Program 91:Search a substring in a given string

Program 91:Search a substring in a given string
 

#include<stdio.h>
#include<string.h>
main()
{
int i,j,flag=0,len1,len2;
char str[100],substr[100];
printf("Enter a string\n");
gets(str);
printf("Enter a substring to be searched\n");
gets(substr);
len1=strlen(str);
len2=strlen(substr);

for(i=0;i<=len1-len2;i++)
{
    for(j=i;j<i+len2;j++)
    {
        flag=1;
        if(str[j]!=substr[j-i])
        {
            flag=0;
            break;
        }
    }
    if(flag==1)
    break;
}
if(flag==1)
{
    printf("Search is Successful\n");
}
else
{
        printf("Search is not Successful\n");
}
}

Explanation:

1. This program starts with initializing :
   • i,j → used as helping variable
     
   • str → To store input from user
   • substr →To store substring to search
   • len1,len2→ To store length of str and sustr
   • flag →To use this for checking whether the given substr is present in str

2. printf("Enter a string\n");
   gets(str);
   printf("Enter a substring to be searched\n");
   gets(substr);

   Taking input from user where str represents string(let's take "all c program") and substr represents substring("lets take "all c" as substring) 

3. for(i=0;i<=len1-len2;i++)
   {

   length of "all c programs" is 14(i.e from 0 to 13) in arrays and "all c" is 5. So len1=14 and len2=5.len1-len2=9.The loop iterates 10 times from 0 to 9.
4. Why len1-len2?
   As substring length is 5 then, if the loop traverses till the last but 5 letters without output i.e from 0 to 9 then remaining letters, after the 9th letter there are 4(all c programs having length of  14 i.e from 0-13 in arrays) letters which will not have enough letters for searching the substring.
5. Main logic
   

   for(j=i;j<i+len2;j++)
       {
           flag=1;
           if(str[j]!=substr[j-i])
           {
               flag=0;
               break;
           }
       }//end of forloop
       if(flag==1)
       break;

   i will traverse from 0 to 9(len1-len2) for each iteration i will be incremented by 1.len1=14,len2=5
   
   • Iteration 1:j=i →j=0,j<0+5 which is true so loop is executed
     •  j=0
     • flag=1 by default when ever the loop is entered
     • now str[0]!=substr[0-0]→a!=a which is false so flag remains same
     • now j is incremented by 1 so j=1
   • Iteration 2:j=1 ,1<0+5 which is true so loop is executed
     •  j=1,flag=1
     • now str[1]!=substr[1-0]→l!=l which is false so flag remains same
     •  now j is incremented by 1 so j=2
   • Iteration 3:j=2 ,2<0+5 which is true so loop is executed
     •  j=2,flag=1
     • now str[2]!=substr[2-0]→l!=l which is false so flag remains same
     •  now j is incremented by 1 so j=3
   • Iteration 4:j=3 ,3<0+5 which is true so loop is executed
     •  j=3,flag=1
     • now str[3]!=substr[3-0]→space!=space which is false so flag remains same
     •  now j is incremented by 1 so j=4
   • Iteration 5:j=4 ,4<0+5 which is true so loop is executed
     •  j=4,flag=1
     • now str[4]!=substr[4-0]→c!=c which is false so flag remains same
     •  now j is incremented by 1 so j=5
   • Iteration 4:j=5,5<0+5 which is false so loop is terminated 
   • So the if part is never executed at all in 5 iterations so flag is remained as 1
   • finally when flag=1 then the main loop for(i=0;i<len1-len2;i++) is terminated as if(flag==1){break;} is there inside the 1st forloop

6. if(flag==1)
   {
       printf("Search is Successful\n");
   }
   else
   {
           printf("Search is not Successful\n");
   }

   As flag ==1 which is true so the if part is executed So Search is successful

 Output: