Number of distinct permutation a String can have

2.2

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;
}

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

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.


Output:

840

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