Count of unique Subsequences of given String with lengths in range [0, N]
Given a string S of length N, the task is to find the number of unique subsequences of the string for each length from 0 to N.
Note: The uppercase letters and lowercase letters are considered different and the result may be large so print it modulo 1000000007.
Examples:
Input: S = “ababd”
Output:
Number of unique subsequences of length 0 is 1
Number of unique subsequences of length 1 is 3
Number of unique subsequences of length 2 is 6
Number of unique subsequences of length 3 is 8
Number of unique subsequences of length 4 is 5
Number of unique subsequences of length 5 is 1
Explanation: 0 length subsequences are 1-> {}
1 length subsequences are 3 -> a, b, d
2 length subsequences are 6 -> ab, aa, ad, ba, bd, bb
3 length subsequences are 8 -> aab, aad, abb, abd, bab, aba, bbd, bad
4 length subsequences are 5 -> aabd, abab, abad, babd, abbd
5 length subsequences are 1 -> ababd
Input: GeeksForGeeks
Output:
Number of unique subsequences of length 0 is 1
Number of unique subsequences of length 1 is 7
Number of unique subsequences of length 2 is 43
Number of unique subsequences of length 3 is 163
Number of unique subsequences of length 4 is 402
Number of unique subsequences of length 5 is 703
Number of unique subsequences of length 6 is 917
Number of unique subsequences of length 7 is 918
Number of unique subsequences of length 8 is 711
Number of unique subsequences of length 9 is 421
Number of unique subsequences of length 10 is 185
Number of unique subsequences of length 11 is 57
Number of unique subsequences of length 12 is 11
Number of unique subsequences of length 13 is 1
Naive Approach: The basic approach to solve the problem is as follows:
Generate every possible subsequence and store it in a set to get unique results. Then traverse each of the subsequence from that set and count it on the basis of their lengths.
Follow the steps to solve the problem:
- Use recursion to generate each subsequence and store it in a set to get unique occurrences.
- Use a map to store the count of subsequences for each possible length.
- Traverse each subsequence, find the length of the subsequence and update the count of the appropriate group of subsequences.
- Traverse map and print both its keys (length of subsequence) & value (count of subsequence).
Below is the implementation for the above approach:
C++14
#include <bits/stdc++.h>
using namespace std;
#define MAX 1000000007
typedef long long ll;
unordered_set<string> sn;
void subsequences( char s[], char op[],
ll i, ll j)
{
if (s[i] == '\0' ) {
op[j] = '\0' ;
sn.insert(op);
return ;
}
else {
op[j] = s[i];
subsequences(s, op, i + 1, j + 1);
subsequences(s, op, i + 1, j);
return ;
}
}
map<ll, ll> getCount()
{
map<ll, ll> freq;
for ( auto x : sn) {
freq[x.length()]
= (freq[x.length()] + 1) % MAX;
}
return freq;
}
void printSubsequences(string& S)
{
ll m = S.length();
char op[m + 1];
subsequences(&S[0], &op[0], 0, 0);
map<ll, ll> ans = getCount();
for ( auto & x : ans)
cout << "Number of unique subsequences of length "
<< x.first << " is " << x.second << endl;
}
int main()
{
string S = "ababd" ;
printSubsequences(S);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static int MAX = 1000000007 ;
static HashSet<String> sn;
static void subsequences(String s, String op, int i)
{
if (i == s.length()) {
sn.add( new String(op));
return ;
}
else {
subsequences(s, op + s.charAt(i), i + 1 );
subsequences(s, op, i + 1 );
return ;
}
}
static HashMap<Long, Long> getCount()
{
HashMap<Long, Long> freq = new HashMap<>();
for (String x : sn) {
long len = x.length();
freq.put(len, (freq.getOrDefault(len, 0L) + 1 )
% MAX);
}
return freq;
}
static void printSubsequences(String S)
