Número de permutaciones de una string en la que todas las ocurrencias de un carácter dado ocurren juntas

Dada una string ‘s’ y un carácter ‘c’, la tarea es encontrar el número de permutaciones de la string en las que todas las ocurrencias del carácter ‘c’ estarán juntas (una tras otra).
Ejemplos: 
 

Entrada: Str = “AKA” ch = ‘A’ 
Salida: 2 
Todas las permutaciones únicas de AKA son: AKA, AAK y KAA 
‘A’ viene consecutivamente solo dos veces. Por lo tanto, la respuesta es 2
Entrada: str= “MISSISSIPPI” ch = ‘S’ 
Salida: 840

Enfoque ingenuo: 
 

• Genera todas las permutaciones de la string dada.
• Para cada permutación, compruebe si todas las apariciones del carácter aparecen juntas.
• La cuenta de las permutaciones del paso anterior es la respuesta.

Enfoque eficiente: Deje que la longitud de la string sea ‘l’ y la frecuencia del carácter en la string sea ‘n’. 
 

• Almacene la frecuencia de cada carácter de la string.
• Si el carácter no está presente en la string, la salida será ‘0’.
• Considere todas las apariciones del personaje como un solo personaje combinado.
• Entonces, ‘l’ se convertirá en ‘l – occ + 1’ donde ‘occ’ es la ocurrencia total del carácter dado y ‘l’ es la nueva longitud de la string.
• Return ( (Factorial de l) / (Producto de los factoriales del n.º de ocurrencias de cada carácter excepto el carácter dado) )

A continuación se muestra la implementación del enfoque anterior: 
 

// 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;
}

   
// 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

# 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

// 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

<?php
// PHP implementation of the approach
 
// Function to return factorial of
// the number passed as argument
function fact($n)
{
    $result = 1;
    for ($i = 1; $i <= $n; $i++)
        $result *= $i;
    return $result;
}
 
// Function to get the total permutations
// which satisfy the given condition
function getResult($str, $ch)
{
    // Create has to store count
    // of each character
    $has = array_fill(0, 26, NULL);
 
    // Store character occurrences
    for ($i = 0; $i < strlen($str); $i++)
        $has[ord($str[$i]) - ord('A')]++;
 
    // 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
    $total = strlen($str);
 
    // 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 = 0; $i < 26; $i++)
    {
        if ($has[$i] > 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
?>

<script>
    // Javascript implementation of the approach
 
// Function to return factorial of
// the number passed as argument
function fact(n)
{
    let result = 1;
    for (let i = 1; i <= n; i++)
        result *= i;
    return result;
}
 
// Function to get the total permutations
// which satisfy the given condition
function getResult(str, ch)
{
    // Create has to store count
    // of each character
    let has = new Array(26).fill(null);
 
    // Store character occurrences
    for (let i = 0; i < str.length; i++)
        has[str.charCodeAt(i) - 'A'.charCodeAt(0)]++;
 
    // Count number of times
    // Particular character comes
    particular = has[ch.charCodeAt(0) - 'A'.charCodeAt(0)];
 
    // If particular character isn't
    // present in the string then return 0
    if (particular == 0)
        return 0;
 
    // Remove count of particular character
    has[ch.charCodeAt(0) - 'A'.charCodeAt(0)] = 0;
 
    // Total length
    // of the string
    let total = str.length;
 
    // Assume all occurrences of
    // particular character as a
    // single character.
    total = total - particular + 1;
 
    // Compute factorial of the length
    let result = fact(total);
 
    // Divide by the factorials of
    // the no. of occurrences of all
    // the characters.
    for (let i = 0; i < 26; i++)
    {
        if (has[i] > 1)
        {
            result = result / fact(has[i]);
        }
    }
 
    // return the result
    return result;
}
 
// Driver Code
let str = "MISSISSIPPI";
 
// Assuming the string and the character
// are all in uppercase
document.write(getResult(str, 'S') + "<br>");
 
// This code is contributed by gfgking
</script>

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