Find the number of strings formed using distinct characters of a given string

Given a string str consisting of lowercase English alphabets, the task is to find the count of all possible string of maximum length that can be formed using the characters of str such that no two characters in the generated string are same.

Examples:

Input: str = “aba”
Output: 2
“ab” and “ba” are the only valid strings.



Input: str = “geeksforgeeks”
Output: 5040

Approach: First, count the number of distinct characters in the string say cnt as no two characters can be same in the resultant string. Now, the total number of strings that can be formed with cnt number of characters is cnt! as every character of str has to be present in the generated string in order to maximise the length and no character should appear more than once.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the factorial of n
int fact(int n)
{
    int fact = 1;
    for (int i = 1; i <= n; i++)
        fact *= i;
  
    return fact;
}
  
// Function to return the count of all
// possible strings that can be formed
// with the characters of the given string
// without repeating characters
int countStrings(string str, int n)
{
  
    // To store the distinct characters
    // of the string str
    set<char> distinct_char;
    for (int i = 0; i < n; i++) {
        distinct_char.insert(str[i]);
    }
  
    return fact(distinct_char.size());
}
  
// Driver code
int main()
{
    string str = "geeksforgeeks";
    int n = str.length();
  
    cout << countStrings(str, n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach 
import java.util.*;
  
class GFG 
{
  
// Function to return the factorial of n
static int fact(int n)
{
    int fact = 1;
    for (int i = 1; i <= n; i++)
        fact *= i;
  
    return fact;
}
  
// Function to return the count of all
// possible strings that can be formed
// with the characters of the given string
// without repeating characters
static int countStrings(String str, int n)
{
  
    // To store the distinct characters
    // of the string str
    Set<Character> distinct_char = new HashSet<>();
    for (int i = 0; i < n; i++) 
    {
        distinct_char.add(str.charAt(i));
    }
  
    return fact(distinct_char.size());
}
  
// Driver code
public static void main(String[] args)
{
    String str = "geeksforgeeks";
    int n = str.length();
  
    System.out.println(countStrings(str, n));
}
}
  
// This code is contributed by PrinciRaj1992

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# Function to return the factorial of n 
def fact(n) : 
  
    fact = 1
    for i in range(1, n + 1) :
        fact *= i; 
  
    return fact; 
  
# Function to return the count of all 
# possible strings that can be formed 
# with the characters of the given string 
# without repeating characters 
def countStrings(string, n) : 
  
    # To store the distinct characters 
    # of the string str 
    distinct_char = set(); 
    for i in range(n) : 
        distinct_char.add(string[i]); 
      
    return fact(len(distinct_char)); 
  
# Driver code 
if __name__ == "__main__"
  
    string = "geeksforgeeks"
    n = len(string); 
  
    print(countStrings(string, n)); 
  
# This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG 
{
  
// Function to return the factorial of n
static int fact(int n)
{
    int fact = 1;
    for (int i = 1; i <= n; i++)
        fact *= i;
  
    return fact;
}
  
// Function to return the count of all
// possible strings that can be formed
// with the characters of the given string
// without repeating characters
static int countStrings(String str, int n)
{
  
    // To store the distinct characters
    // of the string str
    HashSet<char> distinct_char = new HashSet<char>();
    for (int i = 0; i < n; i++) 
    {
        distinct_char.Add(str[i]);
    }
  
    return fact(distinct_char.Count);
}
  
// Driver code
public static void Main(String[] args)
{
    String str = "geeksforgeeks";
    int n = str.Length;
  
    Console.WriteLine(countStrings(str, n));
}
}
  
// This code is contributed by 29AjayKumar

chevron_right


Output:

5040


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.