Dada una palabra en inglés de una longitud máxima de 20 caracteres. Calcula el número de formas de ordenar la palabra de manera que ninguna vocal aparezca junta.
Nota: si el número total de vocales en la palabra dada es uno, entonces el resultado debería ser 0.
Ejemplos:
Input : allahabad Output : 7200 Input : geeksforgeeks Output : 32205600 Input : abcd Output : 0
Ya que la palabra contiene vocales y consonantes. Calcule el número total de formas de organizar la palabra dada y reste el número de formas que tienen todas las vocales juntas. Para calcular el número total de formas, usaremos la siguiente fórmula:
No of ways = (n!) / (r1! * r2! * ... * rk!)
Donde n es el número de caracteres diferentes en la palabra y r1, r2…rk, son la frecuencia de caracteres del mismo tipo.
Para calcular el número de formas en que todas las vocales aparecen juntas, consideramos el grupo de todas las vocales como un solo carácter y, utilizando la fórmula anterior, calcularemos el número total de formas en las que todas las vocales están juntas. Ahora réstelo del número total de formas de obtener el resultado.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ code for above approach
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
// Function to check if a character is vowel or consonent
bool isVowel(char ch)
{
if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u')
return true;
else
return false;
}
// Function to calculate factorial of a number
ll fact(ll n)
{
if (n < 2)
return 1;
return n * fact(n - 1);
}
// Calculating no of ways for arranging vowels
ll only_vowels(map<char, int>& freq)
{
ll denom = 1;
ll cnt_vwl = 0;
// Iterate the map and count the number of vowels and calculate
// no of ways to arrange vowels
for (auto itr = freq.begin(); itr != freq.end(); itr++) {
if (isVowel(itr->first)) {
denom *= fact(itr->second);
cnt_vwl += itr->second;
}
}
return fact(cnt_vwl) / denom;
}
// calculating no of ways to arrange the given word such that all vowels
// come together
ll all_vowels_together(map<char, int>& freq)
{
// calculate no of ways to arrange vowels
ll vow = only_vowels(freq);
// to store denominator of fraction
ll denom = 1;
// count of consonents
ll cnt_cnst = 0;
for (auto itr = freq.begin(); itr != freq.end(); itr++) {
if (!isVowel(itr->first)) {
denom *= fact(itr->second);
cnt_cnst += itr->second;
}
}
// calculate the number of ways to arrange the word such that
// all vowels come together
ll ans = fact(cnt_cnst + 1) / denom;
return (ans * vow);
}
// To calculate total number of permutations
ll total_permutations(map<char, int>& freq)
{
// To store length of the given word
ll cnt = 0;
// denominator of fraction
ll denom = 1;
for (auto itr = freq.begin(); itr != freq.end(); itr++) {
denom *= fact(itr->second);
cnt += itr->second;
}
// return total number of permutations of the given word
return fact(cnt) / denom;
}
// Function to calculate number of permutations such that
// no vowels come together
ll no_vowels_together(string& word)
{
// to store frequency of character
map<char, int> freq;
// count frequency of all characters
for (int i = 0; i < word.size(); i++) {
char ch = tolower(word[i]);
freq[ch]++;
}
// calculate total number of permutations
ll total = total_permutations(freq);
// calculate total number of permutations such that
// all vowels come together
ll vwl_tgthr = all_vowels_together(freq);
// subtract vwl_tgthr from total to get the result
ll res = total - vwl_tgthr;
// return the result
return res;
}
// Driver code
int main()
{
string word = "allahabad";
ll ans = no_vowels_together(word);
cout << ans << endl;
word = "geeksforgeeks";
ans = no_vowels_together(word);
cout << ans << endl;
word = "abcd";
ans = no_vowels_together(word);
cout << ans << endl;
return 0;
}
Java
// Java code for above approach
import java.util.*;
import java.lang.*;
class GFG
{
// Function to check if a character
// is vowel or consonent
static boolean isVowel(char ch)
{
if (ch == 'a' || ch == 'e' ||
ch == 'i' || ch == 'o' ||
ch == 'u')
return true;
else
return false;
}
// Function to calculate factorial of a number
static long fact(long n)
{
if (n < 2)
{
return 1;
}
return n * fact(n - 1);
}
// Calculating no of ways for arranging vowels
static long only_vowels(HashMap<Character, Integer> freq)
{
long denom = 1;
long cnt_vwl = 0;
// Iterate the map and count the number
// of vowels and calculate no of ways
// to arrange vowels
for (Map.Entry<Character, Integer> itr : freq.entrySet())
{
if (isVowel(itr.getKey()))
{
denom *= fact(itr.getValue());
cnt_vwl += itr.getValue();
}
}
return fact(cnt_vwl) / denom;
}
// Calculating no of ways to arrange
// the given word such that all vowels
// come together
static long all_vowels_together(HashMap<Character, Integer> freq)
{
// Calculate no of ways to arrange vowels
long vow = only_vowels(freq);
// To store denominator of fraction
long denom = 1;
