Question

C++ Programming:

Write a function that takes in a C++-style string s (from #include ), as well as an integer n, and then returns true if any substring of length n in s repeats itself, and false otherwise. For example, if s is "toe-to-toe" and n is 3, then the function would return true since "toe" occurs twice within s. However, if n were 4, the function would return false. As a second example, if s is "singing" and n is 2, then the function would return true since, e.g., "in" appears more than once (as does "ng"). In this problem you may use any of the C++ string member functions.

Answer 1

#include <iostream> //for string operstions #include <string> //includiing the namespace standard using namespace std; //function for calculating the substring int ssc (const string& givenst, const string& stsub) if (stsub.length) 0) return 0; int cnt- 0; int n-2 for (size t offset givenst. find(stsub); offset != string: : npos; offset - givenst.find (stsub, offset + stsub.length ())) cnt++; if (cnt--n) return 1; else return 0;

int main() int 1,m; l-ssc("TOE-TO-TOE", "TOE"); m= ssc ("SINGING", "ING"); cout<< "TRUE \n"; TI. else cout<<"FALSE \n"; if (m--1) cout<< "TRUE \n"; TI. else coutくく"FALSE \n"; TI return 0

Default Term sh-4.3$ gt+ -std-c++11 -o main .cpp sh-4.3$ main TRUE TRUE
code:

#include <iostream>
//for string operstions
#include <string>
//includiing the namespace standard
using namespace std;
//function for calculating the substring
int ssc(const string& givenst, const string& stsub)
{
if (stsub.length() == 0) return 0;
int cnt = 0;
int n =2;
for (size_t offset = givenst.find(stsub); offset != string::npos;
   offset = givenst.find(stsub, offset + stsub.length()))
{
cnt++;
  
}
if(cnt==n)
{
return 1;
}
else
{
return 0;
}
  
}

int main()
{
int l,m;
l=ssc("TOE-TO-TOE", "TOE");
m= ssc("SINGING", "ING");
if(l==1)
{
cout<< "TRUE \n";
}
else
{
cout<<"FALSE \n";
}
if(m==1)
{
cout<< "TRUE \n";
}
else
{
cout<<"FALSE \n";
}
return 0;
}