{
int m = S.length();
String op = "" ;
sn = new HashSet<>();
subsequences(S, op, 0 );
HashMap<Long, Long> ans = getCount();
for (Long x : ans.keySet())
System.out.println(
"Number of unique subsequences of length "
+ x + " is " + ans.get(x));
}
public static void main(String[] args)
{
String S = "ababd" ;
printSubsequences(S);
}
}
|
Python3
MAX = 1000000007
sn = set ()
def subsequences(s, op, i):
global sn
if (i = = len (s)):
sn.add("".join(op))
else :
subsequences(s, op + s[i], i + 1 )
subsequences(s, op, i + 1 )
def getCount():
freq = dict ()
for x in sn:
if len (x) in freq:
freq[ len (x)] = (freq[ len (x)] + 1 ) % MAX
else :
freq[ len (x)] = 1 % MAX
return freq
def printSubsequences(s):
global op
m = len (s)
op = ""
subsequences(s, op, 0 )
ans = getCount()
for x in sorted (ans):
print ( "Number of unique subsequences of length" , x, "is" , ans[x])
s = "ababd"
printSubsequences(s)
|
C#
using System;
using System.Collections.Generic;
class GFG {
static int MAX = 1000000007;
static HashSet< string > sn;
static void subsequences( string s, string op, int i)
{
if (i == s.Length) {
sn.Add( new string (op));
return ;
}
else {
subsequences(s, op + s[i], i + 1);
subsequences(s, op, i + 1);
return ;
}
}
static Dictionary< long , long > getCount()
{
Dictionary< long , long > freq
= new Dictionary< long , long >();
foreach ( string x in sn)
{
long len = x.Length;
if (freq.ContainsKey(len))
freq[len] = (freq[len] + 1) % MAX;
else
freq[len] = 1L;
}
return freq;
}
static void printSubsequences( string S)
{
int m = S.Length;
string op = "" ;
sn = new HashSet< string >();
subsequences(S, op, 0);
Dictionary< long , long > ans = getCount();
List< long > keyList = new List< long >(ans.Keys);
keyList.Sort();
foreach ( var key in keyList) Console.WriteLine(
"Number of unique subsequences of length " + key
+ " is " + ans[key]);
}
public static void Main( string [] args)
{
string S = "ababd" ;
printSubsequences(S);
}
}
|
Javascript
let MAX = 1000000007;
let sn = new Set();
function subsequences(s, op, i)
{
if (i == s.length)
sn.add(op);
else
{
subsequences(s, op + s[i], i + 1);
subsequences(s, op, i + 1);
}
}
function getCount()
{
let freq = {};
for (let x of sn)
{
if (freq.hasOwnProperty(x.length))
freq[x.length] = (freq[x.length] + 1) % MAX;
else
freq[x.length] = 1 % MAX;
}
return freq;
}
function printSubsequences(s)
{
let m = s.length;
let op = "" ;
subsequences(s, op, 0);
let ans = getCount();
let keys = Object.keys(ans);
keys.sort();
for (let x of keys)
console.log( "Number of unique subsequences of length" , x, "is" , ans[x]);
}
let s = "ababd" ;
printSubsequences(s);
|
Output
Number of unique subsequences of length 0 is 1
Number of unique subsequences of length 1 is 3
Number of unique subsequences of length 2 is 6
Number of unique subsequences of length 3 is 8
Number of unique subsequences of length 4 is 5
Number of unique subsequences of length 5 is 1
Time Complexity: O(2N)
Auxiliary Space: O(N)
Efficient Approach: The idea to solve the problem using dynamic programming is based on the following observations:
Observations:
Consider a character at ith position to be the end character of the subsequence with length j.
Let the total possible ways be denoted as f(i, j).
This value depends on the values of f(i-1, j) and f(i-1, j-1), i.e. it is the summation of f(i-1, j) and f(i-1, j-1).
So f(i, j) = f(i-1, j) + f(i-1, j-1)
But if the ith character has occurred earlier in an index k, then in the above case, value of f(k-1, j-1) is being considered twice:
- once for f(k, j) and
- next for f(i, j) [as f(k-1, j-1) is a part of f(i-1, j-1) and f(k, j) is part of f(i-1, j)]
and both the time the resulting subsequences are same because the kth and ith character are same.
