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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
… [Trackback]
[…] Find More on on that Topic: geeksforgeeks.org/k-length-words-that-can-be-formed-from-given-characters-without-repetition/ […]
… [Trackback]
[…] Read More on that Topic: geeksforgeeks.org/k-length-words-that-can-be-formed-from-given-characters-without-repetition/ […]
… [Trackback]
[…] Read More on to that Topic: geeksforgeeks.org/k-length-words-that-can-be-formed-from-given-characters-without-repetition/ […]