Common character count giving unexpected problems - c++

Question

I have a simple problem where you get two strings where you need to see how many characters are common.
For s1 = "aabcc" and s2 = "dacaa", the output should be
solution(s1, s2) = 3.
I've seen a lot of people solve these types of problems, but I don't want to copy and paste. I want to learn it my way with my solution and improve upon that. I think my solution makes sense but it obviously doesn't, because it doesn't work.
This is my solution:
#include <iostream>
using namespace std;
int main()
{
int counter {};
string s1 = "aabcc";
string s2 = "dacaa";
for (int i {}; i < s1.size(); ++i)
{
for (int j {}; j < s2.size(); ++j)
{
if (s1.at(i) == s2.at(j));
{
cout << "Found character " << s1.at(i) << " at index " << i << " and " << s2.at(j) << " at index " << j << endl;
counter++;
s2.erase(s2.begin()+j);
break;
}
}
cout << s2 << endl;
}
cout << counter << endl;
return 0;
}
This is the output I get (I did a lot of "cout" for debugging purposes.
Found character a at index 0 and d at index 0
acaa
Found character a at index 1 and a at index 0
caa
Found character b at index 2 and c at index 0
aa
Found character a at index 3 and a at index 0
Found character a at index 4 and a at index 0
5
Let's talk about the first iteration. What I don't understand is how s1.at(i) is equal to s2.at(j) at index 0? s1.at(0) is "a" and s2.at(0) is "d". Since when was a = d?
This problem is driving me crazy and I would appreciate if you could explain why my solution acts like this.

Answer 1

There is a typo
if (s1.at(i) == s2.at(j));
Remove the semicolon.
In general the approach is bad because it changes one of the source strings.
I can suggest the following solution by means of using an additional container.
#include <iostream>
#include <string>
#include <string_view>
#include <utility>
#include <map>
#include <iterator>
#include <algorithm>
#include <numeric>
size_t common_letters( std::string_view s1, std::string_view s2 )
{
std::map<char, std::pair<size_t, size_t>> m;
for (char c : s1)
{
if ( m.count( c ) == 0 && s2.find( c ) != std::string_view::npos )
{
m[c].first = std::count( std::begin( s1 ), std::end( s1 ), c );
m[c].second = std::count( std::begin( s2 ), std::end( s2 ), c );
}
}
return std::accumulate( std::begin( m ), std::end( m ), ( size_t )0,
[]( const auto &acc, const auto &item )
{
return acc + std::min( item.second.first, item.second.second );
} );
}
int main()
{
std::string s1 = "aabcc";
std::string s2 = "dacaa";
std::cout << "common_letters( s1, s2 ) = " << common_letters(s1, s2) << '\n';
const char *s3 = "aabcc";
const char *s4 = "dacaa";
std::cout << "common_letters( s3, s4 ) = " << common_letters( s3, s4 ) << '\n';
}
The program output is
common_letters( s1, s2 ) = 3
common_letters( s3, s4 ) = 3
The if statement within the range-based for loop can be rewritten the following way to make the function more efficient
for (char c : s1)
{
if ( auto pos = std::string_view::npos;
m.count( c ) == 0 && ( pos = s2.find( c ) ) != std::string_view::npos )
{
m[c].first = std::count( std::begin( s1 ), std::end( s1 ), c );
m[c].second = std::count( std::next( std::begin( s2 ), pos ), std::end( s2 ), c );
}
}
Or if the compiler supports the C++ 20 then
for ( size_t i = 0; char c : s1)
{
if ( auto pos = std::string_view::npos;
m.count( c ) == 0 && ( pos = s2.find( c ) ) != std::string_view::npos )
{
m[c].first = std::count( std::next( std::begin( s1 ), i ), std::end( s1 ), c );
m[c].second = std::count( std::next( std::begin( s2 ), pos ), std::end( s2 ), c );
}
i++;
}
Or instead of the range-based for loop there can be used an ordinary for loop with iterators. That is the function can look the following way
size_t common_letters( std::string_view s1, std::string_view s2 )
{
std::map<char, std::pair<size_t, size_t>> m;
for ( auto first = std::begin( s1 ), last = std::end( s1 ); first != last; ++first )
{
if ( auto pos = std::string_view::npos;
m.count( *first ) == 0 && ( pos = s2.find( *first ) ) != std::string_view::npos )
{
m[*first] = { std::count( first , last, *first ),
std::count( std::next( std::begin( s2 ), pos ), std::end( s2 ), *first ) };
}
}
return std::accumulate( std::begin( m ), std::end( m ), ( size_t )0,
[]( const auto &acc, const auto &item )
{
return acc + std::min( item.second.first, item.second.second );
} );
}

Answer 2

There is a small typo in your code as Vlad from Moscow said. The reason why that small semicolon ruins the result is that then the code:
if (s1.at(i) == s2.at(j));
{ // From here
cout << "Found character " << s1.at(i) << " at index " << i << " and " << s2.at(j) << " at index " << j << endl;
counter++;
s2.erase(s2.begin()+j);
break;
} // To here
...is not considered as the part of the if statement because a semicolon refers to the end of a statement. So the semicolon ended the if statement, and as the code in the brackets is after the if statement has been ended by the semicolon, it is not considered as a part of the if statement. I hope the explanation is clear :).