Divine Words Solution

Problem

Chef has a certain liking towards certain English characters. He considers these characters beautiful. You are given a string $S$ consisting of all the characters Chef finds beautiful.

Chef considers a word $divine$ if each and every character present in the word is beautiful. Given a list of $N$ words, determine which of them are $divine$.

Input Format

• The first line of the input contains a string $S$, consisting of beautiful characters. Each character in $S$ will appear only once.
• The second line of the input contains an integer $N$.
• Each of the following $N$ lines contains a single string $P_i$, denoting the $i^{th}$ word in the list.

Output Format

For each of the $N$ words, output "Yes" (without quotes) if the word is divine and "No" (without quotes) otherwise.

Constraints

• $1 \leq |S| \leq 26$
• $1 \leq N \leq 1000$
• $1 \leq |P_i| \leq 12$
• Each character in $S$ will appear only once.
• $S$, $P_i$ consist only of lowercase English alphabets.

Subtasks

• Subtask $1$ (30 points): $|S|$ = $1$
• Subtask $2$ (70 points) : Original constraints

Sample 1:

Input

Output

rested
3
terp
dest
nop

No
Yes
No

Explanation:

Test Case $1$: Character $p$ isn't beautiful. Therefore this word is not divine.

Test Case $2$: Every character in the word $d, e, s, t$ is beautiful. Therefore, this word is divine.

Program :

 #include <stdio.h>

#include<string.h> 

int main(){

    char s[26]; 

    scanf("%s",s); 

    int t; 

    scanf("%d\n",&t); 

    while(t-->0) 

    {  

        char a[26]; 

        scanf("%s",a); 

        int i,j,count=0; 

        for(i=0;i<strlen(a);i++) 

        {

            for(j=0;j<strlen(s);j++) 

            {

                if(a[i]!=s[j])

                {

                    count==0; 

                    }

                    else 

                    {

                        count++; 

                        break; 

                        }

                        }  

                        }

                        if(count==strlen(a)) 

                        {

                            printf("Yes\n"); 

                            }

                            else

                            {

                                printf("No\n"); }  

                                } 

    return 0; 

}