Set

A set container stores only unique values in a sorted fashion. A set organizes the values using the value as a key. The set container is immutable, that is, the values stored in a set can't be modified; however, the values can be deleted. A set generally uses a red-black tree data structure, which is a form of balanced BST. The time complexity of set operations are guaranteed to be O ( log N ). 

Let's write a simple program using a set:

#include <iostream>
#include <set>
#include <vector>
#include <iterator>
#include <algorithm>
using namespace std;

int main( ) {
    set<int> s1 = { 1, 3, 5, 7, 9 };
    set<int> s2 = { 2, 3, 7, 8, 10 };

    vector<int> v( s1.size() + s2.size() );
        
    cout << "\nFirst set values are ..." << endl;
    copy ( s1.begin(), s1.end(), ostream_iterator<int> ( cout, "\t" ) );
    cout << endl;

    cout << "\nSecond set values are ..." << endl;
    copy ( s2.begin(), s2.end(), ostream_iterator<int> ( cout, "\t" ) );
    cout << endl;

    auto pos = set_difference ( s1.begin(), s1.end(), s2.begin(), s2.end(), v.begin() ); 
    v.resize ( pos - v.begin() );

    cout << "\nValues present in set one but not in set two are ..." << endl;
    copy ( v.begin(), v.end(), ostream_iterator<int> ( cout, "\t" ) );
    cout << endl; 

    v.clear();

    v.resize ( s1.size() + s2.size() );

    pos = set_union ( s1.begin(), s1.end(), s2.begin(), s2.end(), v.begin() );

    v.resize ( pos - v.begin() );

    cout << "\nMerged set values in vector are ..." << endl;
    copy ( v.begin(), v.end(), ostream_iterator<int> ( cout, "\t" ) );
    cout << endl; 

    return 0;
}

The output can be viewed with the following command:

./a.out

The output of the program is as follows:

First set values are ...
1   3   5   7   9

Second set values are ...
2   3   7   8   10

Values present in set one but not in set two are ...
1   5   9

Merged values of first and second set are ...
1   2   3   5   7   8   9  10