How to compare and display how many numbers in two one-dimensional arrays are equal [C / C++]?

Question

C++ / C#

There are two arrays arr[size] and arr_2[size] /* size = 7 */ Please tell me how to compare these two arrays using C / C++ and display the number of equal numbers in them. For example: arr{1, 2, 3, 4, 5, 6, 7}, and arr_2{1, 2, 3, 44, 55, 66, 77} And in cout outputs "There are 3 equal numbers in these arrays". Help, I will be very grateful

Answer 1

The simplest and most understandable are two loops nested inside each other. Those. You take the zero element of the first array, compare with each element of the second. Then the first one, etc.
In c ++, you can pervert and write it in one
construction

inline int count_cpp()
{
    return count_if(begin(a), end(a), [](int c) {static set<int> s(begin(b), end(b)); return  s.find(c) != s.end();});
}


inline int count_c()
{
    int c = 0;
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            if(a[i] == b[j]) {
                ++ c;
                break;
            }
        }
    }
    return c;
}


inline int count_map()
{
    int c = 0;
    map<int, bool> bm;
    for (auto i : b) {
        bm[i] = true;
    }
    for (auto i : a) {
        if (bm.find(i) != bm.end()) {
            ++ c;
        }
    }
    return c;
}

Execution time for 10000 random elements in the range 0-10000

[cpp method]      Time:   4006 us.
[pure c method]   Time:   71019 us.
[map method]     Time:   5178 us.

Answer 2

A good method that works in O(n log(n)): https://ru.cppreference.com/w/cpp/algorithm/set_in... (see code example at the very bottom)
The main thing is not to forget to pre-sort the arrays , because this function itself only compares.