Check if the string satisfies the given condition
Last Updated :
12 Oct, 2022
Given a string str, the task is to check whether the count of vowels in the given string is prime or not. If it’s prime then print YES else print NO.
Examples:
Input: str = “geeksforgeeks”
Output: YES
Count of vowels is 5 (e, e, o, e and e) which is prime.
Input: str = “aeioua”
Output: NO
Approach: Initialize a variable count = 0 and traverse the string character by character, if the current character is a vowel, then update count = count + 1. In the end, if count is prime then print YES else print NO.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
bool prime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
bool isVowel(char c)
{
c = tolower(c);
if (c == 'a'
|| c == 'e'
|| c == 'i'
|| c == 'o'
|| c == 'u')
return true;
return false;
}
bool isValidString(string word)
{
int cnt = 0;
for (int i = 0; i < word.length(); i++) {
if (isVowel(word[i]))
cnt++;
}
if (prime(cnt))
return true;
else
return false;
}
int main()
{
string s = "geeksforgeeks";
if (isValidString(s))
cout << "YES";
else
cout << "NO";
return 0;
}
|
Java
import java.util.*;
class GFG
{
static boolean prime(int n)
{
if (n <= 1)
{
return false;
}
if (n <= 3)
{
return true;
}
if (n % 2 == 0 || n % 3 == 0)
{
return false;
}
for (int i = 5; i * i <= n; i = i + 6)
{
if (n % i == 0 || n % (i + 2) == 0)
{
return false;
}
}
return true;
}
static boolean isVowel(char c)
{
c = Character.toLowerCase(c);
if (c == 'a' || c == 'e' ||
c == 'i' || c == 'o' || c == 'u')
{
return true;
}
return false;
}
static boolean isValidString(String word)
{
int cnt = 0;
for (int i = 0; i < word.length(); i++)
{
if (isVowel(word.charAt(i)))
{
cnt++;
}
}
if (prime(cnt))
{
return true;
} else
{
return false;
}
}
public static void main(String args[])
{
String s = "geeksforgeeks";
if (isValidString(s))
System.out.println("YES");
else
System.out.println("NO");
}
}
|
Python3
def prime(n):
if (n <= 1):
return False
if (n <= 3):
return True
if (n % 2 == 0 or n % 3 == 0):
return False
i = 5
while i * i <= n:
if (n % i == 0 or n % (i + 2) == 0):
return False
i += 6
return True
def isVowel(c):
c = c.lower()
if (c == 'a' or c == 'e' or
c == 'i' or c == 'o' or
c == 'u'):
return True
return False
def isValidString(word):
cnt = 0;
for i in range(len(word)):
if (isVowel(word[i])):
cnt += 1
if (prime(cnt)):
return True
else:
return False
if __name__=="__main__":
s = "geeksforgeeks"
if (isValidString(s)):
print("YES")
else:
print("NO")
|
C#
using System;
class GFG
{
static Boolean prime(int n)
{
if (n <= 1)
{
return false;
}
if (n <= 3)
{
return true;
}
if (n % 2 == 0 || n % 3 == 0)
{
return false;
}
for (int i = 5; i * i <= n; i = i + 6)
{
if (n % i == 0 || n % (i + 2) == 0)
{
return false;
}
}
return true;
}
static Boolean isVowel(char c)
{
c = char.ToLower(c);
if (c == 'a' || c == 'e' ||
c == 'i' || c == 'o' || c == 'u')
{
return true;
}
return false;
}
static Boolean isValidString(String word)
{
int cnt = 0;
for (int i = 0; i < word.Length; i++)
{
if (isVowel(word[i]))
{
cnt++;
}
}
if (prime(cnt))
{
return true;
} else
{
return false;
}
}
public static void Main(String []args)
{
String s = "geeksforgeeks";
if (isValidString(s))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
|
Javascript
<script>
function prime(n)
{
if (n <= 1)
{
return false;
}
if (n <= 3)
{
return true;
}
if (n % 2 == 0 || n % 3 == 0)
{
return false;
}
for (let i = 5; i * i <= n; i = i + 6)
{
if (n % i == 0 || n % (i + 2) == 0)
{
return false;
}
}
return true;
}
function isVowel(c)
{
c = c.toLowerCase();
if (c == 'a' || c == 'e' ||
c == 'i' || c == 'o' || c == 'u')
{
return true;
}
return false;
}
function isValidString(word)
{
let cnt = 0;
for (let i = 0; i < word.length; i++)
{
if (isVowel(word[i]))
{
cnt++;
}
}
if (prime(cnt))
{
return true;
} else
{
return false;
}
}
let s = "geeksforgeeks";
if (isValidString(s))
document.write("YES<br>");
else
document.write("NO<br>");
</script>
|
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1)
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