unorderd_set in STL and its applications

unordered_set is implemented using hash table where keys are hashed into indices of this hash table so it is not possible to maintain an order. All operation on unordered_set takes constant time O(1) on an average which can go up to linear time in worst case which depends on the internally used hash function but practically they perform very well and generally provide constant time lookup operation.
The unordered-set can contain key of any type – predefined or user-defined data structure but when we define key of type user define type, we need to specify our comparison function according to which keys will be compared.

set vs unordered_set
Set set is an ordered sequence of unique keys whereas unordered_set is a set in which key can be stored in any order, so unordered.
Set is implemented as balanced tree structure that is why it is possible to maintain an order between the elements (by specific tree traversal). Time complexity of set operations is O(Log n) while for unordered_set, it is O(1).

Methods on unordered_set
For unordered_set many function are defined among which most useful are size and empty for capacity, find for searching a key, insert and erase for modification.
The Unordered_set allows only unique keys, for duplicate keys unordered_multiset should be used.

Example of declaration, find, insert and iteration in unordered_set is given below :

// C++ program to demonstrate various function of unordered_set
#include <bits/stdc++.h>
using namespace std;

int main()
    // declaring set for storing string data-type
    unordered_set<string> stringSet;

    // inserting various string, same string will be stored
    // once in set

    string key = "slow";

    //     find returns end iterator if key is not found,
    //  else it returns iterator to that key
    if (stringSet.find(key) == stringSet.end())
        cout << key << " not found\n\n";
        cout << "Found " << key << endl << endl;

    key = "c++";
    if (stringSet.find(key) == stringSet.end())
        cout << key << " not found\n";
        cout << "Found " << key << endl;

    // now iterating over whole set and printing its
    // content
    cout << "\nAll elements : ";
    unordered_set<string> :: iterator itr;
    for (itr = stringSet.begin(); itr != stringSet.end(); itr++)
        cout << (*itr) << endl;

Output :

slow not found

Found c++

All elements : 

Find, insert and erase take constant amount of time on average. The find() function returns an iterator to end() if key is not there in set otherwise iterator to key position is returned. The iterator works as pointer to key values so we can get key by dereferencing them by *.

A practical problem based on unordered_set – given a set of integer number, find duplicate among them.

Input  : arr[] = {1, 5, 2, 1, 4, 3, 1, 7, 2, 8, 9, 5}
Output : Duplicate item are : 5 2 1 

Below is C++ solution using unordered_set.

// C++ program to find duplicate from an array using
// unordered_set
#include <bits/stdc++.h>
using namespace std;

// Print duplicates in arr[0..n-1] using unordered_set
void printDuplicates(int arr[], int n)
    // declaring unordered sets for checking and storing
    // duplicates
    unordered_set<int> intSet;
    unordered_set<int> duplicate;

    // looping through array elements
    for (int i = 0; i < n; i++)
        // if element is not there then insert that
        if (intSet.find(arr[i]) == intSet.end())

        // if element is already there then insert into
        // duplicate set

    // printing the result
    cout << "Duplicate item are : ";
    unordered_set<int> :: iterator itr;

    // iterator itr loops from begin() till end()
    for (itr = duplicate.begin(); itr != duplicate.end(); itr++)
        cout << *itr << " ";

// Driver code
int main()
    int arr[] = {1, 5, 2, 1, 4, 3, 1, 7, 2, 8, 9, 5};
    int n = sizeof(arr) / sizeof(int);

    printDuplicates(arr, n);
    return 0;

Output :

Duplicate item are : 5 2 1 

Recent articles on unordered_set

This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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Improved By : Sai Harish Pathuri

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