# Number of distinct permutation a String can have

We are given a string having only lowercase alphabets. The task is to find out total number of distinct permutation can be generated by that string.

Examples:

Input : aab Output : 3 Different permutations are "aab", "aba" and "baa". Input : ybghjhbuytb Output : 1663200

A **simple solution** is to find all the distinct permutation and count them.

We can find the count **without finding all permutation**. Idea is to find all the characters that is getting repeated, i.e., frequency of all the character. Then, we divide the factorial of the length of string by multiplication of factorial of frequency of characters.

In second example, number of character is 11 and here **h** and **y** are repeated **2** times whereas **g** is repeated **3** times.

So, number of permutation is **11! / (2!2!3!) = 1663200**

Below is the implementation of above idea.

## C++

`// C++ program to find number of distinct ` `// permutations of a string. ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` `const` `int` `MAX_CHAR = 26; ` ` ` `// Utility function to find factorial of n. ` `int` `factorial(` `int` `n) ` `{ ` ` ` `int` `fact = 1; ` ` ` `for` `(` `int` `i = 2; i <= n; i++) ` ` ` `fact = fact * i; ` ` ` `return` `fact; ` `} ` ` ` `// Returns count of distinct permutations ` `// of str. ` `int` `countDistinctPermutations(string str) ` `{ ` ` ` `int` `length = str.length(); ` ` ` ` ` `int` `freq[MAX_CHAR]; ` ` ` `memset` `(freq, 0, ` `sizeof` `(freq)); ` ` ` ` ` `// finding frequency of all the lower case ` ` ` `// alphabet and storing them in array of ` ` ` `// integer ` ` ` `for` `(` `int` `i = 0; i < length; i++) ` ` ` `if` `(str[i] >= ` `'a'` `) ` ` ` `freq[str[i] - ` `'a'` `]++; ` ` ` ` ` `// finding factorial of number of appearances ` ` ` `// and multiplying them since they are ` ` ` `// repeating alphabets ` ` ` `int` `fact = 1; ` ` ` `for` `(` `int` `i = 0; i < MAX_CHAR; i++) ` ` ` `fact = fact * factorial(freq[i]); ` ` ` ` ` `// finding factorial of size of string and ` ` ` `// dividing it by factorial found after ` ` ` `// multiplying ` ` ` `return` `factorial(length) / fact; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `string str = ` `"fvvfhvgv"` `; ` ` ` `printf` `(` `"%d"` `, countDistinctPermutations(str)); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// Java program to find number of distinct ` `// permutations of a string. ` `public` `class` `GFG { ` ` ` ` ` `static` `final` `int` `MAX_CHAR = ` `26` `; ` ` ` ` ` `// Utility function to find factorial of n. ` ` ` `static` `int` `factorial(` `int` `n) ` ` ` `{ ` ` ` `int` `fact = ` `1` `; ` ` ` `for` `(` `int` `i = ` `2` `; i <= n; i++) ` ` ` `fact = fact * i; ` ` ` `return` `fact; ` ` ` `} ` ` ` ` ` `// Returns count of distinct permutations ` ` ` `// of str. ` ` ` `static` `int` `countDistinctPermutations(String str) ` ` ` `{ ` ` ` `int` `length = str.length(); ` ` ` ` ` `int` `[] freq = ` `new` `int` `[MAX_CHAR]; ` ` ` ` ` `// finding frequency of all the lower case ` ` ` `// alphabet and storing them in array of ` ` ` `// integer ` ` ` `for` `(` `int` `i = ` `0` `; i < length; i++) ` ` ` `if` `(str.charAt(i) >= ` `'a'` `) ` ` ` `freq[str.charAt(i) - ` `'a'` `]++; ` ` ` ` ` `// finding factorial of number of appearances ` ` ` `// and multiplying them since they are ` ` ` `// repeating alphabets ` ` ` `int` `fact = ` `1` `; ` ` ` `for` `(` `int` `i = ` `0` `; i < MAX_CHAR; i++) ` ` ` `fact = fact * factorial(freq[i]); ` ` ` ` ` `// finding factorial of size of string and ` ` ` `// dividing it by factorial found after ` ` ` `// multiplying ` ` ` `return` `factorial(length) / fact; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `String str = ` `"fvvfhvgv"` `; ` ` ` `System.out.println(countDistinctPermutations(str)); ` ` ` `} ` `} ` `// This code is contributed by Sumit Ghosh ` |

