可以由给定字符不重复构成的 K 长度单词
原文:https://www . geesforgeks . org/k-length-words-可以由给定字符不重复构成的单词/
给定一个整数 k 和一个由小写英文字母组成的字符串 str ，任务是在不允许重复的情况下，统计 str 的字符可以组成多少个 k 字符单词(有无意义)。 举例:
输入: str = "cat "，k = 3 输出: 6 必选词为“cat”“CTA”“act”“ATC”“TCA”“tac”。 输入: str = "geeksforgeeks "，k = 3 输出: 840
进场:统计 str 中不同字符的个数并存储在 cnt 中，现在的任务是将 cnt 字符中的 k 字符排列出来，即 n P r = n！/(n–r)！。 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 required count
int findPermutation(string str, int k)
{
bool has[26] = { false };
// To store the count of distinct characters in str
int cnt = 0;
// Traverse str character by character
for (int i = 0; i < str.length(); i++) {
// If current character is appearing
// for the first time in str
if (!has[str[i] - 'a']) {
// Increment the distinct character count
cnt++;
// Update the appearance of the current character
has[str[i] - 'a'] = true;
}
}
long long int ans = 1;
// Since P(n, r) = n! / (n - r)!
for (int i = 2; i <= cnt; i++)
ans *= i;
for (int i = cnt - k; i > 1; i--)
ans /= i;
// Return the answer
return ans;
}
// Driver code
int main()
{
string str = "geeksforgeeks";
int k = 4;
cout << findPermutation(str, k);
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 solution
{
// Function to return the required count
static int findPermutation(String str, int k)
{
boolean[] has = new boolean[26];
Arrays.fill(has,false);
// To store the count of distinct characters in str
int cnt = 0;
// Traverse str character by character
for (int i = 0; i < str.length(); i++) {
// If current character is appearing
// for the first time in str
if (!has[str.charAt(i) - 'a'])
{
// Increment the distinct character count
cnt++;
// Update the appearance of the current character
has[str.charAt(i) - 'a'] = true;
}
}
int ans = 1;
// Since P(n, r) = n! / (n - r)!
for (int i = 2; i <= cnt; i++)
ans *= i;
for (int i = cnt - k; i > 1; i--)
ans /= i;
// Return the answer
return ans;
}
// Driver code
public static void main(String args[])
{
String str = "geeksforgeeks";
int k = 4;
System.out.println(findPermutation(str, k));
}
}
// This code is contributed by
// Sanjit_prasad
Python 3
# Python3 implementation of the approach
import math as mt
# Function to return the required count
def findPermutation(string, k):
has = [False for i in range(26)]
# To store the count of distinct
# characters in str
cnt = 0
# Traverse str character by character
for i in range(len(string)):
# If current character is appearing
# for the first time in str
if (has[ord(string[i]) - ord('a')] == False):
# Increment the distinct
# character count
cnt += 1
# Update the appearance of the
# current character
has[ord(string[i]) - ord('a')] = True
ans = 1
# Since P(n, r) = n! / (n - r)!
for i in range(2, cnt + 1):
ans *= i
for i in range(cnt - k, 1, -1):
ans //= i
# Return the answer
return ans
# Driver code
string = "geeksforgeeks"
k = 4
print(findPermutation(string, k))
# This code is contributed
# by Mohit kumar 29
C
// C# implementation of the approach
using System;
class solution
{
// Function to return the required count
static int findPermutation(string str, int k)
{
bool []has = new bool[26];
for (int i = 0; i < 26 ; i++)
has[i] = false;
// To store the count of distinct characters in str
int cnt = 0;
// Traverse str character by character
for (int i = 0; i < str.Length; i++) {
// If current character is appearing
// for the first time in str
if (!has[str[i] - 'a'])
{
// Increment the distinct character count
cnt++;
// Update the appearance of the current character
has[str[i] - 'a'] = true;
}
}
int ans = 1;
// Since P(n, r) = n! / (n - r)!
for (int i = 2; i <= cnt; i++)
ans *= i;
for (int i = cnt - k; i > 1; i--)
ans /= i;
// Return the answer
return ans;
}
// Driver code
public static void Main()
{
string str = "geeksforgeeks";
int k = 4;
Console.WriteLine(findPermutation(str, k));
}
// This code is contributed by Ryuga
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the required count
function findPermutation($str, $k)
{
$has = array();
for ($i = 0; $i < 26; $i++)
{
$has[$i]= false;
}
// To store the count of distinct characters in $str
$cnt = 0;
// Traverse $str character by character
for ($i = 0; $i < strlen($str); $i++)
{
// If current character is appearing
// for the first time in $str
if ($has[ord($str[$i]) - ord('a')] == false)
{
// Increment the distinct character count
$cnt++;
// Update the appearance of the current character
$has[ord($str[$i]) - ord('a')] = true;
}
}
$ans = 1;
// Since P(n, r) = n! / (n - r)!
for ($i = 2; $i <= $cnt; $i++)
$ans *= $i;
for ($i = $cnt - $k; $i > 1; $i--)
$ans /= $i;
// Return the answer
return $ans;
}
// Driver code
$str = "geeksforgeeks";
$k = 4;
echo findPermutation($str, $k);
// This code is contributed by ihritik
?>
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the required count
function findPermutation(str, k) {
var has = new Array(26);
for (var i = 0; i < 26; i++) has[i] = false;
// To store the count of distinct characters in str
var cnt = 0;
// Traverse str character by character
for (var i = 0; i < str.length; i++) {
// If current character is appearing
// for the first time in str
if (!has[str[i].charCodeAt(0) - "a".charCodeAt(0)]) {
// Increment the distinct character count
cnt++;
// Update the appearance of the current character
has[str[i].charCodeAt(0) - "a".charCodeAt(0)] = true;
}
}
var ans = 1;
// Since P(n, r) = n! / (n - r)!
for (var i = 2; i <= cnt; i++) ans *= i;
for (var i = cnt - k; i > 1; i--) ans /= i;
// Return the answer
return ans;
}
// Driver code
var str = "geeksforgeeks";
var k = 4;
document.write(findPermutation(str, k));
</script>
Output:
840
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