std::set_intersection in C++

The intersection of two sets is formed only by the elements that are present in both sets. The elements copied by the function come always from the first range, in the same order. The elements in the both the ranges shall already be ordered.

Examples:

Input :
5 10 15 20 25
50 40 30 20 10

Output :
The intersection has 2 elements :
10 20 



1. Comparing elements using “<":
Syntax :

Template :
OutputIterator set_intersection (InputIterator1 first1, InputIterator1 last1,
                                 InputIterator2 first2, InputIterator2 last2,
                                 OutputIterator result);

Parameters :

first1, last1
Input iterators to the initial and final positions of the first
sorted sequence. The range used is [first1, last1), which contains
all the elements between first1 and last1, including the element
pointed by first1 but not the element pointed by last1.
first2, last2
Input iterators to the initial and final positions of the second
sorted sequence. The range used is [first2, last2).

result
Output iterator to the initial position of the range where the
resulting sequence is stored.
The pointed type shall support being assigned the value of an
element from the first range.

Return Type :
An iterator to the end of the constructed range.
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// CPP program to illustrate
// std :: set_intersection
  
#include <bits/stdc++.h>
  
int main()
{
    int first[] = { 5, 10, 15, 20, 25 };
    int second[] = { 50, 40, 30, 20, 10 };
    int n = sizeof(first) / sizeof(first[0]);
  
    std::vector<int> v1(5);
    std::vector<int> v2(5);
    std::vector<int>::iterator it, ls;
  
    std::sort(first, first + 5);
    std::sort(second, second + 5);
  
    // Print elements
    std::cout << "First array :";
    for (int i = 0; i < n; i++)
        std::cout << " " << first[i];
    std::cout << "\n";
  
    // Print elements
    std::cout << "Second array :";
    for (int i = 0; i < n; i++)
        std::cout << " " << second[i];
    std::cout << "\n\n";
  
    // std :: set_intersection
    ls = std::set_intersection(first, first + 5, second, second + 5, v1.begin());
  
    std::cout << "The intersection has " << (ls - v1.begin()) << " elements:";
    for (it = v1.begin(); it != ls; ++it)
        std::cout << ' ' << *it;
    std::cout << "\n";
  
    return 0;
}

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Output:

First array : 5 10 15 20 25
Second array : 10 20 30 40 50

The intersection has 2 elements: 10 20

2. By comparing using a pre-defined function:
Synatax :

Template :
OutputIterator set_intersection (InputIterator1 first1, InputIterator1 last1,
                                 InputIterator2 first2, InputIterator2 last2,
                                 OutputIterator result, Compare comp);

Parameters :

first1, last1, first2, last2, result are same as mentioned above.

comp
Binary function that accepts two arguments of the types pointed
by the input iterators, and returns a value convertible to bool.
The function shall not modify any of its arguments.
This can either be a function pointer or a function object.
It follows the strict weak ordering to order the elements.

Return Type :
An iterator to the end of the constructed range.
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// CPP program to illustrate
// std :: set_intersection
  
#include <bits/stdc++.h>
  
bool comp(int a, int b)
{
    return a < b;
}
  
int main()
{
    int first[] = { 5, 10, 15, 20, 25 };
    int second[] = { 50, 40, 30, 20, 10 };
    int n = sizeof(first) / sizeof(first[0]);
  
    std::vector<int> v1(5);
    std::vector<int> v2(5);
    std::vector<int>::iterator it, ls;
  
    std::sort(first, first + 5);
    std::sort(second, second + 5);
  
    // Print elements
    std::cout << "First array :";
    for (int i = 0; i < n; i++)
        std::cout << " " << first[i];
    std::cout << "\n";
  
    // Print elements
    std::cout << "Second array :";
    for (int i = 0; i < n; i++)
        std::cout << " " << second[i];
    std::cout << "\n\n";
  
    // std :: set_intersection
    ls = std::set_intersection(first, first + 5, second, second + 5, v1.begin(), comp);
  
    std::cout << "The intersection has " << (ls - v1.begin()) << " elements:";
    for (it = v1.begin(); it != ls; ++it)
        std::cout << ' ' << *it;
    std::cout << "\n";
  
    return 0;
}

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Output:

First array : 5 10 15 20 25
Second array : 10 20 30 40 50
The intersection has 2 elements: 10 20

Possible Application : It is used to find the elements that are present only in both the sets.

1. It can be used to find the list of students that are present in both the classes.

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// CPP program to demonstrate use of
// std :: set_intersection
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
  
using namespace std;
  
// Driver code
int main()
{
    string first[] = { "Sachin", "Rakesh", "Sandeep", "Serena" };
    string second[] = { "Vaibhav", "Sandeep", "Rakesh", "Neha" };
    int n = sizeof(first) / sizeof(first[0]);
  
    // Print students of first list
    cout << "Students in first class :";
    for (int i = 0; i < n; i++)
        cout << " " << first[i];
    cout << "\n";
  
    // Print students of second list
    cout << "Students in second class :";
    for (int i = 0; i < n; i++)
        cout << " " << second[i];
    cout << "\n\n";
  
    vector<string> v(10);
    vector<string>::iterator it, st;
  
    // Sorting both the list
    sort(first, first + n);
    sort(second, second + n);
  
    // Using default operator<
    it = set_intersection(first, first + n, second, second + n, v.begin());
  
    cout << "Students attending both the classes only are :\n";
    for (st = v.begin(); st != it; ++st)
        cout << ' ' << *st;
    cout << '\n';
  
    return 0;
}

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OUTPUT :

Students in first class : Sachin Rakesh Sandeep Serena
Students in second class : Vaibhav Sandeep Rakesh Neha

Students attending both classes only are :
 Rakesh Sandeep

2. It can also be used to find the common elements in both the list.
Program is given above.
3.Complexity
Complexity is linear in the distance between [first1, last1) and [first2, last2): performs up to
2*(count1+count2)-1 comparisons. Where count1 = last1- first1 and count2 = last2- first2.
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