How to Check If a Set Contains an Element in C++?

In C++, the std::set container stores the unique elements in the sorted order. In this article, we will learn how to check if the set contains a given element or not.

Examples

Input: s = {1, 3, 5, 7, 9}, x = 5
Output: Element found
Explanation: The element 5 is present in the set.

Input: s = {10, 20, 30, 40}, x = 25
Output: Element not found
Explanation: The element 25 is not present in the set.

Following are the 4 different methods to check if a set contains an element or not in C++:

Using std::set::find() Method

C++ STL provides the std::set::find() method that searches the set for a given element and returns an iterator to the element if it is found. If it is not found, std::set::end() is returned. It is the member function of std::set() container so we can directly use it with any set.

Example

// C++ program to check if a set contains
// an element using set::find() method
#include <bits/stdc++.h>
using namespace std;

int main() {
    set<int> s = {1, 3, 5, 7, 9};
    int x = 5;

    // Check if the element is found
    if (s.find(x) != s.end()) {
        cout << "Element found";
    } else {
        cout << "Element not found";
    }
    return 0;
}

The Value 45 is Present

Time Complexity: O(log n), where n is the number of elements in the set.
Auxiliary Space: O(1)

Using std::set::count() Method

We can also use std::set::count() method to check whether the std::set contains an element or not. If the element present, it returns 1, otherwise, returns 0. It is also a member function of std::set container.

Example

// C++ program to check if a set contains
// an element using set::count() method
#include <bits/stdc++.h>
using namespace std;

int main() {
    set<int> s = {1, 3, 5, 7, 9};
    int x = 4;

    // Check if the element exists using count()
    if (s.count(x) > 0) {
        cout << "Element found";
    } else {
        cout << "Element not found";
    }
    return 0;
}

The Value 45 is Present

Time Complexity: O(log n), where n is the number of element in the set container.
Auxiliary Space: O(1)

Using set::contains() Method (C++ 20)

Since C++ 2, we can also use std::set::contains() method to check whether the std::set contains an element or not. If the element present, it returns true, otherwise, returns false.

Example

// C++ program to check if a set contains
// an element using set::contains()
#include <bits/stdc++.h>
using namespace std;

int main() {
    set<int> s = {1, 3, 5, 7, 9};
    int x = 7;

    // Check if element exists using
  	// set::contains()
    if (s.contains(x)) {
        cout << "Element found";
    } else {
        cout << "Element not found";
    }
    return 0;
}

Output

The Value 45 is Present

Time Complexity: O(log n), where n is the number of element in the set container
Auxiliary Space: O(1)

Manually Using Loop

Another way to check if a set contains an element is by manually iterating through the std::set container using a loop with iterators. By comparing each element with the value inside if statement, you can check whether the element is present in the set.

Example

// C++ program to check if a set contains
// an element using a manual loop
#include <bits/stdc++.h>
using namespace std;

int main() {
    set<int> s = {1, 3, 5, 7, 9};
    int x = 3;
    bool f = false;

    // Manually iterate over set to find x
    for (int i : s) {
        if (i == x) {
            f = true;
            break;
        }
    }

  	// Check if element exists
    if (f) {
        cout << "Element found";
    } else {
        cout << "Element not found";
    }

    return 0;
}

The Value 45 is Present

Time Complexity: O(n), where n is the number of element in the set container
Auxiliary Space: O(1)