Minimum Number of Distinct Elements after Removing M items

Approach

The idea is to store each element's occurrence in a hash and then count the occurrence of each frequency in a hash again. Then we will sort the frequency of occurrences of each element in increasing order and will start removing the element whose frequency is less or equal to m.

Algorithm

1. Create a hashmap to store the occurrence of elements.
2. Count the occurrence of all elements and store them in the hash.
3. Sort the hash in increasing order.
4. Remove the elements whose frequency is less than m from the hashmap.
5. Print the number of elements present in the hash.

Implementation in C++

#include <bits/stdc++.h>
#include <algorithm>
using namespace std;

#define pb push_back

int hash_fun(int arr[], int n, int m)
{
    int temp = 0;
    // hashmap for storing the occurence
    unordered_map<int, int> mp;
    vector<pair<int, int>> v;

    for (int i = 0; i < n; i++)
        mp[arr[i]]++;

    // use vector to store the pair key with its occurrence
    for (auto i = mp.begin(); i != mp.end(); i++)
        v.pb(make_pair(i->second, i->first));

    sort(v.begin(), v.end()); // sorting the vector

    int vec_size = v.size();

    for (int i = 0; i < vec_size; i++)
    {
        // Remove the value if it is less than or equal to m
        // else return the remaining size
        if (v[i].first <= m)
        {
            m = m - v[i].first;
            temp++;
        }
        else
            return vec_size - temp;
    }
    return vec_size - temp;
}

int main()
{
    int m;
    int arr[] = {3, 1, 2, 2, 3, 3};
    int size = sizeof(arr) / sizeof(arr[0]);

    cout << "Enter the value of m : ";
    cin >> m;
    cout << endl;

    cout << hash_fun(arr, size, m);
    return 0;
}

You can also try this code with Online C++ Compiler

Run Code

Input

Enter the value of m : 3

Output

1

Time Complexity

The time complexity of the above approach is O(NlogN) because we have used sorting, which marks the upper bound for time complexity.

Space Complexity

The space complexity of the approach is O(N), as we are storing the elements as a key in the hashmap.

You can also read about the Longest Consecutive Sequence.

Frequently Asked Questions

What is HashMap data structure?

A HashMap is a data structure that can map certain keys to a specific value. The values and keys could be anything.
 

Is sort in C++ STL?

A C++ Standard Template Library(STL) includes a built-in function called sort(). The elements in the range can be sorted using this function in either ascending or descending order.
 

What is the time complexity of sort function in C++ STL?

The worst-case time complexity of sort() is in O(n log n), where n is the number of sorted elements.
 

Also Read - Strong number in c

Conclusion

In this article, we have extensively discussed the problem of printing the minimum number of distinct elements after removing m items. We solved the problem using the hashmap data structure and discussed its time as well as space complexity.

We hope that this blog has helped you enhance your knowledge about the count of minimum distinct elements and if you like to learn more, check out our articles Implementation of HashMap, Sorting Based Problems, STL in C++, and many more on our Website.

Recommended Problems - 

• Sort 0 1 2 
• Wiggle sort
• Find first and last position in sorted array
   

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