Encuentre la cantidad de strings formadas usando caracteres distintos de una string dada

Dada una string str que consta de alfabetos ingleses en minúsculas, la tarea es encontrar el recuento de todas las strings posibles de longitud máxima que se pueden formar utilizando los caracteres de str de modo que no haya dos caracteres iguales en la string generada.
Ejemplos: 
 

Entrada: str = “aba” 
Salida: 2 
“ab” y “ba” son las únicas strings válidas.
Entrada: str = «geeksforgeeks» 
Salida: 5040 
 

Enfoque: primero, cuente la cantidad de caracteres distintos en la string, digamos cnt , ya que no hay dos caracteres que puedan ser iguales en la string resultante. Ahora, el número total de strings que se pueden formar con cnt número de caracteres es cnt! ya que cada carácter de str debe estar presente en la string generada para maximizar la longitud y ningún carácter debe aparecer más de una vez.
A continuación se muestra la implementación del enfoque anterior: 
 

// 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;
}

// 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

# 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

// 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

<script>
 
    // Javascript implementation of the approach
     
    // Function to return the factorial of n
    function fact(n)
    {
        let fact = 1;
        for (let 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
    function countStrings(str, n)
    {
       
        // To store the distinct characters
        // of the string str
        let distinct_char = new Set();
        for (let i = 0; i < n; i++) {
            distinct_char.add(str[i]);
        }
       
        return fact(distinct_char.size);
    }
     
    let str = "geeksforgeeks";
    let n = str.length;
   
    document.write(countStrings(str, n));
     
</script>

5040