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# Chef and Happy String CodeChef Solution

## Problem – Chef and Happy String CodeChef Solution

Chef has a string S with him. Chef is happy if the string contains a contiguous substring of length strictly greater than 2 in which all its characters are vowels.

Determine whether Chef is happy or not.

Note that, in english alphabet, vowels are `a``e``i``o`, and `u`.

### Input Format

• First line will contain T, number of test cases. Then the test cases follow.
• Each test case contains of a single line of input, a string S.

### Output Format

For each test case, if Chef is happy, print `HAPPY` else print `SAD`.

You may print each character of the string in uppercase or lowercase (for example, the strings `hAppY``Happy``haPpY`, and `HAPPY` will all be treated as identical).

### Constraints

• 1≤T≤1000
• 3≤∣S∣≤1000, where ∣S∣ is the length of S.
• S will only contain lowercase English letters.

### Sample 1:

``````Input:
4
aeiou
abxy
aebcdefghij
abcdeeafg
Output:
Happy
Happy``````

### Explanation:

Test case 1: Since the string `aeiou` is a contiguous substring and consists of all vowels and has a length >2, Chef is happy.

Test case 2: Since none of the contiguous substrings of the string consist of all vowels and have a length >2, Chef is sad.

Test case 3: Since none of the contiguous substrings of the string consist of all vowels and have a length >2, Chef is sad.

Test case 4: Since the string `eea` is a contiguous substring and consists of all vowels and has a length >2, Chef is happy.

## Chef and Happy String CodeChef Solution in Java

``````/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
int t;
Scanner scan=new Scanner(System.in);
t=scan.nextInt();
while(t-->0){
String str;
int count=0;
str=scan.next();
int j=str.length();
for(int i=0;i<j;i++){
if(str.charAt(i)=='a'||str.charAt(i)=='e'||str.charAt(i)=='i'||str.charAt(i)=='o'||str.charAt(i)=='u'){
count++;
}
else{
count=0;
}

if(count==3){
System.out.println("Happy");
break;
}
}
if(count!=3){
}
}
}
}
``````

## Chef and Happy String CodeChef Solution in C++17

``````#include <iostream>
#include<string>
using namespace std;

int isVowel(char ch){
return ch=='a'|| ch=='e' || ch=='i' || ch=='o' || ch=='u';
}

int main() {
int t;
cin>>t;
while(t--)
{
string s;
int flag=0;
cin>>s;
for(int i=0;i<s.length()-2;i++){
if(isVowel(s[i])&&isVowel(s[i+1])&&isVowel(s[i+2])){
flag=1;
break;}
}
if(flag){
cout<<"Happy"<<endl;
}
else{
}

}
return 0;
}
``````

## Chef and Happy String CodeChef Solution in Pyth 3

``````n = int(input())
while(n):
list1 = ['a', 'e', 'i', 'o', 'u']
n = n-1
count = 0
str1 = input()
for i in str1:
if i in list1:
count = count+1
if count > 2:
break
else:
count = 0
if count > 2:
print("Happy")
else:
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