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

  • Difficulty Level : Hard
  • Last Updated : 15 Mar, 2021

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:
“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++




// 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




// 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




# 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#




// 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

Javascript




<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>
Output: 
5040

 

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