// Count of consonents
long cnt_cnst = 0;
for (Map.Entry<Character, Integer> itr : freq.entrySet())
{
if (!isVowel(itr.getKey()))
{
denom *= fact(itr.getValue());
cnt_cnst += itr.getValue();
}
}
// Calculate the number of ways to
// arrange the word such that
// all vowels come together
long ans = fact(cnt_cnst + 1) / denom;
return (ans * vow);
}
// To calculate total number of permutations
static long total_permutations(HashMap<Character, Integer> freq)
{
// To store length of the given word
long cnt = 0;
// Denominator of fraction
long denom = 1;
for (Map.Entry<Character, Integer> itr : freq.entrySet())
{
denom *= fact(itr.getValue());
cnt += itr.getValue();
}
// Return total number of
// permutations of the given word
return fact(cnt) / denom;
}
// Function to calculate number of permutations
// such that no vowels come together
static long no_vowels_together(String word)
{
// To store frequency of character
HashMap<Character, Integer> freq = new HashMap<>();
// Count frequency of all characters
for(int i = 0; i < word.length(); i++)
{
char ch = Character.toLowerCase(word.charAt(i));
if (freq.containsKey(ch))
{
freq.put(ch, freq.get(ch)+1);
}
else
{
freq.put(ch, 1);
}
}
// Calculate total number of permutations
long total = total_permutations(freq);
// Calculate total number of permutations
// such that all vowels come together
long vwl_tgthr = all_vowels_together(freq);
// Subtract vwl_tgthr from total
// to get the result
long res = total - vwl_tgthr;
// Return the result
return res;
}
// Driver code
public static void main(String[] args) {
String word = "allahabad";
long ans = no_vowels_together(word);
System.out.println(ans);
word = "geeksforgeeks";
ans = no_vowels_together(word);
System.out.println(ans);
word = "abcd";
ans = no_vowels_together(word);
System.out.println(ans);
}
}
// This code is contributed by divyesh072019
Python3
# Python3 code for above approach # Function to check if a character is # vowel or consonent def isVowel(ch): if (ch == 'a' or ch == 'e' or ch == 'i' or ch == 'o' or ch == 'u') : return True else: return False # Function to calculate factorial of a number def fact(n): if (n < 2): return 1 return n * fact(n - 1) # Calculating no of ways for arranging vowels def only_vowels(freq): denom = 1 cnt_vwl = 0 # Iterate the map and count the number of # vowels and calculate no of ways to arrange vowels for itr in freq: if (isVowel(itr)): denom *= fact(freq[itr]) cnt_vwl += freq[itr] return fact(cnt_vwl) // denom # calculating no of ways to arrange the given word # such that vowels come together def all_vowels_together(freq): # calculate no of ways to arrange vowels vow = only_vowels(freq) # to store denominator of fraction denom = 1 # count of consonents cnt_cnst = 0 for itr in freq: if (isVowel(itr) == False): denom *= fact(freq[itr]) cnt_cnst += freq[itr] # calculate the number of ways to arrange # the word such that vowels come together ans = fact(cnt_cnst + 1) // denom return (ans * vow) # To calculate total number of permutations def total_permutations(freq): # To store length of the given word cnt = 0 # denominator of fraction denom = 1 for itr in freq: denom *= fact(freq[itr]) cnt += freq[itr] # return total number of permutations # of the given word return fact(cnt) // denom # Function to calculate number of permutations # such that no vowels come together def no_vowels_together(word): # to store frequency of character freq = dict() # count frequency of all characters for i in word: ch = i.lower() freq[ch] = freq.get(ch, 0) + 1 # calculate total number of permutations total = total_permutations(freq) # calculate total number of permutations # such that vowels come together vwl_tgthr = all_vowels_together(freq) # subtract vwl_tgthr from total # to get the result res = total - vwl_tgthr # return the result return res # Driver code word = "allahabad" ans = no_vowels_together(word) print(ans) word = "geeksforgeeks" ans = no_vowels_together(word) print(ans) word = "abcd" ans = no_vowels_together(word) print(ans) # This code is contributed by Mohit Kumar
C#
// C# code for above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to check if a character
// is vowel or consonent
static bool isVowel(char ch)
{
if (ch == 'a' || ch == 'e' ||
ch == 'i' || ch == 'o' ||
ch == 'u')
return true;
else
return false;
}
// Function to calculate factorial of a number
static long fact(long n)
{
if (n < 2)
{
return 1;
}
return n * fact(n - 1);
}
// Calculating no of ways for arranging vowels
static long only_vowels(Dictionary<char, int> freq)
{
long denom = 1;
long cnt_vwl = 0;
// Iterate the map and count the number
// of vowels and calculate no of ways
// to arrange vowels
foreach(KeyValuePair<char, int> itr in freq)
{
if (isVowel(itr.Key))
{
denom *= fact(itr.Value);
cnt_vwl += itr.Value;
}
}
return fact(cnt_vwl) / denom;
}
// Calculating no of ways to arrange
