CODEWITHSUNDEEP
c
Arjun
Arjun

Reputation: 3

Check vowels occurence in a string

This is the problem I'm trying to solve:

Input:
First line contains N, the size of the string. Second line contains the letters (only lowercase).

Output:
Print YES if all vowels are found in the string, NO otherwise.

Constraints:
The size of the string will not be greater than 10,000. 1 ≤ N ≤ 10000

The following code I wrote is always showing NO.

 #include <stdio.h>
 #include<conio.h>

 int main()
 {
   int a,b,c=0,d=0,e=0,f=0,g=0,i;
   char string[10000];
   scanf("%d",&a);
   scanf("%s",string);
   for(i=0;i<a;a++)
   {
     if(string[i]==('a'))
       c=1;
     if(string[i]==('e'))
       d=1;
     if(string[i]==('i'))
       e=1;
     if(string[i]==('o'))
       f=1;
     if(string[i]==('u'))
       g=1; 
   }
   if((c==1)&&(d==1)&&(e==1)&&(f==1)&&(g==1))
     printf("YES");
   else
     printf("NO");
   return 0;
   getch ();
 }

Upvotes: 0

Views: 4525

Answers (3)

Sandeep Kumar
Sandeep Kumar

Reputation: 336

This program can be done in more simpler way other as below.

#include <stdio.h>
#include<conio.h>

int main()
{
    int i, flag = 0;
    char string[10000], *ptr;
    char cmp[] = "aeiou";

    printf("Please enter string = " );
    scanf("%s", string);
    i = 0;
    while(cmp[i])
    {
        ptr = string;
        while(*ptr)
        {
            if(cmp[i] == *ptr)
            break;
            ptr++;
        }
        if(*ptr != cmp[i++])
        {
            flag = 1;
            break;
        }
    }

    if(flag == 1)
        printf("NO");
    else
        printf("YES");
}

In this program I have used just one flag instead of 5 flags. Always try to write simple code rather then using unnecessary variable and flags.

Upvotes: 1

mtszkw
mtszkw

Reputation: 2783

Here is an infinite loop that causes a problem:

for(i=0;i<a;a++)

You should increment i, instead of a (length of a string). If you fix this one char in loop statement, the program will run well at all. Anyway, I changed your code a bit to be more readable. Take a look if you want, just for your information, sir:

 #include <stdio.h>
 #include <stdlib.h>

 int main()
 {
   int len, a=0, e=0, i=0, o=0, u=0, it;
   char string[10000];

   scanf("%d", &len);
   scanf("%s", string);

   for(it=0;it<len;it++)
   {
     if(string[it]=='a') a = 1;
     else if(string[it]=='e') e = 1;
     else if(string[it]=='i') i = 1;
     else if(string[it]=='o') o = 1;
     else if(string[it]=='u') u = 1;
   }
   if(a && e && i && o && u) printf("YES\n");
   else printf("NO\n");

   system("PAUSE");
   return 0;
 }

I assume you are running your program under Windows, so instead of conio's getch() try to use system("PAUSE") or just even better way to do this (for both Windows for UNIX): getchar()

Upvotes: 2

anjruu
anjruu

Reputation: 1264

I've renamed all of your variables, but otherwise left the code the same.

 #include <stdio.h>
 #include<conio.h>

 int main()
 {
   int foundA = 0, foundE = 0, foundI = 0, foundO = 0, foundU = 0;
   int i, length;
   char string[10000];

   scanf("%d", &length);
   scanf("%s", string);

   for(i=0; i<length; length++)
   {
     if(string[i]==('a'))
       foundA=1;
     else if(string[i]==('e'))
       foundE=1;
     else if(string[i]==('i'))
       foundI=1;
     else if(string[i]==('o'))
       foundO=1;
     else if(string[i]==('u'))
       foundU=1; 
   }
   if((foundA==1)&&(foundE==1)&&(foundI==1)&&(foundO==1)&&(foundU==1))
     printf("YES");
   else
     printf("NO");
   return 0;

   getch ();
 }

Looking the the for-loop condition for(i=0; i<length; length++), I think it's pretty clear what's wrong. Instead of incrementing the counter, you're incrementing the length of the string. Eventually, the counter overflows to a negative number, and the loop terminates without ever looking at a character besides the first one. The lesson here is to name your variables properly.

If you want to be picky, then signed integer overflow is undefined behavior, but for most systems, INT_MAX + 1 will be INT_MIN.

Upvotes: 1

Related Questions