查找使用给定字符串的不同字符形成的字符串数

给定一个由小写英文字母组成的字符串str ，任务是找到所有可能的最大长度字符串的计数，该字符串可以使用str的字符形成，使得生成的字符串中没有两个字符相同。
例子：

Input: str = “aba”
Output: 2
“ab” and “ba” are the only valid strings.
Input: str = “geeksforgeeks”
Output: 5040

编程需要懂一点英语

方法：首先，计算字符串中不同字符的数量，比如cnt ，因为结果字符串中没有两个字符可以相同。现在，可以用cnt个字符组成的字符串总数是cnt！因为str的每个字符都必须出现在生成的字符串中以最大化长度，并且任何字符都不应出现多次。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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 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 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 distinct_char = new HashSet();
    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

输出：