Algorithm
problem Link : https://www.codechef.com/problems/HAPPYSTR
Problem
Chef has a string with him. Chef is happy if the string contains a contiguous substring of length strictly greater than 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 , number of test cases. Then the test cases follow.
- Each test case contains of a single line of input, a string .
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
- , where is the length of .
- will only contain lowercase English letters.
Sample 1:
4 aeiou abxy aebcdefghij abcdeeafg
Happy Sad Sad Happy
Explanation:
Test case : Since the string aeiou is a contiguous substring and consists of all vowels and has a length , Chef is happy.
Test case : Since none of the contiguous substrings of the string consist of all vowels and have a length , Chef is sad.
Test case : Since none of the contiguous substrings of the string consist of all vowels and have a length , Chef is sad.
Test case : Since the string eea is a contiguous substring and consists of all vowels and has a length , Chef is happy.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include<stdio.h>
#include < string.h>
int vcheck(char ch)
{
if(ch=='a' || ch=='e' || ch=='i' || ch=='o' || ch=='u') return 1;
else return 0;
}
int main()
{
int t,l,i;
char str[1000];
scanf("%d",&t);
while(t--)
{
scanf("%s",str);
l=strlen(str);
int count=0;
for(i=0;i<l;i++)
{
if(vcheck(str[i])==1 && vcheck(str[i+1])==1 && vcheck(str[i+2])==1) count++;
}
if(count==0) printf("Sad\n");
else printf("Happy\n">;
}
} Input
aeiou
abxy
aebcdefghij
abcdeeafg
Output
Sad
Sad
Happy
Demonstration
CodeChef solution HAPPYSTR - Chef and Happy String Codechef solution in C,C++