Number of permutations of a string in which all the occurrences of a given character occurs together

Given a string ‘s’ and a character ‘c’, the task is to find the number of permutations of the string in which all the occurrences of the character ‘c’ will be together (one after another).

Examples :

Input: Str = “AKA” ch = ‘A’
Output: 2
All the unique permutations of AKA are : AKA, AAK and KAA
‘A’ comes consecutively only twice. Hence, the answer is 2

Input: str= “MISSISSIPPI” ch = ‘S’
Output: 840

Naive Approach:

  • Generate all the permutations of the given string.
  • For each permutation check whether all occurrences of the character appear together.
  • Count of the permutations from the previous step is the answer.

Efficient Approach: Let the length of the string be ‘l’ and frequency of the character in the string be ‘n’.

  • Store the frequency of every character of the string.
  • If the character is not present in the string then the output will be ‘0’.
  • Consider all the occurrences of the character as a combined single character.
  • So, ‘l’ will become ‘l – occ + 1’ where ‘occ’ is the total occurrence of the given character and ‘l’ is the new length of the string.
  • Return ( (Factorial of l) / (Product of the factorials of the no. of occurrences of each character except the given character) )

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return factorial
// of the number passed as argument
long long int fact(int n)
{
    long long result = 1;
    for (int i = 1; i <= n; i++)
        result *= i;
    return result;
}
  
// Function to get the total permutations
// which satisfy the given condition
int getResult(string str, char ch)
{
    // Create has to store count
    // of each character
    int has[26] = { 0 };
  
    // Store character occurrences
    for (int i = 0; i < str.length(); i++)
        has[str[i] - 'A']++;
  
    // Count number of times
    // Particular character comes
    int particular = has[ch - 'A'];
  
    // If particular character isn't
    // present in the string then return 0
    if (particular == 0)
        return 0;
  
    // Remove count of particular character
    has[ch - 'A'] = 0;
  
    // Total length
    // of the string
    int total = str.length();
  
    // Assume all occurrences of
    // particular character as a
    // single character.
    total = total - particular + 1;
  
    // Compute factorial of the length
    long long int result = fact(total);
  
    // Divide by the factorials of
    // the no. of occurrences of all
    // the characters.
    for (int i = 0; i < 26; i++) {
        if (has[i] > 1) {
            result = result / fact(has[i]);
        }
    }
  
    // return the result
    return result;
}
  
// Driver Code
int main()
{
    string str = "MISSISSIPPI";
  
    // Assuming the string and the character
    // are all in uppercase
    cout << getResult(str, 'S') << endl;
  
  return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

    
// Java implementation of above approach
import java.util.*;
class solution
{
  
// Function to return factorial
// of the number passed as argument
 static int fact(int n)
{
     int result = 1;
    for (int i = 1; i <= n; i++)
        result *= i;
    return result;
}
  
// Function to get the total permutations
// which satisfy the given condition
static int getResult(String str, char ch)
{
    // Create has to store count
    // of each character
    int has[] = new int[26];
      
    for(int i=0;i<26;i++)
    has[i]=0;
  
    // Store character occurrences
    for (int i = 0; i < str.length(); i++)
        has[str.charAt(i) - 'A']++;
  
    // Count number of times
    // Particular character comes
    int particular = has[ch - 'A'];
  
    // If particular character isn't
    // present in the string then return 0
    if (particular == 0)
        return 0;
  
    // Remove count of particular character
    has[ch - 'A'] = 0;
  
    // Total length
    // of the string
    int total = str.length();
  
    // Assume all occurrences of
    // particular character as a
    // single character.
    total = total - particular + 1;
  
    // Compute factorial of the length
     int result = fact(total);
  
    // Divide by the factorials of
    // the no. of occurrences of all
    // the characters.
    for (int i = 0; i < 26; i++) {
        if (has[i] > 1) {
            result = result / fact(has[i]);
        }
    }
  
