Count Distinct Subsequences

Given a string, find the count of distinct subsequences of it. 

Examples: 

Input: str = “gfg”
Output: 7
Explanation: The seven distinct subsequences are “”, “g”, “f”, “gf”, “fg”, “gg” and “gfg”

Input: str = “ggg”
Output: 4
Explanation: The four distinct subsequences are “”, “g”, “gg” and “ggg”

The problem of counting distinct subsequences is easy if all characters of input string are distinct. The count is equal to ⁿC₀ + ⁿC₁ + ⁿC₂ + … ⁿC_n = 2ⁿ.
How to count distinct subsequences when there can be repetition in input string? 
A Simple Solution to count distinct subsequences in a string with duplicates is to generate all subsequences. For every subsequence, store it in a hash table if it doesn’t exist already. The time complexity of this solution is exponential and it requires exponential extra space.

Method 1(Naive Approach): Using a set (without Dynamic Programming)

Approach: Generate all the possible subsequences of a given string. The subsequences of a string can be generated in the following manner: 

1. Include a particular element(say i^th) in the output array and recursively call the function for the rest of the input string. This results in the subsequences of a string having i^th character. 
2. Exclude a particular element(say i^th) and recursively call the function for the rest of the input string. This contains all the subsequences which don’t have the i^th character.
   Once we have generated a subsequence, in the base case of the function we insert that generated subsequence in an unordered set. An unordered Set is a Data structure, that stores distinct elements in an unordered manner. This way we insert all the generated subsequences in the set and print the size of the set as our answer because at last, the set will contain only distinct subsequences. 

Implementation:

4

Time Complexity: O(2^n)
Auxiliary Space: O(2^n)
where n is the length of the string.

Method 2(Efficient Approach): Using Dynamic Programming

An Efficient Solution doesn’t require the generation of subsequences.   

Let countSub(n) be count of subsequences of 
first n characters in input string. We can
recursively write it as below. 
countSub(n) = 2*Count(n-1) - Repetition
If current character, i.e., str[n-1] of str has
not appeared before, then 
   Repetition = 0
Else:
   Repetition  =  Count(m)
   Here m is index of previous occurrence of
   current character. We basically remove all
   counts ending with previous occurrence of
   current character.

How does this work? 
If there are no repetitions, then count becomes double of count for n-1 because we get count(n-1) more subsequences by adding current character at the end of all subsequences possible with n-1 length. 
If there are repetitions, then we find a count of all distinct subsequences ending with the previous occurrence. This count can be obtained by recursively calling for an index of the previous occurrence. 
Since the above recurrence has overlapping subproblems, we can solve it using Dynamic Programming. 

Below is the implementation of the above idea.  

7

Time Complexity: O(n) 
Auxiliary Space: O(n) 

Method 3: Using Map

Idea:

Let’s say we have 2 variables : `allCount` which adds up total distinct subsequence count and `levelCount` which stores the count of subsequences ending at index i. To find repetitions we will store the most recent levelCount for each character. Finally we will see how we can determine `allCount` using the `levelCount` variable.

Below is the steps to solve the problem:

• Declare a map and store all the characters by initializing -1.
• Start a loop to iterate through the characters of the input string s.
• Inside the loop, when i (the current index) is 0, this is the first character in the string.
  • Set allCount to 1 since the first character is always unique.
  • Update the map mp with the index 1 for the first character c.
• For characters at positions other than 0:
  • Calculate the current levelCount as allCount + 1, representing the number of unique substrings at the current level.
• • If mp is less than 0, it means the character is new (has not been seen before in this substring). In this case:
  • Increment allCount by levelCount to account for the new character.
  • If mp is not less than 0, it means the character has been seen before in this substring. In this case:
  • Adjust allCount by adding levelCount – mp to account for the fact that some substrings may have been counted already.
• Update the map mp with the current levelCount for the character c since this is the latest level of uniqueness.