Minimum number of distinct elements after removing m items

2.1

Given an array of items, an i-th index element denotes the item id’s and given a number m, the task is to remove m elements such that there should be minimum distinct id’s left.Print the number of distinct id’s.

Examples:

Input : arr[] = { 2, 2, 1, 3, 3, 3} 
            m = 3
Output : 1
Remove 1 and both 2's.So, only 3 will be 
left that's why distinct id is 1.

Input : arr[] = { 2, 4, 1, 5, 3, 5, 1, 3} 
            m = 2
Output : 3
Remove 2 and 4 completely. So, remaining ids 
are 1, 3 and 5 i.e. 3

Asked in : Morgan Stanley

1- Count the occurrence of elements and store in the hash.
2- Sort the hash.
3- Start removing elements from hash.
4- Return the number of values left in the hash.

C++

// C++ program for above implementation
#include <bits/stdc++.h>
using namespace std;

// Function to find distintc id's
int distinctIds(int arr[], int n, int mi)
{
    unordered_map<int, int> m;
    vector<pair<int, int> > v;
    int count = 0;

    // Store the occurrence of ids
    for (int i = 0; i < n; i++)
        m[arr[i]]++;

    // Store into the vector second as first and vice-versa
    for (auto it = m.begin(); it != m.end(); it++)
        v.push_back(make_pair(it->second, it->first));

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

    int size = v.size();

    // Start removing elements from the beginning
    for (int i = 0; i < size; i++) {

        // Remove if current value is less than 
        // or equal to mi
        if (v[i].first <= mi) {
            mi -= v[i].first;
            count++;
        }

        // Return the remaining size
        else
            return size - count;
    }
    return size - count;
}

// Driver code
int main()
{
    int arr[] = { 2, 3, 1, 2, 3, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);

    int m = 3;

    cout << distinctIds(arr, n, m);
    return 0;
}

Java

//Java program for Minimum number of
//distinct elements after removing m items
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;

public class DistinctIds
{
    // Function to find distintc id's
    static int distinctIds(int arr[], int n, int mi)
    {

        Map<Integer, Integer> m = new HashMap<Integer, Integer>();
        int count = 0;
        int size = 0;

        // Store the occurrence of ids
        for (int i = 0; i < n; i++)
        {

            // If the key is not add it to map
            if (m.containsKey(arr[i]) == false)
            {
                m.put(arr[i], 1);
                size++;
            }

            // If it is present then increase the value by 1
            else m.put(arr[i], m.get(arr[i]) + 1);
        }

        // Start removing elements from the beginning
        for (Entry<Integer, Integer> mp:m.entrySet())
        {
            // Remove if current value is less than
            // or equal to mi
            if (mp.getKey() <= mi)
            {
                mi -= mp.getKey();
                count++;
            }
            // Return the remaining size
            else return size - count;
        }

        return size - count;
    }

    //Driver method to test above function
    public static void main(String[] args)
    {
        // TODO Auto-generated method stub
        int arr[] = {2, 3, 1, 2, 3, 3};
        int m = 3;

        System.out.println(distinctIds(arr, arr.length, m));
    }
}
//This code is contributed by Sumit Ghosh


Output:

1

Time Complexity : O(n log n)

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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