找出使用给定字符串的不同字符形成的字符串数量
原文:https://www . geeksforgeeks . org/find-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，这些字符串可以使用 str 的字符形成，这样生成的字符串中没有两个字符相同。 例:
输入: str = "aba" 输出:2 “ab”和“ba”是唯一有效的字符串。 输入: str = "geeksforgeeks" 输出: 5040
方法:首先，计算字符串中不同字符的数量，说 cnt ，因为在结果字符串中没有两个字符可以相同。现在可以用 cnt 字符数组成的字符串总数是 cnt！由于字符串的每个字符都必须出现在生成的字符串中，以便最大化长度，因此任何字符都不应出现一次以上。 以下是上述方法的实施:
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<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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