.charAt(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.charAt(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.charAt(i) == ` `'c'` `)` ` ` `cCount = (bCount + ` `2` `* cCount);` ` ` `}` ` ` `return` `cCount;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `String s = ` `"abbc"` `;` ` ` `System.out.println(countSubsequences(s));` ` ` `}` `}` `// This code is contributed by Sumit Ghosh` |

## Python 3

`# Python 3 program to count` `# subsequences of the form` `# a ^ i b ^ j c ^ k` `# Returns count of subsequences` `# of the form a ^ i b ^ j c ^ k` `def` `countSubsequences(s):` ` ` `# Initialize counts of different` ` ` `# subsequences caused by different` ` ` `# combination of 'a'` ` ` `aCount ` `=` `0` ` ` `# Initialize counts of different` ` ` `# subsequences caused by different` ` ` `# combination of 'a' and different` ` ` `# combination of 'b'` ` ` `bCount ` `=` `0` ` ` `# Initialize counts of different` ` ` `# subsequences caused by different` ` ` `# combination of 'a', 'b' and 'c'.` ` ` `cCount ` `=` `0` ` ` `# Traverse all characters` ` ` `# of given string` ` ` `for` `i ` `in` `range` `(` `len` `(s)):` ` ` ` ` `# 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.` ` ` `elif` `(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.` ` ` `elif` `(s[i] ` `=` `=` `'c'` `):` ` ` `cCount ` `=` `(bCount ` `+` `2` `*` `cCount)` ` ` `return` `cCount` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `s ` `=` `"abbc"` ` ` `print` `(countSubsequences(s))` `# This code is contributed` `# by ChitraNayal` |

## C#

`// C# program to count subsequences` `// of the form a^i b^j c^k` `using` `System;` `public` `class` `GFG {` ` ` `// Returns count of subsequences` ` ` `// of the form a^i b^j c^k` ` ` `static` `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` `(` `int` `i = 0; i < s.Length; 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` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `String s = ` `"abbc"` `;` ` ` `Console.Write(countSubsequences(s));` ` ` `}` `}` `// This code is contributed by Nitin Mittal.` |

## PHP

`<?php` `// PHP program to count subsequences` `// of the form a^i b^j c^k` `// Returns count of subsequences` `// of the form a^i b^j c^k` `function` `countSubsequences(` `$s` `)` `{` ` ` ` ` `// Initialize counts of` ` ` `// different subsequences` ` ` `// caused by different` ` ` `// combination of 'a'` ` ` `$aCount` `= 0;` ` ` `// Initialize counts of` ` ` `// different subsequences` ` ` `// caused by different` ` ` `// combination of 'a' and` ` ` `// different combination of 'b'` ` ` `$bCount` `= 0;` ` ` `// Initialize counts of` ` ` `// different subsequences` ` ` `// caused by different` ` ` `// combination of 'a', 'b'` ` ` `// and 'c'.` ` ` `$cCount` `= 0;` ` ` `// Traverse all characters` ` ` `// of given string` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `strlen` `(` `$s` `); ` `$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. */` ` ` `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` `$cCount` `;` `}` ` ` `// Driver Code` ` ` `$s` `= ` `"abbc"` `;` ` ` `echo` `countSubsequences(` `$s` `) ;` `// This code is contributed by nitin mittal.` `?>` |

## Javascript

`<script>` `// JavaScript program for the above approach` ` ` `// Returns count of subsequences` ` ` `// of the form a^i b^j c^k` ` ` `function` `countSubsequences(s)` ` ` `{` ` ` `// Initialize counts of different` ` ` `// subsequences caused by different` ` ` `// combination of 'a'` ` ` `let aCount = 0;` ` ` ` ` `// Initialize counts of different` ` ` `// subsequences caused by different` ` ` `// combination of 'a' and` ` ` `// different combination of 'b'` ` ` `let bCount = 0;` ` ` ` ` `// Initialize counts of different` ` ` `// subsequences caused by different` ` ` `// combination of 'a', 'b' and 'c'` ` ` `let cCount = 0;` ` ` ` ` `// Traverse all characters of given string` ` ` `for` `(let i = 0; i < s.length; 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` ` ` ` ` `let s = ` `"abbc"` `;` ` ` `document.write(countSubsequences(s));` `</script>` |

**Output:**

3

**Complexity Analysis:**

**Time Complexity:**O(n).

One traversal of the string is needed.**Auxiliary Space:**O(1).

No extra space is needed.

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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