Count Distinct Subsequences

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 nC0 + nC1 + nC2 + … nCn = 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: 
a) Include a particular element(say ith) 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 ith character. 
b) Exclude a particular element(say ith) and recursively call the function for the rest of the input string. This contains all the subsequences which don’t have the ith 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. 
 

C++

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// C++ program to print distinct
// subsequences of a given string
#include <bits/stdc++.h>
using namespace std;
 
// Create an empty set to store the subsequences
unordered_set<string> sn;
 
// Function for generating the subsequences
void subsequences(char s[], char op[], int i, int j)
{
 
    // Base Case
    if (s[i] == '\0') {
        op[j] = '\0';
 
        // Insert each generated
        // subsequence into the set
        sn.insert(op);
        return;
    }
 
    // Recursive Case
    else {
        // When a particular character is taken
        op[j] = s[i];
        subsequences(s, op, i + 1, j + 1);
 
        // When a particular character isn't taken
        subsequences(s, op, i + 1, j);
        return;
    }
}
 
// Driver Code
int main()
{
    char str[] = "ggg";
    int m = sizeof(str) / sizeof(char);
    int n = pow(2, m) + 1;
 
    // Output array for storing
    // the generating subsequences
    // in each call
    char op[n];
 
    // Function Call
    subsequences(str, op, 0, 0);
 
    // Output will be the number
    // of elements in the set
    cout << sn.size();
    sn.clear();
    return 0;
 
    // This code is contributed by Kishan Mishra
}

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Python3

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# Python3 program to print
# distinct subsequences of
# a given string
import math
 
# Create an empty set
# to store the subsequences
sn = []
global m
m = 0
 
# Function for generating
# the subsequences
def subsequences(s, op, i, j):
 
    # Base Case
    if(i == m):
        op[j] = None
        temp = "".join([i for i in op if i] )
 
        # Insert each generated
        # subsequence into the set
        sn.append(temp)
        return
      
    # Recursive Case
    else:
 
        # When a particular
        # character is taken
        op[j] = s[i]
 
        subsequences(s, op,
                     i + 1, j + 1)
 
        # When a particular
        # character isn't taken
        subsequences(s, op,
                     i + 1, j)
        return
 
# Driver Code
str = "ggg"
m = len(str)
n = int(math.pow(2, m) + 1)
 
# Output array for storing
# the generating subsequences
# in each call
op = [None for i in range(n)]
 
# Function Call
subsequences(str, op, 0, 0)
 
# Output will be the number
#of elements in the set
print(len(set(sn)))
 
# This code is contributed by avanitrachhadiya2155

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Output: 

4






 

Time Complexity: O(2^n) 
Axillary Space: O(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 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. 
 

C++

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// 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

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// 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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Python3

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# Python3 program to count number of
# distinct subseqences of a given string
 
MAX_CHAR = 256
 
def countSub(ss):
 
    # create an array to store index of last
    last = [-1 for i in range(MAX_CHAR + 1)]
     
    # length of input string
    n = len(ss)
     
    # dp[i] is going to store count of
    # discount subsequence of length of i
    dp = [-2 for i in range(n + 1)]
      
    # empty substring has only
    # one subseqence
    dp[0] = 1
     
    # Traverse through all lengths
    # from 1 to n
    for i in range(1, n + 1):
         
        # number of subseqence with
        # substring str[0...i-1]
        dp[i] = 2 * dp[i - 1]
 
        # if current character has appeared
        # before, then remove all subseqences
        # ending with previous occurrence.
        if last[ord(ss[i - 1])] != -1:
            dp[i] = dp[i] - dp[last[ord(ss[i - 1])]]
        last[ord(ss[i - 1])] = i - 1
     
    return dp[n]
     
# Driver code
print(countSub("gfg"))
 
# This code is contributed
# by mohit kumar 29

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C#

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// C# program to count number of distinct
// subsequences of a given string.
using System;
 
public class Count_Subsequences {
 
    static readonly 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];
 
        for (int i = 0; i < MAX_CHAR; i++)
            last[i] = -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[i - 1]] != -1)
                dp[i] = dp[i] - dp[last[(int)str[i - 1]]];
 
            // Mark occurrence of current character
            last[(int)str[i - 1]] = (i - 1);
        }
        return dp[n];
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        Console.WriteLine(countSub("gfg"));
    }
}
 
// This code is contributed 29AjayKumar

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Output: 
 

7






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

Method 3: Without extra space

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.