So, in that case, we need to consider the value only once. Therefore f(i, j) = f(i-1, j) + f(i-1, j-1) – f(k-1, j-1)
So there are two cases:
- f(i, j) = f(i-1, j) + f(i-1, j-1) when ith character does not occur earlier
- f(i, j) = f(i-1, j) + f(i-1, j-1) – f(k-1, j-1) when ith character occurs earlier at kth index.
Follow the below steps to solve the problem:
- Create a 2-dimensional array dp[][] where dp[i][j] represents the number of unique subsequences of S until i-th element in string and subsequences are of length j.
- Create an array (say last[]) to store the previous occurrence of a character.
- Use the transition function shown above to calculate the value of dp[i][j].
- Base cases are dp[0][0]=1 and dp[i][j]=0 for every j>i.
Below is the implementation of the above approach:
C++14
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mod 1000000007
void findSubsequences(string& s)
{
int n = s.length();
ll dp[n + 2][n + 2];
memset (dp, 0, sizeof (dp));
vector<ll> last(256, -1);
dp[0][0] = 1;
for ( int i = 0; i < n + 1; i++) {
for ( int j = i + 1; j < n + 1; j++)
dp[i][j] = 0;
}
for ( int i = 1; i <= n; i++) {
for ( int j = 0; j < n + 1; j++) {
dp[i][j] = (dp[i][j]
+ dp[i - 1][j])
% mod;
if (j >= 1) {
dp[i][j]
= (dp[i][j]
+ dp[i - 1][j - 1])
% mod;
if (last[s[i - 1]] != -1) {
dp[i][j]
= (dp[i][j]
- dp[last[s[i - 1]]][j - 1])
% mod;
}
}
}
last[s[i - 1]] = (i - 1) % mod;
}
cout << "Number of unique subsequences of length 0 is 1"
<< endl;
for ( int i = 1; i <= n; i++)
cout << "Number of unique subsequences of length "
<< i << " is " << dp[n][i] << endl;
}
int main()
{
string S = "ababd" ;
findSubsequences(S);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG
{
static int mod = 1000000007 ;
static void findSubsequences(String s)
{
int n = s.length();
int [][] dp = new int [n + 2 ][n + 2 ];
for ( int i = 0 ; i < n + 2 ; i++) {
for ( int j = 0 ; j < n + 2 ; j++)
dp[i][j] = 0 ;
}
int [] last = new int [ 256 ];
for ( int i = 0 ; i < 256 ; i++) {
last[i] = - 1 ;
}
dp[ 0 ][ 0 ] = 1 ;
for ( int i = 0 ; i < n + 1 ; i++) {
for ( int j = i + 1 ; j < n + 1 ; j++)
dp[i][j] = 0 ;
}
for ( int i = 1 ; i <= n; i++) {
for ( int j = 0 ; j < n + 1 ; j++) {
dp[i][j] = (dp[i][j]
+ dp[i - 1 ][j])
% mod;
if (j >= 1 ) {
dp[i][j]
= (dp[i][j]
+ dp[i - 1 ][j - 1 ])
% mod;
if (last[s.charAt(i - 1 )] != - 1 ) {
dp[i][j]
= (dp[i][j]
- dp[last[s.charAt(i - 1 )]][j - 1 ])
% mod;
}
}
}
last[s.charAt(i - 1 )] = (i - 1 ) % mod;
}
System.out.println( "Number of unique subsequences of length 0 is 1" );
for ( int i = 1 ; i <= n; i++)
System.out.println( "Number of unique subsequences of length "
+ i + " is " + dp[n][i]);
}
public static void main(String args[])
{
String S = "ababd" ;
findSubsequences(S);
}
}
|
Python3
mod = 1000000007
def findSubsequences(s):
n = len (s);
dp = [];
for i in range ( 0 , n + 2 ):
dp.append([ 0 ] * (n + 2 ))
last = [ - 1 ] * 256
dp[ 0 ][ 0 ] = 1 ;
