Dada una string str de alfabetos ingleses en minúsculas, la tarea es verificar si la string es una vocal prima o no. Se dice que una string es vocal prima si todas las vocales de la string aparecen solo en índices primos.
Ejemplos:
Entrada: str = “geeksforgeeks”
Salida: No
str[1] = ‘e’ es una vocal pero 1 no es primo.
Entrada: str = «bcae»
Salida: Sí
Todas las vocales están en índices primos, es decir, 2 y 3.
Enfoque: use la criba de Eratóstenes para encontrar todos los números primos menores que N, de modo que se pueda verificar la primalidad de todos los índices de la string. Ahora, si hay algún índice no primo tal que el carácter en esa posición es una vocal, entonces la string no es una vocal prima sino lo es.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if c is a vowel
bool isVowel(char c)
{
if (c == 'a' || c == 'e' || c == 'i'
|| c == 'o' || c == 'u')
return true;
return false;
}
// Function that returns true if all the vowels in
// the given string are only at prime indices
bool isVowelPrime(string str, int n)
{
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
bool prime[n];
memset(prime, true, sizeof(prime));
// 0 and 1 are not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p < n; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p] == true) {
// Update all multiples of p greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i < n; i += p)
prime[i] = false;
}
}
// For every character of the given string
for (int i = 0; i < n; i++) {
// If current character is vowel
// and the index is not prime
if (isVowel(str[i]) && !prime[i])
return false;
}
return true;
}
// Driver code
int main()
{
string str = "geeksforgeeks";
int n = str.length();
if (isVowelPrime(str, n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function that returns true
// if c is a vowel
static boolean isVowel(char c)
{
if (c == 'a' || c == 'e' ||
c == 'i' || c == 'o' ||
c == 'u')
return true;
return false;
}
// Function that returns true if all the vowels in
// the given string are only at prime indices
static boolean isVowelPrime(String str, int n)
{
// Create a boolean array "prime[0..n]" and
// initialize all entries it as true.
// A value in prime[i] will finally be false
// if i is Not a prime, else true.
boolean []prime = new boolean[n];
Arrays.fill(prime, true);
// 0 and 1 are not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p < n; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i < n; i += p)
prime[i] = false;
}
}
// For every character of the given string
for (int i = 0; i < n; i++)
{
// If current character is vowel
// and the index is not prime
if (isVowel(str.charAt(i)) && !prime[i])
return false;
}
return true;
}
// Driver code
public static void main(String[] args)
{
String str = "geeksforgeeks";
int n = str.length();
if (isVowelPrime(str, n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Princi Singh
Python3
# Python3 implementation of the approach
# Function that returns true if c is a vowel
def isVowel(c):
if (c == 'a' or c == 'e' or
c == 'i' or c == 'o' or
c == 'u'):
return True
return False
# Function that returns true if
# all the vowels in the given string
# are only at prime indices
def isVowelPrime(Str, n):
# Create a boolean array "prime[0..n]"
# and initialize all entries in it as true.
# A value in prime[i] will finally be false
# if i is Not a prime, else true.
prime = [True for i in range(n)]
# 0 and 1 are not prime
prime[0] = False
prime[1] = False
for p in range(2, n):
if p * p > n:
break
# If prime[p] is not changed,
# then it is a prime
if (prime[p] == True):
# Update all multiples of p greater than or
# equal to the square of it
# numbers which are multiple of p and are
# less than p^2 are already been marked.
for i in range(2 * p, n, p):
prime[i] = False
# For every character of the given String
for i in range(n):
# If current character is vowel
# and the index is not prime
if (isVowel(Str[i]) and
prime[i] == False):
return False
return True
# Driver code
Str= "geeksforgeeks";
n = len(Str)
if (isVowelPrime(Str, n)):
print("Yes")
else:
print("No")
# This code is contributed by Mohit Kumar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true
// if c is a vowel
static Boolean isVowel(char c)
{
if (c == 'a' || c == 'e' ||
c == 'i' || c == 'o' ||
c == 'u')
return true;
return false;
}
// Function that returns true if all the vowels in
// the given string are only at prime indices
static Boolean isVowelPrime(String str, int n)
{
// Create a boolean array "prime[0..n]" and
// initialize all entries it as true.
// A value in prime[i] will finally be false
// if i is Not a prime, else true.
Boolean []prime = new Boolean[n];
for(int i = 0; i < n; i++)
prime[i] = true;
// 0 and 1 are not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p < n; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p greater than
// or equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i < n; i += p)
prime[i] = false;
}
}
// For every character of the given string
for (int i = 0; i < n; i++)
{
// If current character is vowel
// and the index is not prime
if (isVowel(str[i]) && !prime[i])
return false;
}
return true;
}
// Driver code
public static void Main(String[] args)
{
String str = "geeksforgeeks";
int n = str.Length;
if (isVowelPrime(str, n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// Javascript implementation of the approach
// Function that returns true if c is a vowel
function isVowel(c)
{
if (c == 'a' || c == 'e' || c == 'i'
|| c == 'o' || c == 'u')
return true;
return false;
}
// Function that returns true if all the vowels in
// the given string are only at prime indices
function isVowelPrime(str, n)
{
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
var prime = Array(n).fill(true);
// 0 and 1 are not prime
prime[0] = false;
prime[1] = false;
for (var p = 2; p * p < n; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p] == true) {
// Update all multiples of p greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for (var i = p * p; i < n; i += p)
prime[i] = false;
}
}
// For every character of the given string
for (var i = 0; i < n; i++) {
// If current character is vowel
// and the index is not prime
if (isVowel(str[i]) && !prime[i])
return false;
}
return true;
}
// Driver code
var str = "geeksforgeeks";
var n = str.length;
if (isVowelPrime(str, n))
document.write( "Yes");
else
document.write( "No");
</script>
Producción:
No