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Number of words that can be made using exactly P consonants and Q vowels from the given string

  • Difficulty Level : Medium
  • Last Updated : 01 Apr, 2021

Given a string str and two integers P and Q. The task is to find the total count of words that can be formed by choosing exactly P consonants and Q vowels from the given string.
Examples: 
 

Input: str = “geek”, P = 1, Q = 1 
Output:
“ge”, “ge”, “eg”, “ek”, “eg”, “ek”, 
“ke” and “ke” are the possible words.
Input: str = “crackathon”, P = 4, Q = 3 
Output: 176400 
 

 

Approach: Since P consonants and Q vowels has to be chosen from the original count of consonants and vowels in the given string. So, binomial coefficient can be used to calculate the combinations of choosing these characters and the characters chosen can be arranged in themselves using the factorial of their count.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
#define lli long long int
 
// Function to return the value of nCk
lli binomialCoeff(lli n, lli k)
{
    if (k == 0 || k == n)
        return 1;
 
    return binomialCoeff(n - 1, k - 1)
           + binomialCoeff(n - 1, k);
}
 
// Function to return the factorial of n
lli fact(lli n)
{
    if (n >= 1)
        return n * fact(n - 1);
    else
        return 1;
}
 
// Function that returns true if ch is a vowel
bool isVowel(char ch)
{
    if (ch == 'a' || ch == 'e' || ch == 'i'
        || ch == 'o' || ch == 'u') {
        return true;
    }
 
    return false;
}
 
// Function to return the number of words possible
lli countWords(string s, int p, int q)
{
 
    // To store the count of vowels and
    // consonanats in the given string
    lli countc = 0, countv = 0;
    for (int i = 0; i < s.length(); i++) {
 
        // If current character is a vowel
        if (isVowel(s[i]))
            countv++;
        else
            countc++;
    }
 
    // Find the total possible words
    lli a = binomialCoeff(countc, p);
    lli b = binomialCoeff(countv, q);
    lli c = fact(p + q);
    lli ans = (a * b) * c;
    return ans;
}
 
// Driver code
int main()
{
    string s = "crackathon";
    int p = 4, q = 3;
 
    cout << countWords(s, p, q);
 
    return 0;
}

Java




// Java implementation of the above approach
class GFG
{
     
    // Function to return the value of nCk
    static long binomialCoeff(long n, long k)
    {
        if (k == 0 || k == n)
            return 1;
     
        return binomialCoeff(n - 1, k - 1) +
               binomialCoeff(n - 1, k);
    }
     
    // Function to return the factorial of n
    static long fact(long n)
    {
        if (n >= 1)
            return n * fact(n - 1);
        else
            return 1;
    }
     
    // Function that returns true if ch is a vowel
    static boolean isVowel(char ch)
    {
        if (ch == 'a' || ch == 'e' || ch == 'i' ||
                         ch == 'o' || ch == 'u')
        {
            return true;
        }
     
        return false;
    }
     
    // Function to return the number of words possible
    static long countWords(String s, int p, int q)
    {
     
        // To store the count of vowels and
        // consonanats in the given string
        long countc = 0, countv = 0;
        for (int i = 0; i < s.length(); i++)
        {
     
            // If current character is a vowel
            if (isVowel(s.charAt(i)))
                countv++;
            else
                countc++;
        }
     
        // Find the total possible words
        long a = binomialCoeff(countc, p);
        long b = binomialCoeff(countv, q);
        long c = fact(p + q);
        long ans = (a * b) * c;
        return ans;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        String s = "crackathon";
        int p = 4, q = 3;
     
        System.out.println(countWords(s, p, q));
    }
}
 
// This Code is contributed by AnkitRai01

Python3




# Python3 implementation of the approach
 
# Function to return the value of nCk
def binomialCoeff(n, k):
    if (k == 0 or k == n):
        return 1
 
    return binomialCoeff(n - 1, k - 1) + \
           binomialCoeff(n - 1, k)
 
# Function to return the factorial of n
def fact(n):
    if (n >= 1):
        return n * fact(n - 1)
    else:
        return 1
 
# Function that returns true if ch is a vowel
def isVowel(ch):
 
    if (ch == 'a' or ch == 'e' or
        ch == 'i' or ch == 'o' or ch == 'u'):
        return True
 
    return False
 
# Function to return the number of words possible
def countWords(s, p, q):
 
    # To store the count of vowels and
    # consonanats in the given string
    countc = 0
    countv = 0
    for i in range(len(s)):
 
        # If current character is a vowel
        if (isVowel(s[i])):
            countv += 1
        else:
            countc += 1
 
    # Find the total possible words
    a = binomialCoeff(countc, p)
    b = binomialCoeff(countv, q)
    c = fact(p + q)
    ans = (a * b) * c
    return ans
 
# Driver code
s = "crackathon"
p = 4
q = 3
 
print(countWords(s, p, q))
 
# This code is contributed by Mohit Kumar

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
     
class GFG
{
     
    // Function to return the value of nCk
    static long binomialCoeff(long n, long k)
    {
        if (k == 0 || k == n)
            return 1;
     
        return binomialCoeff(n - 1, k - 1) +
               binomialCoeff(n - 1, k);
    }
     
    // Function to return the factorial of n
    static long fact(long n)
    {
        if (n >= 1)
            return n * fact(n - 1);
        else
            return 1;
    }
     
    // Function that returns true if ch is a vowel
    static bool isVowel(char ch)
    {
        if (ch == 'a' || ch == 'e' || ch == 'i' ||
                         ch == 'o' || ch == 'u')
        {
            return true;
        }
        return false;
    }
     
    // Function to return the number of words possible
    static long countWords(String s, int p, int q)
    {
     
        // To store the count of vowels and
        // consonanats in the given string
        long countc = 0, countv = 0;
        for (int i = 0; i < s.Length; i++)
        {
     
            // If current character is a vowel
            if (isVowel(s[i]))
                countv++;
            else
                countc++;
        }
     
        // Find the total possible words
        long a = binomialCoeff(countc, p);
        long b = binomialCoeff(countv, q);
        long c = fact(p + q);
        long ans = (a * b) * c;
        return ans;
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        String s = "crackathon";
        int p = 4, q = 3;
     
        Console.WriteLine(countWords(s, p, q));
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function to return the value of nCk
function binomialCoeff(n, k)
{
    if (k == 0 || k == n)
        return 1;
 
    return binomialCoeff(n - 1, k - 1)
           + binomialCoeff(n - 1, k);
}
 
// Function to return the factorial of n
function fact(n)
{
    if (n >= 1)
        return n * fact(n - 1);
    else
        return 1;
}
 
// Function that returns true if ch is a vowel
 
function isVowel(ch)
{
    if (ch == 'a' || ch == 'e' || ch == 'i'
        || ch == 'o' || ch == 'u') {
        return true;
    }
 
    return false;
}
 
// Function to return the number of words possible
 
function countWords(s, p, q)
{
 
    // To store the count of vowels and
    // consonanats in the given string
    var countc = 0, countv = 0;
    for (var i = 0; i < s.length; i++) {
 
        // If current character is a vowel
        if (isVowel(s[i]))
            countv++;
        else
            countc++;
    }
 
    // Find the total possible words
    var a = binomialCoeff(countc, p);
    var b = binomialCoeff(countv, q);
    var c = fact(p + q);
    var ans = (a * b) * c;
    return ans;
}
 
// Driver code
var s = "crackathon";
var p = 4, q = 3;
document.write(countWords(s, p, q));
 
</script>
Output: 
176400

 

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