Number of distinct permutation a String can have
Last Updated :
12 Jul, 2022
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 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 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.
// 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.
<script>
// Javascript program to find number of distinct
// permutations of a string.
let MAX_CHAR = 26;
// Utility function to find factorial of n.
function factorial(n)
{
let fact = 1;
for (let i = 2; i <= n; i++)
fact = fact * i;
return fact;
}
// Returns count of distinct permutations
// of str.
function countDistinctPermutations(str)
{
let length = str.length;
let freq = new Array(MAX_CHAR);
freq.fill(0);
// finding frequency of all the lower case
// alphabet and storing them in array of
// integer
for (let i = 0; i < length; i++)
if (str[i].charCodeAt() >= 'a'.charCodeAt())
freq[str[i].charCodeAt() - 'a'.charCodeAt()]++;
// finding factorial of number of appearances
// and multiplying them since they are
// repeating alphabets
let fact = 1;
for (let 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 parseInt(factorial(length) / fact, 10);
}
let str = "fvvfhvgv";
document.write(countDistinctPermutations(str));
// This code is contributed by vaibhavrabadiya117.
</script>
Time Complexity: O(n), where n is the length of the string
Auxiliary Space: O(n), where n is the length of the string
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