for i in range ( 1 , n + 1 ):
for j in range ( 0 , n + 1 ):
dp[i][j] = (dp[i][j] + dp[i - 1 ][j]) % mod;
if (j > = 1 ):
dp[i][j] = (dp[i][j] + dp[i - 1 ][j - 1 ]) % mod;
if (last[ ord (s[i - 1 ])] ! = - 1 ):
dp[i][j] = (dp[i][j] - dp[last[ ord (s[i - 1 ])]][j - 1 ]) % mod
last[ ord (s[i - 1 ])] = (i - 1 ) % mod
print ( "Number of unique subsequences of length 0 is 1" )
for i in range ( 1 , n + 1 ):
print ( "Number of unique subsequences of length" , i, "is" , dp[n][i])
if __name__ = = '__main__' :
S = "ababd" ;
findSubsequences(S)
|
C#
using System;
class GFG
{
static int mod = 1000000007;
static void findSubsequences(String s)
{
int n = s.Length;
int [,] dp = new int [n + 2, n + 2];
for ( int i = 0; i < n + 2; i++)
{
for ( int j = 0; j < n + 2; j++)
dp[i, j] = 0;
}
int [] last = new int [256];
for ( int i = 0; i < 256; i++)
{
last[i] = -1;
}
dp[0, 0] = 1;
for ( int i = 0; i < n + 1; i++)
{
for ( int j = i + 1; j < n + 1; j++)
dp[i, j] = 0;
}
for ( int i = 1; i <= n; i++)
{
for ( int j = 0; j < n + 1; j++)
{
dp[i, j] = (dp[i, j]
+ dp[i - 1, j])
% mod;
if (j >= 1)
{
dp[i, j]
= (dp[i, j]
+ dp[i - 1, j - 1])
% mod;
if (last[s[i - 1]] != -1)
{
dp[i, j]
= (dp[i, j]
- dp[last[s[i - 1]], j - 1])
% mod;
}
}
}
last[s[i - 1]] = (i - 1) % mod;
}
Console.WriteLine( "Number of unique subsequences of length 0 is 1" );
for ( int i = 1; i <= n; i++)
Console.WriteLine( "Number of unique subsequences of length "
+ i + " is " + dp[n, i]);
}
public static void Main()
{
String S = "ababd" ;
findSubsequences(S);
}
}
|
Javascript
const mod = 1000000007;
function findSubsequences(s)
{
var n = s.length;
s = s.split( "" );
for ( var i = 0; i < n; i++)
s[i] = s[i].charCodeAt(0);
var dp = [];
for ( var i = 0; i < n + 2; i++)
dp.push( new Array(n + 2));
for ( var i = 0; i < n + 2; i++) {
for ( var j = 0; j < n + 2; j++)
dp[i][j] = 0;
}
var last = new Array(256).fill(-1);
dp[0][0] = 1;
for ( var i = 0; i < n + 1; i++) {
for ( var j = i + 1; j < n + 1; j++)
dp[i][j] = 0;
}
for ( var i = 1; i <= n; i++) {
for ( var j = 0; j < n + 1; j++) {
dp[i][j] = (dp[i][j]
+ dp[i - 1][j])
% mod;
if (j >= 1) {
dp[i][j]
= (dp[i][j]
+ dp[i - 1][j - 1])
% mod;
if (last[s[i - 1]] != -1) {
dp[i][j]
= (dp[i][j]
- dp[last[s[i - 1]]][j - 1])
% mod;
}
}
}
last[s[i - 1]] = (i - 1) % mod;
}
console.log( "Number of unique subsequences of length 0 is 1" );
for ( var i = 1; i <= n; i++)
console.log( "Number of unique subsequences of length "
+ i + " is " + dp[n][i]);
}
var S = "ababd" ;
findSubsequences(S);
|
Output
Number of unique subsequences of length 0 is 1
Number of unique subsequences of length 1 is 3
Number of unique subsequences of length 2 is 6
Number of unique subsequences of length 3 is 8
Number of unique subsequences of length 4 is 5
Number of unique subsequences of length 5 is 1
Time Complexity: O(N2)
Auxiliary Space: O(N)
Last Updated :
08 Jul, 2022
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