给定字符串的每个可能长度的不同排列计数

给定一个字符串S，任务是计算给定字符串的每个可能长度的不同排列。

注意：字符串不允许重复字符。

Input: S = “abc”
Output: 15
Explanation:
Possible Permutations of every length are:
{“a”, “b”, “c”, “ab”, “bc”, “ac”, “ba”, “ca”, “cb”, “abc”, “acb”, “bac”, “bca”, “cab”, “cba”}

Input: S = “xz”
Output: 4

方法：想法是找到字符串的每个可能长度的组合的计数，它们的和是不同长度可能的不同排列的总数。因此，对于N个长度的字符串，不同长度的不同排列的总数为：

Total Combinations possible: ⁿP₁ + ⁿP₂ + ⁿP₃ + ⁿP₄ + …… + ⁿP_n

下面是上述方法的实现：

C++

// C++ implementation of the
// above approach
  
#include 
#include 
using namespace std;
  
// Function to find the factorial
// of a number
int fact(int a)
{
    int i, f = 1;
  
    // Loop to find the factorial
    // of the given number
    for (i = 2; i <= a; i++)
        f = f * i;
    return f;
}
  
// Function to find the number
// of permutations possible
// for a given string
int permute(int n, int r)
{
    int ans = 0;
    ans = (fact(n) / fact(n - r));
    return ans;
}
  
// Function to find the total
// number of combinations possible
int findPermutations(int n)
{
    int sum = 0, P;
    for (int r = 1; r <= n; r++) {
        P = permute(n, r);
        sum = sum + P;
    }
    return sum;
}
  
// Driver Code
int main()
{
    string str = "xz";
    int result, n;
    n = str.length();
  
    cout << findPermutations(n);
    return 0;
}

Java

// Java implementation of the
// above approach
class GFG{
  
// Function to find the factorial
// of a number
static int fact(int a)
{
    int i, f = 1;
  
    // Loop to find the factorial
    // of the given number
    for(i = 2; i <= a; i++)
        f = f * i;
      
    return f;
}
  
// Function to find the number
// of permutations possible
// for a given String
static int permute(int n, int r)
{
    int ans = 0;
    ans = (fact(n) / fact(n - r));
    return ans;
}
  
// Function to find the total
// number of combinations possible
static int findPermutations(int n)
{
    int sum = 0, P;
    for(int r = 1; r <= n; r++)
    {
        P = permute(n, r);
        sum = sum + P;
    }
    return sum;
}
  
// Driver Code
public static void main(String[] args)
{
    String str = "xz";
    int result, n;
    n = str.length();
  
    System.out.print(findPermutations(n));
}
}
  
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach
  
# Function to find the factorial
# of a number
def fact(a):
  
    f = 1
  
    # Loop to find the factorial
    # of the given number
    for i in range(2, a + 1):
        f = f * i
  
    return f
  
# Function to find the number
# of permutations possible
# for a given string
def permute(n, r):
  
    ans = 0
    ans = fact(n) // fact(n - r)
  
    return ans
  
# Function to find the total
# number of combinations possible
def findPermutations(n):
  
    sum = 0
    for r in range(1, n + 1):
        P = permute(n, r)
        sum = sum + P
  
    return sum
  
# Driver Code
str = "xz"
n = len(str)
  
# Function call
print(findPermutations(n))
  
# This code is contributed by Shivam Singh

C#

// C# implementation of the
// above approach
using System;
  
class GFG{
  
// Function to find the factorial
// of a number
static int fact(int a)
{
    int i, f = 1;
  
    // Loop to find the factorial
    // of the given number
    for(i = 2; i <= a; i++)
        f = f * i;
      
    return f;
}
  
// Function to find the number
// of permutations possible
// for a given String
static int permute(int n, int r)
{
    int ans = 0;
    ans = (fact(n) / fact(n - r));
    return ans;
}
  
// Function to find the total
// number of combinations possible
static int findPermutations(int n)
{
    int sum = 0, P;
    for(int r = 1; r <= n; r++)
    {
        P = permute(n, r);
        sum = sum + P;
    }
    return sum;
}
  
// Driver Code
public static void Main(String[] args)
{
    String str = "xz";
    int n;
    n = str.Length;
  
    Console.Write(findPermutations(n));
}
}
  
// This code is contributed by amal kumar choubey

输出：

4