Given a string, find the 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 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 ^{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; ` `} ` |

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## 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 ` |

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Output:

7

Time Complexity : O(n)

Auxiliary Space : O(n)

This article is contributed by **Shival Agrawal**. 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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