*chevron_right*

*filter_none*

## Python

`# Python program to find number of distinct ` `# permutations of a string. ` ` ` `MAX_CHAR ` `=` `26` ` ` `# Utility function to find factorial of n. ` `def` `factorial(n) : ` ` ` ` ` `fact ` `=` `1` `; ` ` ` `for` `i ` `in` `range` `(` `2` `, n ` `+` `1` `) : ` ` ` `fact ` `=` `fact ` `*` `i; ` ` ` `return` `fact ` ` ` `# Returns count of distinct permutations ` `# of str. ` `def` `countDistinctPermutations(st) : ` ` ` ` ` `length ` `=` `len` `(st) ` ` ` `freq ` `=` `[` `0` `] ` `*` `MAX_CHAR ` ` ` ` ` `# finding frequency of all the lower ` ` ` `# case alphabet and storing them in ` ` ` `# array of integer ` ` ` `for` `i ` `in` `range` `(` `0` `, length) : ` ` ` `if` `(st[i] >` `=` `'a'` `) : ` ` ` `freq[(` `ord` `)(st[i]) ` `-` `97` `] ` `=` `freq[(` `ord` `)(st[i]) ` `-` `97` `] ` `+` `1` `; ` ` ` ` ` `# finding factorial of number of ` ` ` `# appearances and multiplying them ` ` ` `# since they are repeating alphabets ` ` ` `fact ` `=` `1` ` ` `for` `i ` `in` `range` `(` `0` `, MAX_CHAR) : ` ` ` `fact ` `=` `fact ` `*` `factorial(freq[i]) ` ` ` ` ` `# finding factorial of size of string ` ` ` `# and dividing it by factorial found ` ` ` `# after multiplying ` ` ` `return` `factorial(length) ` `/` `fact ` ` ` `# Driver code ` `st ` `=` `"fvvfhvgv"` `print` `(countDistinctPermutations(st)) ` ` ` `# This code is contributed by Nikita Tiwari. ` |

*chevron_right*

*filter_none*

## C#

`// C# program to find number of distinct ` `// permutations of a string. ` `using` `System; ` ` ` `public` `class` `GFG { ` ` ` ` ` `static` `int` `MAX_CHAR = 26; ` ` ` ` ` `// Utility function to find factorial of n. ` ` ` `static` `int` `factorial(` `int` `n) ` ` ` `{ ` ` ` `int` `fact = 1; ` ` ` `for` `(` `int` `i = 2; i <= n; i++) ` ` ` `fact = fact * i; ` ` ` ` ` `return` `fact; ` ` ` `} ` ` ` ` ` `// Returns count of distinct permutations ` ` ` `// of str. ` ` ` `static` `int` `countDistinctPermutations(String str) ` ` ` `{ ` ` ` `int` `length = str.Length; ` ` ` ` ` `int` `[] freq = ` `new` `int` `[MAX_CHAR]; ` ` ` ` ` `// finding frequency of all the lower case ` ` ` `// alphabet and storing them in array of ` ` ` `// integer ` ` ` `for` `(` `int` `i = 0; i < length; i++) ` ` ` `if` `(str[i] >= ` `'a'` `) ` ` ` `freq[str[i] - ` `'a'` `]++; ` ` ` ` ` `// finding factorial of number of appearances ` ` ` `// and multiplying them since they are ` ` ` `// repeating alphabets ` ` ` `int` `fact = 1; ` ` ` `for` `(` `int` `i = 0; i < MAX_CHAR; i++) ` ` ` `fact = fact * factorial(freq[i]); ` ` ` ` ` `// finding factorial of size of string and ` ` ` `// dividing it by factorial found after ` ` ` `// multiplying ` ` ` `return` `factorial(length) / fact; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main(String []args) ` ` ` `{ ` ` ` `String str = ` `"fvvfhvgv"` `; ` ` ` ` ` `Console.Write(countDistinctPermutations(str)); ` ` ` `} ` `} ` ` ` `// This code is contributed by parashar. ` |

*chevron_right*

*filter_none*

Output:

840

This article is contributed by Aditya Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Recommended Posts:

- Find total number of distinct years from a string
- Find the number of strings formed using distinct characters of a given string
- Lexicographically smallest permutation with distinct elements using minimum replacements
- Find distinct characters in distinct substrings of a string
- Permutation of a string with maximum number of characters greater than its adjacent characters
- Minimum length of string having all permutation of given string.
- Lexicographically n-th permutation of a string
- String which when repeated exactly K times gives a permutation of S
- Print all permutation of a string using ArrayList
- Find n-th lexicographically permutation of a string | Set 2
- Lexicographically smallest permutation of a string with given subsequences
- Maximum even length sub-string that is permutation of a palindrome
- Distinct permutations of the string | Set 2
- Check if given string can be split into four distinct strings
- Convert given string so that it holds only distinct characters