Given an integer k and a string str consisting of lowercase English alphabets, the task is to count how many k-character words (with or without meaning) can be formed from the characters of str when repetition is not allowed.
Examples:
Input: str = “cat”, k = 3
Output: 6
Required words are “cat”, “cta”, “act”, “atc”, “tca” and “tac”.Input: str = “neveropen”, k = 3
Output: 840
Approach: Count the number of distinct characters in str and store it in cnt, now the task is to arrange k characters out of cnt characters i.e. nPr = n! / (n – r)!.
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 required countint 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 codeint main(){ string str = "neveropen"; int k = 4; cout << findPermutation(str, k); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class solution{// Function to return the required countstatic 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 codepublic static void main(String args[]){ String str = "neveropen"; int k = 4; System.out.println(findPermutation(str, k));}}// This code is contributed by// Sanjit_prasad |
Python3
# Python3 implementation of the approachimport math as mt # Function to return the required countdef 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 codestring = "neveropen"k = 4print(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 = "neveropen"; int k = 4; Console.WriteLine(findPermutation(str, k)); } // This code is contributed by Ryuga} |
PHP
<?php// PHP implementation of the approach// Function to return the required countfunction 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 = "neveropen";$k = 4;echo findPermutation($str, $k);// This code is contributed by ihritik?> |
Javascript
<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 = "neveropen"; var k = 4; document.write(findPermutation(str, k)); </script> |
Output
840
Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1)
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