Find k most frequent in linear time

Given an array of integers, we need to print k most frequent elements. If there is a tie, we need to prefer the elements whose first appearance is first.

Examples:

Input : arr[] = {10, 5, 20, 5, 10, 10, 30}, k = 2
Output : 10 5

Input : arr[] = {7, 7, 6, 6, 6, 7, 5, 4, 4, 10, 5}, k = 3
Output : 7 6 5
Explanation :
In this example, 7 and 6 have same frequencies. We print 7 first because first appearance of 7 is first. Similarly, 5 and 4 have same frequencies. We prefer 5 because 5’s first appearance is first.

We have discussed two methods in the below post
Find k numbers with most occurrences in the given array



Let us first talk about a simple solution that prints in any order in case of tie. Then we will discuss the solution that takes of the order.

The idea is to use hashing with frequency indexing. We first store counts in a hash. Then we traverse through the hash and use frequencies as index to store elements with given frequencies. The important factor here is, maximum frequency can be n. So an array of size n+1 would be good.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation to find k numbers with most
// occurrences in the given array
#include <bits/stdc++.h>
using namespace std;
  
// funnction to print the k numbers with most occurrences
void print_N_mostFrequentNumber(int arr[], int n, int k)
{
    // unordered_map 'um' implemented as frequency
    // hash table
    unordered_map<int, int> um;
    for (int i = 0; i < n; i++)
        um[arr[i]]++;
  
    // Use frequencies as indexes and put
    // elements with given frequency in a
    // vector (related to a frequency)
    vector<int> freq[n + 1];
    for (auto x : um)
        freq[x.second].push_back(x.first);
  
    // Initialize count of items printed
    int count = 0;
  
    // Traverse the frequency array from
    // right side as we need the most
    // frequent items.
    for (int i = n; i >= 0; i--) {
  
        // Print items of current frequency
        for (int x : freq[i]) {
            cout << x << " ";
            count++;
            if (count == k)
                return;
        }
    }
}
  
// Driver program to test above
int main()
{
    int arr[] = { 3, 1, 4, 4, 5, 2, 6, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;
    print_N_mostFrequentNumber(arr, n, k);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation to find k elements with max occurence.
import java.util.*;
public class KFrequentNumbers {
    static void print_N_mostFrequentNumber(int[] arr, int n, int k)
    {
        Map<Integer, Integer> mp = new HashMap<Integer, Integer>();
  
        // Put count of all the distinct elements in Map
        // with element as the key & count as the value.
        for (int i = 0; i < n; i++) {
  
            // Get the count for the element if already
            // present in the Map or get the default value
            // which is 0.
            mp.put(arr[i], mp.getOrDefault(arr[i], 0) + 1);
        }
  
        // Initialize an array list of array lists
        List<List<Integer> > freq = new ArrayList<List<Integer> >();
        for (int i = 0; i <= n; i++)
            freq.add(new ArrayList<Integer>());
  
        // Use frequencies as indexes and add corresponding
        // values to the list
        for (Map.Entry<Integer, Integer> x : mp.entrySet())
            freq.get(x.getValue()).add(x.getKey());
  
        // Traverse freq[] from right side.
        int count = 0;
        for (int i = n; i >= 0; i--) {
            for (int x : freq.get(i)) {
                System.out.println(x);
                count++;
                if (count == k)
                    return;
            }
        }
    }
  
    // Driver Code to test the code.
    public static void main(String[] args)
    {
        int arr[] = { 3, 1, 4, 4, 5, 2, 6, 1 };
        int n = arr.length;
        int k = 2;
        print_N_mostFrequentNumber(arr, n, k);
    }
}

chevron_right


Output:

4 1

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

Printing according to first appearance. To keep the required order, we traverse the original array instead of map. To avoid duplicates, we need to mark processed entries as -1 in map.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation to find k numbers with most
// occurrences in the given array
#include <bits/stdc++.h>
using namespace std;
  
// function to print the k numbers with most occurrences
void print_N_mostFrequentNumber(int arr[], int n, int k)
{
    // unordered_map 'um' implemented as frequency
    // hash table
    unordered_map<int, int> um;
    for (int i = 0; i < n; i++)
        um[arr[i]]++;
  
    // Use frequencies as indexes and put
    // elements with given frequency in a
    // vector (related to a frequency)
    vector<int> freq[n + 1];
    for (int i = 0; i < n; i++) {
        int f = um[arr[i]];
        if (f != -1) {
            freq[f].push_back(arr[i]);
            um[arr[i]] = -1;
        }
    }
  
    // Initialize count of items printed
    int count = 0;
  
    // Traverse the frequency array from
    // right side as we need the most
    // frequent items.
    for (int i = n; i >= 0; i--) {
  
        // Print items of current frequency
        for (int x : freq[i]) {
            cout << x << " ";
            count++;
            if (count == k)
                return;
        }
    }
}
  
// Driver program to test above
int main()
{
    int arr[] = { 3, 1, 4, 4, 5, 2, 6, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    print_N_mostFrequentNumber(arr, n, k);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation to find k elements with max occurence.
import java.util.*;
public class KFrequentNumbers {
    static void print_N_mostFrequentNumber(int[] arr, int n, int k)
    {
        Map<Integer, Integer> mp = new HashMap<Integer, Integer>();
  
        // Put count of all the distinct elements in Map
        // with element as the key & count as the value.
        for (int i = 0; i < n; i++) {
  
            // Get the count for the element if already
            // present in the Map or get the default value
            // which is 0.
            mp.put(arr[i], mp.getOrDefault(arr[i], 0) + 1);
        }
  
        // Initialize an array list of array lists
        List<List<Integer> > freq = new ArrayList<List<Integer> >();
        for (int i = 0; i <= n; i++)
            freq.add(new ArrayList<Integer>());
  
        // Use frequencies as indexes and add corresponding
        // values to the list
        for (int i = 0; i < n; i++) {
            int f = mp.get(arr[i]);
            if (f != -1) {
                freq.get(f).add(arr[i]);
                mp.put(arr[i], -1);
            }
        }
  
        // Traverse freq[] from right side.
        int count = 0;
        for (int i = n; i >= 0; i--) {
            for (int x : freq.get(i)) {
                System.out.println(x);
                count++;
                if (count == k)
                    return;
            }
        }
    }
  
    // Driver Code to test the code.
    public static void main(String[] args)
    {
        int arr[] = { 3, 1, 4, 4, 5, 2, 6, 1 };
        int n = arr.length;
        int k = 3;
        print_N_mostFrequentNumber(arr, n, k);
    }
}

chevron_right


Output:

1 4 3

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.


Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.