Here is a representation of above statement:



Say s = “abab”

Lets initialize variables as

map = {a:-1,b:-1}

levelCount=0;

allCount = 0;

Now iterating through each character

1st iteration ‘a’ 

Distinct Subsequences ending with ‘a’ is “a” . So the levelCount = 1

 levelCount = 1; allCount = 1;  map = {a:1,b:-1}

2nd iteration ‘b’

Distinct Subsequences ending with ‘b’ are “ab”,”b”. So the levelCount = 2. 

Also total subsequences we found till now are 3. allCount = 3.

Here we can see that levelCount can be determined from last allCount variable

levelCount = allCount+1

In case this is a distinct character allCount can also be determined easily as

allCount = allCount + levelCount;

allCount = 3

We also update the map with the current character’s levelCount. map{a:1,b:2}

3rd iteration ‘a’  

Now we have a repetition. 



Distinct Subsequences ending with ‘a’ are now “aa”,”ba”,”aba”,”a”.  So our levelCount is now 4 which can also be written as allCount+1 = 3+1 = 4.

if this was a distinct character allcount would have been 7 (allCount = allCount+levelCount = 3+4) but we will have remove the repetition which is  map.get(thisCharacter), so now allCount is 7-1 = 6

Here note that we have essentially remove the results of our 1st iteration which was duplicate subsequence “a”.

map is now updated with new levelCount of a to {a:4,b:2}

In case of duplication out allCount calculation changes as

allCount = allCount + levelCount – map.get(currentCharacter);

allCount = 3+4-1 = 6

4th iteration ‘b’

Again a repetition. 

Subsequences ending with ‘b’ are now “abb”,”bb”,”aab”,”bab”,”abab”,”ab” ,”b” which count is same as levelCount = allCount+1 = 6+1 = 7 .

allCount will be = allCount+levelCount – map.get(‘b’) =  6+7-2 = 11

Total number of distinct subsequences is allCount.

If empty String is also included then our answer is allCount+1.
Below is the implementation of the above approach.

Java

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// Java Program for above approach
import java.io.*;
import java.util.*;
class SubsequenceCount
{
 
  // Returns count of distinct
  // subsequences of str.
  public static int countSub(String s)
  {
    HashMap<Character,
             Integer> map = new HashMap<Character,
                                        Integer>();
 
    // Iterate from 0 to s.length()
    for(int i = 0; i < s.length(); i++)
    {
      map.put(s.charAt(i), -1);
    }
     
    int allCount = 0;
    int levelCount = 0;
     
    // Iterate from 0 to s.length()
    for(int i=0;i<s.length();i++)
    {
      char c = s.charAt(i);
       
      // Check if i equal to 0
      if(i==0)
      {
        allCount = 1;
        map.put(c,1);
        levelCount = 1;
        continue;
      }
       
      // Replace levelCount withe
      // allCount + 1
      levelCount = allCount + 1;
       
      // If map is less than 0
      if(map.get(c)<0)
      {
        allCount = allCount + levelCount;
      }
      else
      {
        allCount = allCount + levelCount - map.get(c);
      }
      map.put(c,levelCount);
    }
     
    // Return answer
    return allCount;
 
  }
   
  // Driver Code
  public static void main(String[] args)
  {
    List<String> list = Arrays.asList("abab","gfg");
     
    for(String s : list)
    {
      int cnt = countSub(s);
      int withEmptyString = cnt+1;
      System.out.println("With empty string count for " +
                         s +" is " + withEmptyString);
      System.out.println("Without empty string count for " +
                         s + " is " + cnt);
    }
  }
}

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Time Complexity: O(n)

Space Complexity: O(1)

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