Finding no.of times small string is in large string (c++) - c++

Question

I have to implement c++ code to find out the number of occurrences of a small string in a large string with overlapping allowed. For eg- if large string is acacab and small string is aca, then answer should be 2.
I'm not getting correct answer by making this code:
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
int i, j, k, c=0;
char lstr[30],sstr[10],tstr[10];
cout<<"Enter large string:"<<endl;
cin>>lstr;
cout<<"Enter small string:"<<endl;
cin>>sstr;
for(i=0;i<strlen(lstr);i++)
{
if(lstr[i]==sstr[0])
{
j=i;
for(k=0;k<strlen(sstr);k++,j++)
{
tstr[k]=lstr[j];
}
}
if(strcmp(tstr,sstr)==0)
c++;
}
cout<<c;
return 0;
}

Answer 1

It's much easier if you use C++ strings instead of C ones:
int c=0;
string lstr, sstr;
cout<<"Enter large string:"<<endl;
cin>>lstr;
cout<<"Enter small string:"<<endl;
cin>>sstr;
for (size_t pos = 0;
(pos = lstr.find(sstr, pos)) != string::npos;
pos++)
{
c++;
}
cout<<c;

Answer 2

This should work, add a '\0' to terminate c-style string, and move if(strcmp(tstr,sstr)==0) in the if(lstr[i]==sstr[0]) statement(otherwise you will continuing increment c when lstr[i] != sstr[0]):
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
int i, j, k, c=0;
char lstr[30],sstr[10],tstr[10];
cout<<"Enter large string:"<<endl;
cin>>lstr;
cout<<"Enter small string:"<<endl;
cin>>sstr;
for(i=0;i<strlen(lstr);i++)
{
if(lstr[i]==sstr[0])
{
j=i;
for(k=0;k<strlen(sstr) && j<strlen(lstr);k++,j++)
{
tstr[k]=lstr[j];
}
tstr[k] = 0;
// ^^^^^^^^^^^^
if(strcmp(tstr,sstr)==0)
c++;
}
}
cout<<c;
return 0;
}

Answer 3

This will solve your problem more efficiently in O(n*m) complexity, where n is large string size and m is small string size.
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
int i, j, k, c=0;
char lstr[30],sstr[10];
cout<<"Enter large string:"<<endl;
cin>>lstr;
cout<<"Enter small string:"<<endl;
cin>>sstr;
for(i=0; lstr[i]; i++) // You don't need to calculate strlen each time. This loop will break when lstr reaches null ('\0') character
{
if(lstr[i]==sstr[0])
{
for(k=0,j=i; sstr[k] && lstr[j]; k++,j++) // don't need to calculate strlen each time
{
if(sstr[k]!=lstr[j]) // Break if not match
break;
}
if(k==strlen(sstr)) // Whole string matched
c++;
}
}
cout<<c<<"\n";
return 0;
}
If you want to solve this problem in O(n) complexity you can use KMP algorithm. Which is the best choice for this kind of problems.