    // return the result
    return result;
}
  
// Driver Code
public static void main(String args[])
{
    String str = "MISSISSIPPI";
  
    // Assuming the string and the character
    // are all in uppercase
    System.out.println( getResult(str, 'S') );
  
}
}
//contributed by Arnab Kundu

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of 
# the approach 
  
# Function to return factorial 
# of the number passed as argument 
def fact(n) :
  
    result = 1
  
    for i in range(1, n + 1) :
        result *= i
  
    return result
  
# Function to get the total permutations 
# which satisfy the given condition 
def getResult(string, ch):
  
    # Create has to store count 
    # of each character
    has = [0] * 26
  
    # Store character occurrences 
    for i in range(len(string)) :
        has[ord(string[i]) - ord('A')] += 1
  
    # Count number of times 
    # Particular character comes 
    particular = has[ord(ch) - ord('A')]
  
    # If particular character isn't 
    # present in the string then return 0
    if particular == 0 :
        return 0
  
    # Remove count of particular character 
    has[ord(ch) - ord('A')] = 0
  
    # Total length 
    # of the string 
    total = len(string)
  
    # Assume all occurrences of 
    # particular character as a 
    # single character. 
    total = total - particular + 1
  
    # Compute factorial of the length 
    result = fact(total)
  
    # Divide by the factorials of 
    # the no. of occurrences of all 
    # the characters. 
    for i in range(26) :
  
        if has[i] > 1 :
            result /= fact(has[i])
  
    # return the result 
    return result
  
  
# Driver code
if __name__ == "__main__" :
  
    string = "MISSISSIPPI"
  
    # Assuming the string and the character 
    # are all in uppercase 
    print(getResult(string,'S'))
  
# This code is contributed 
# by ANKITRAI1

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of above approach
using System;
  
class GFG
{
  
// Function to return factorial
// of the number passed as argument
static int fact(int n)
{
    int result = 1;
    for (int i = 1; i <= n; i++)
        result *= i;
    return result;
}
  
// Function to get the total permutations
// which satisfy the given condition
static int getResult(string str, char ch)
{
    // Create has to store count
    // of each character
    int []has = new int[26];
      
    for(int i = 0; i < 26; i++)
    has[i] = 0;
  
    // Store character occurrences
    for (int i = 0; i < str.Length; i++)
        has[str[i] - 'A']++;
  
    // Count number of times
    // Particular character comes
    int particular = has[ch - 'A'];
  
    // If particular character 
    // isn't present in the string 
    // then return 0
    if (particular == 0)
        return 0;
  
    // Remove count of particular character
    has[ch - 'A'] = 0;
  
    // Total length of the string
    int total = str.Length;
  
    // Assume all occurrences of
    // particular character as a
    // single character.
    total = total - particular + 1;
  
    // Compute factorial of the length
    int result = fact(total);
  
    // Divide by the factorials of
    // the no. of occurrences of all
    // the characters.
    for (int i = 0; i < 26; i++) 
    {
        if (has[i] > 1) 
        {
            result = result / fact(has[i]);
        }
    }
  
    // return the result
    return result;
}
  
// Driver Code
public static void Main()
{
    string str = "MISSISSIPPI";
  
    // Assuming the string and the 
    // character are all in uppercase
    Console.WriteLine(getResult(str, 'S') );
}
}
  
// This code is contributed by anuj_67

chevron_right


PHP

1)
{
$result = $result / fact($has[$i]);
}
}

// return the result
return $result;
}

// Driver Code
$str = “MISSISSIPPI”;

// Assuming the string and the character
// are all in uppercase
echo getResult($str, ‘S’).”\n” ;

// This code is contributed by ita_c
?>

Output:

840


My Personal Notes arrow_drop_up

All powers are within you You can do anything and everything Believe in that

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 Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : andrew1234, vt_m, AnkitRai01, Ita_c