Related Articles

# K length words that can be formed from given characters without repetition

• Difficulty Level : Easy
• Last Updated : 13 May, 2021

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:
Required words are “cat”, “cta”, “act”, “atc”, “tca” and “tac”.
Input: str = “geeksforgeeks”, 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 ``using` `namespace` `std;` `// Function to return the required count``int` `findPermutation(string str, ``int` `k)``{``    ``bool` `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[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

 `// 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`

## Python3

 `# 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``;``    ` `    ``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``}`

## PHP

 ` 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``?>`

## Javascript

 ``
Output:
`840`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up