// the given word such that all vowels
// come together
static long all_vowels_together(Dictionary<char, int> freq)
{
// Calculate no of ways to arrange vowels
long vow = only_vowels(freq);
// To store denominator of fraction
long denom = 1;
// Count of consonents
long cnt_cnst = 0;
foreach(KeyValuePair<char, int> itr in freq)
{
if (!isVowel(itr.Key))
{
denom *= fact(itr.Value);
cnt_cnst += itr.Value;
}
}
// Calculate the number of ways to
// arrange the word such that
// all vowels come together
long ans = fact(cnt_cnst + 1) / denom;
return (ans * vow);
}
// To calculate total number of permutations
static long total_permutations(Dictionary<char, int> freq)
{
// To store length of the given word
long cnt = 0;
// Denominator of fraction
long denom = 1;
foreach(KeyValuePair<char, int> itr in freq)
{
denom *= fact(itr.Value);
cnt += itr.Value;
}
// Return total number of
// permutations of the given word
return fact(cnt) / denom;
}
// Function to calculate number of permutations
// such that no vowels come together
static long no_vowels_together(string word)
{
// To store frequency of character
Dictionary<char,
int> freq = new Dictionary<char,
int>();
// Count frequency of all characters
for(int i = 0; i < word.Length; i++)
{
char ch = Char.ToLower(word[i]);
if (freq.ContainsKey(ch))
{
freq[ch]++;
}
else
{
freq[ch] = 1;
}
}
// Calculate total number of permutations
long total = total_permutations(freq);
// Calculate total number of permutations
// such that all vowels come together
long vwl_tgthr = all_vowels_together(freq);
// Subtract vwl_tgthr from total
// to get the result
long res = total - vwl_tgthr;
// Return the result
return res;
}
// Driver Code
static void Main()
{
string word = "allahabad";
long ans = no_vowels_together(word);
Console.WriteLine(ans);
word = "geeksforgeeks";
ans = no_vowels_together(word);
Console.WriteLine(ans);
word = "abcd";
ans = no_vowels_together(word);
Console.WriteLine(ans);
}
}
// This code is contributed by divyeshrabadiya07
Javascript
<script>
// Javascript code for above approach
// Function to check if a character
// is vowel or consonent
function isVowel(ch)
{
if (ch == 'a' || ch == 'e' ||
ch == 'i' || ch == 'o' ||
ch == 'u')
return true;
else
return false;
}
// Function to calculate factorial of a number
function fact(n)
{
if (n < 2)
{
return 1;
}
return n * fact(n - 1);
}
// Calculating no of ways for arranging vowels
function only_vowels(freq)
{
let denom = 1;
let cnt_vwl = 0;
// Iterate the map and count the number
// of vowels and calculate no of ways
// to arrange vowels
for (let [key, value] of freq.entries())
{
if (isVowel(key))
{
denom *= fact(value);
cnt_vwl += value;
}
}
return Math.floor(fact(cnt_vwl) / denom);
}
// Calculating no of ways to arrange
// the given word such that all vowels
// come together
function all_vowels_together(freq)
{
// Calculate no of ways to arrange vowels
let vow = only_vowels(freq);
// To store denominator of fraction
let denom = 1;
// Count of consonents
let cnt_cnst = 0;
for (let [key, value] of freq.entries())
{
if (!isVowel(key))
{
denom *= fact(value);
cnt_cnst += value;
}
}
// Calculate the number of ways to
// arrange the word such that
// all vowels come together
let ans = Math.floor(fact(cnt_cnst + 1) / denom);
return (ans * vow);
}
// To calculate total number of permutations
function total_permutations(freq)
{
// To store length of the given word
let cnt = 0;
// Denominator of fraction
let denom = 1;
for (let [key, value] of freq.entries())
{
denom *= fact(value);
cnt += value;
}
// Return total number of
// permutations of the given word
return Math.floor(fact(cnt) / denom);
}
// Function to calculate number of permutations
// such that no vowels come together
function no_vowels_together(word)
{
// To store frequency of character
let freq = new Map();
// Count frequency of all characters
for(let i = 0; i < word.length; i++)
{
let ch = word[i].toLowerCase();
if (freq.has(ch))
{
freq.set(ch, freq.get(ch)+1);
}
else
{
freq.set(ch, 1);
}
}
// Calculate total number of permutations
let total = total_permutations(freq);
// Calculate total number of permutations
// such that all vowels come together
let vwl_tgthr = all_vowels_together(freq);
// Subtract vwl_tgthr from total
// to get the result
let res = total - vwl_tgthr;
// Return the result
return res;
}
// Driver code
let word = "allahabad";
let ans = no_vowels_together(word);
document.write(ans+"<br>");
word = "geeksforgeeks";
ans = no_vowels_together(word);
document.write(ans+"<br>");
word = "abcd";
ans = no_vowels_together(word);
document.write(ans+"<br>");
// This code is contributed by unknown2108
</script>
Producción:
7200 32205600 0
Complejidad del tiempo: O(n) (n = longitud de la string)
Espacio Auxiliar : O(1)
Publicación traducida automáticamente
Artículo escrito por Vivek.Pandit y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA