Find Maximum and Minimum element in a Set in C++ STL

Given a Set, the task is to find the maximum and minimum element of this set in C++ STL. Examples:

Input: set={1, 6, 15, 10, 5}
Output: max = 15, min = 1

Input: set={10, 20, 30, 40, 50, 60}
Output: max = 60, min = 10

Using set.begin() and set.end() methods Approach: Elements in a set are stored in sorted order. So the minimum element of the set will reside in the first element and the maximum element in the last element. Therefore, this first and last element can be fetched with the help of set.begin() and set.end() methods respectively. Program:

CPP

#include <bits/stdc++.h> 

using namespace std; 
 
// Function to print the set 

void printSet(set<int> my_set) 
{ 
 
    // Print the set 

    cout << "Set: "; 

    for (auto i : my_set) 

        cout << i << " "; 
 
    cout << '\n'; 
} 
 
// Function to find the maximum element 

int findMax(set<int> my_set) 
{ 
 
    // Get the maximum element 

    int max_element; 

    if (!my_set.empty()) 

        max_element = *(my_set.rbegin()); 
 
    // return the maximum element 

    return max_element; 
} 
 
// Function to find the minimum element 

int findMin(set<int> my_set) 
{ 
 
    // Get the minimum element 

    int min_element; 

    if (!my_set.empty()) 

        min_element = *my_set.begin(); 
 
    // return the minimum element 

    return min_element; 
} 
 
int main() 
{ 
 
    // Get the set 

    set<int> my_set; 
 
    // Add the elements in the set 

    my_set.insert(1); 

    my_set.insert(6); 

    my_set.insert(15); 

    my_set.insert(10); 

    my_set.insert(5); 
 
    // Print the set 

    printSet(my_set); 
 
    // Get the minimum element 

    cout << "Minimum element: "

        << findMin(my_set) 

        << endl; 
 
    // Get the maximum element 

    cout << "Maximum element: "

        << findMax(my_set) 

        << endl; 
}

Output:

Set: 1 5 6 10 15 
Minimum element: 1
Maximum element: 15

Time Complexity: O(n)
Auxiliary Space: O(n)

Using set.rbegin() and set.rend() methods Approach: Elements in a set are stored in sorted order. So the minimum element of the set will reside in the first element and the maximum element in the last element. Therefore, this first and last element can be fetched with the help of set.rend() and set.rbegin() methods respectively. Program:

CPP

#include <bits/stdc++.h> 

using namespace std; 
 
// Function to print the set 

void printSet(set<int> my_set) 
{ 
 
    // Print the set 

    cout << "Set: "; 

    for (auto i : my_set) 

        cout << i << " "; 
 
    cout << '\n'; 
} 
 
// Function to find the maximum element 

int findMax(set<int> my_set) 
{ 
 
    // Get the maximum element 

    int max_element; 

    if (!my_set.empty()) 

        max_element = *my_set.rbegin(); 
 
    // return the maximum element 

    return max_element; 
} 
 
// Function to find the minimum element 

int findMin(set<int> my_set) 
{ 
 
    // Get the minimum element 

    int min_element; 

    if (!my_set.empty()) 

        min_element = *(--my_set.rend()); 
 
    // return the minimum element 

    return min_element; 
} 
 
int main() 
{ 
 
    // Get the set 

    set<int> my_set; 
 
    // Add the elements in the set 

    my_set.insert(1); 

    my_set.insert(6); 

    my_set.insert(15); 

    my_set.insert(10); 

    my_set.insert(5); 
 
    // Print the set 

    printSet(my_set); 
 
    // Get the minimum element 

    cout << "Minimum element: "

        << findMin(my_set) 

        << endl; 
 
    // Get the maximum element 

    cout << "Maximum element: "

        << findMax(my_set) 

        << endl; 
}

Output:

Set: 1 5 6 10 15 
Minimum element: 1
Maximum element: 15

Time Complexity: O(n)
Auxiliary Space: O(n)

Time Complexity: Both the methods are of constant time complexity. In set the maximum element is stored at last so we can return the last element using rbegin() method in O(1) time. Similarly, for minimum element using begin() method in O(1) time.

Article Tags :

C++

C++ Programs

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STL