Question

Example :

S1 : abcde
S2: cdef

Answer : cde

Answer 1

It’s reasonable to assume the set of characters is small and finite compared to the potential string length. So, handling 8-bit characters for example (in roughly-C++ pseudo-code):

bool in_a[256] = { false };
bool in_b[256] = { false };
for (int i = 0; i < a.size(); ++i)
    in_a[a[i]] = true;
for (int i = 0; i < b.size(); ++i)
    in_b[b[i]] = true;
// and in_a and in_b
for (int i = 0; i < b.size(); ++i)
    if (in_a[i] && in_b[i])
    {
        std::cout << i;
        if (isprint(i)) std::cout << '\'' << (char)i << '\'';
        std::cout << ' ';
    }
std::cout << '\n';

Note that using a hash table rather than an array is a huge waste of time (unless handling say a 32-bit character representation).

An implementation of the simplification suggested in Yordan’s comment follows. This avoids the final loop from 0..255 and needs only one tracking array. The order of results is unsorted.

#include <vector>
#include <iostream>

int main()
{
    std::string a = "abdcdefg";
    std::string b = "byfdz";
    bool track[256] = { false };
    for (int i = 0; i < a.size(); ++i)
        track[a[i]] = true;
    for (int i = 0; i < b.size(); ++i)
        if (track[b[i]])
        {
            track[b[i]] = false; // don't match again
            std::cout << (int)b[i];
            if (isprint(b[i])) std::cout << " \'" << (char)(b[i]) << "\'";
            std::cout << ' ';
        }
    std::cout << '\n';
}