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

Examples:

Input : str = "gfg" Output : 7 The seven distinct subsequences are "", "g", "f", "gf", "fg", "gg" and "gfg" Input : str = "ggg" Output : 4 The six 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 ^{n}C_{0} + ^{n}C_{1} + ^{n}C_{2} + … ^{n}C_{n} = 2^{n}.

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. Time complexity of this solution is exponential and it requires exponential extra space.

An Efficient Solution doesn’t require 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) - RepetitionIfcurrent character, i.e., str[n-1] of str has not appeared before, then Repetition = 0Else:Repetition = Count(m) Heremis 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 repetitions, then we find count of all distinct subsequences ending with previous occurrence. This count can be obtained be recursively calling for index of previous occurrence.

Since above recurrence has overlapping subproblems, we can solve it using Dynamic Programming.

Below is C++ implementation of above idea.

## C++

// C++ program to count number of distinct // subsequences of a given string. #include <bits/stdc++.h> using namespace std; const int MAX_CHAR = 256; // Returns count of distinct sunsequences of str. int countSub(string str) { // Create an array to store index // of last vector<int> last(MAX_CHAR, -1); // Length of input string int n = str.length(); // dp[i] is going to store count of distinct // subsequences of length i. int dp[n+1]; // Empty substring has only one subsequence dp[0] = 1; // Traverse through all lengths from 1 to n. for (int i=1; i<=n; i++) { // Number of subsequences with substring // str[0..i-1] dp[i] = 2*dp[i-1]; // If current character has appeared // before, then remove all subsequences // ending with previous occurrence. if (last[str[i-1]] != -1) dp[i] = dp[i] - dp[last[str[i-1]]]; // Mark occurrence of current character last[str[i-1]] = (i-1); } return dp[n]; } // Driver code int main() { cout << countSub("gfg"); return 0; }

## Java

// Java program to count number of distinct // subsequences of a given string. import java.util.ArrayList; import java.util.Arrays; public class Count_Subsequences { static final int MAX_CHAR = 256; // Returns count of distinct sunsequences of str. static int countSub(String str) { // Create an array to store index // of last int[] last = new int[MAX_CHAR]; Arrays.fill(last, -1); // Length of input string int n = str.length(); // dp[i] is going to store count of distinct // subsequences of length i. int[] dp = new int[n+1]; // Empty substring has only one subsequence dp[0] = 1; // Traverse through all lengths from 1 to n. for (int i=1; i<=n; i++) { // Number of subsequences with substring // str[0..i-1] dp[i] = 2*dp[i-1]; // If current character has appeared // before, then remove all subsequences // ending with previous occurrence. if (last[(int)str.charAt(i-1)] != -1) dp[i] = dp[i] - dp[last[(int)str.charAt(i-1)]]; // Mark occurrence of current character last[(int)str.charAt(i-1)] = (i-1); } return dp[n]; } // Driver code public static void main(String args[]) { System.out.println(countSub("gfg")); } } // This code is contributed by Sumit Ghosh

Output:

7

Time Complexity : O(n)

Auxiliary Space : O(n)

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