Compare vector<string> and print yes or no using c++ - c++

Question

I have two string vectors a and b with a[0]="hello",a[1]="world",b[0]="heyyy"and b[1]="namaste" and i have to decide whether the following a[0] and b[0] has anything matching and a[1] and b[1] has anything matching. If a[0] and b[0] has one or more characters matching then print "YES", else print "NO". Similarly if a[1] and b[1] has one or more characters matching then print "YES", else print "NO". For example from the above information, a[0] and b[0] have 'h' and 'e' as matching and a[1] and b[1] has no character matching. At the end, the expected output is
"YES"
"NO"
Based on the above information, being a beginner in C++ i developed a C++ program which is not even partially correct. It would be great if someone solve this. Thanks in advance.
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main()
{
std::vector<string>a,b;
a={"hello", "world"};
b={"heyyy", "namaste"};
const char* acstr;
const char* bcstr;
acstr = a[0].c_str();
bcstr = b[0].c_str();
for(int i=0;i<sizeof(acstr);i++)
{
for(int j=0;j<sizeof(bcstr);j++)
{
if(acstr[i]==bcstr[j])
{
cout << "YES";
}
else{
continue;
}
}
}
return 0;
}

Answer 1

There is no need to convert an object of the type std::string to pointer like
acstr = a[0].c_str();
Also in this loop
for(int i=0;i<sizeof(acstr);i++)
the expression sizeof(acstr) does not give the length of the pointed string. It gives the size of the pointer itself that depending on the used system can be equal to 4 or 8 bytes independent on the length of the pointed string.
In general the vectors can have different number of elements. So you need to use the size of the smallest vector in a loop.
To determine whether a character is present in a string you can use method find of the class std::string.
Here is a demonstrative program.
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
int main()
{
std::vector<std::string> v1 = { "hello", "world" };
std::vector<std::string> v2 = { "heyyy", "namaste" };
std::vector<std::string>::size_type n = std::min( v1.size(), v2.size() );
for ( std::vector<std::string>::size_type i = 0; i < n; i++ )
{
std::string::size_type j = 0;
while ( j < v1[i].size() && v2[i].find( v1[i][j] ) == std::string::npos )
{
++j;
}
std::cout << ( j == v1[i].size() ? "NO" : "YES" ) << '\n';
}
return 0;
}
Its output is
YES
NO

Answer 2

Maybe you want like:
#include <bits/stdc++.h> // do not use this header
using namespace std;
int main()
{
std::vector<string>a, b;
a = { "hello", "world" };
b = { "heyyy", "namaste" };
for (int i = 0; i < (int)a.size(); i++)
{
bool bFound = false;
for (int j = 0; j < min(a[i].length(), b[i].length()); j++)
{
if (a[i][j] == b[i][j])
{
bFound = true;
}
}
cout << (bFound ? "YES" : "NO") << endl;
}
return 0;
}
Some advices:
You don't have to use c_str() to do character comparison.
sizeof(const char*) doesn't give you length of string. use strlen(const char*) if you want.

Answer 3

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main()
{
int flag = 0;
std::vector<string>a,b;
a={"hello", "world"};
b={"heyyy", "namaste"};
for(int k=0;k<2;k++){ //You can find and put the length of a or b, instead of using 2
for(int i=0;i<a[k].length();i++)
{
for(int j=0;j<b[k].length();j++)
{
if(a[k][i]==b[k][j])
{
flag = 1;
}
}
}
if(flag == 1){
cout << "YES\n";
}else{
cout << "NO\n";
}
flag = 0;
}
return 0;
}