Find top k (or most frequent) numbers in a stream

Given an array of n numbers. Your task is to read numbers from the array and keep at-most K numbers at the top (According to their decreasing frequency) every time a new number is read. We basically need to print top k numbers sorted by frequency when input stream has included k distinct elements, else need to print all distinct elements sorted by frequency.

Examples:

Input : arr[] = {5, 2, 1, 3, 2}
k = 4
Output : 5 2 5 1 2 5 1 2 3 5 2 1 3 5
Explanation:



  1. After reading 5, there is only one element 5 whose frequency is max till now.
    so print 5.
  2. After reading 2, we will have two elements 2 and 5 with the same frequency.
    As 2, is smaller than 5 but their frequency is the same so we will print 2 5.
  3. After reading 1, we will have 3 elements 1, 2 and 5 with the same frequency,
    so print 1 2 5.
  4. Similarly after reading 3, print 1 2 3 5
  5. After reading last element 2 since 2 has already occurred so we have now a
    frequency of 2 as 2. So we keep 2 at the top and then rest of the element
    with the same frequency in sorted order. So print, 2 1 3 5.

Input : arr[] = {5, 2, 1, 3, 4}
k = 4
Output : 5 2 5 1 2 5 1 2 3 5 1 2 3 4
Explanation:

  1. After reading 5, there is only one element 5 whose frequency is max till now.
    so print 5.
  2. After reading 2, we will have two elements 2 and 5 with the same frequency.
    As 2, is smaller than 5 but their frequency is the same so we will print 2 5.
  3. After reading 1, we will have 3 elements 1, 2 and 5 with the same frequency,
    so print 1 2 5.
    Similarly after reading 3, print 1 2 3 5
  4. After reading last element 4, All the elements have same frequency
    So print, 1 2 3 4.

Approach: The idea is to store the top k elements with maximum frequency. To store them a vector or an array can be used. To keep the track of frequencies of elements create a HashMap to store element-frequency pair. Given a stream of numbers, when a new element appears in the stream update the frequency of that element in HashMap and put that element at the end of the list of K numbers (total k+1 elements) now compare adjacent elements of the list and swap if higher frequency element is stored next to it.

Algorithm:

  1. Create a Hashmap hm, and an array of k + 1 length.
  2. Traverse the input array from start to end.
  3. Insert the element at k+1 th position of the array, update the frequency of that element in HashMap.
  4. Now, traverse the temp array from start to end – 1
  5. For very element, compare the frequency and swap if higher frequency element is stored next to it, if the frequency is same then swap is the next element is greater.
  6. print the top k element in each traversal of original array.

Implementation:

C++

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// C++ program to find top k elements in a stream
#include <bits/stdc++.h>
using namespace std;
  
// Function to print top k numbers
void kTop(int a[], int n, int k)
{
    // vector of size k+1 to store elements
    vector<int> top(k + 1);
  
    // array to keep track of frequency
    unordered_map<int, int> freq;
  
    // iterate till the end of stream
    for (int m = 0; m < n; m++) {
        // increase the frequency
        freq[a[m]]++;
  
        // store that element in top vector
        top[k] = a[m];
  
        // search in top vector for same element
        auto it = find(top.begin(), top.end() - 1, a[m]);
  
        // iterate from the position of element to zero
        for (int i = distance(top.begin(), it) - 1; i >= 0; --i) {
            // compare the frequency and swap if higher
            // frequency element is stored next to it
            if (freq[top[i]] < freq[top[i + 1]])
                swap(top[i], top[i + 1]);
  
            // if frequency is same compare the elements
            // and swap if next element is high
            else if ((freq[top[i]] == freq[top[i + 1]])
                     && (top[i] > top[i + 1]))
                swap(top[i], top[i + 1]);
            else
                break;
        }
  
        // print top k elements
        for (int i = 0; i < k && top[i] != 0; ++i)
            cout << top[i] << ' ';
    }
    cout << endl;
}
  
// Driver program to test above function
int main()
{
    int k = 4;
    int arr[] = { 5, 2, 1, 3, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    kTop(arr, n, k);
    return 0;
}

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Java

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import java.io.*;
import java.util.*;
class GFG {
  
    // function to search in top vector for element
    static int find(int[] arr, int ele)
    {
        for (int i = 0; i < arr.length; i++)
            if (arr[i] == ele)
                return i;
        return -1;
    }
  
    // Function to print top k numbers
    static void kTop(int[] a, int n, int k)
    {
        // vector of size k+1 to store elements
        int[] top = new int[k + 1];
  
        // array to keep track of frequency
        HashMap<Integer, Integer> freq = new HashMap<>();
        for (int i = 0; i < k + 1; i++)
            freq.put(i, 0);
  
        // iterate till the end of stream
        for (int m = 0; m < n; m++) {
            // increase the frequency
            if (freq.containsKey(a[m]))
                freq.put(a[m], freq.get(a[m]) + 1);
            else
                freq.put(a[m], 1);
  
            // store that element in top vector
            top[k] = a[m];
  
            // search in top vector for same element
            int i = find(top, a[m]);
            i -= 1;
  
            // iterate from the position of element to zero
            while (i >= 0) {
                // compare the frequency and swap if higher
                // frequency element is stored next to it
                if (freq.get(top[i]) < freq.get(top[i + 1])) {
                    int temp = top[i];
                    top[i] = top[i + 1];
                    top[i + 1] = temp;
                }
  
                // if frequency is same compare the elements
                // and swap if next element is high
                else if ((freq.get(top[i]) == freq.get(top[i + 1])) && (top[i] > top[i + 1])) {
                    int temp = top[i];
                    top[i] = top[i + 1];
                    top[i + 1] = temp;
                }
  
                else
                    break;
                i -= 1;
            }
  
            // print top k elements
            for (int j = 0; j < k && top[j] != 0; ++j)
                System.out.print(top[j] + " ");
        }
        System.out.println();
    }
  
    // Driver program to test above function
    public static void main(String args[])
    {
        int k = 4;
        int[] arr = { 5, 2, 1, 3, 2 };
        int n = arr.length;
        kTop(arr, n, k);
    }
}
  
// This code is contributed by rachana soma

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Python

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# Python program to find top k elements in a stream
  
# Function to print top k numbers
def kTop(a, n, k):
  
    # list of size k + 1 to store elements
    top = [0 for i in range(k + 1)]
   
    # dictionary to keep track of frequency
    freq = {i:0 for i in range(k + 1)}
  
    # iterate till the end of stream
    for m in range(n):
  
        # increase the frequency
        if a[m] in freq.keys():
            freq[a[m]] += 1
        else:
            freq[a[m]] = 1
  
        # store that element in top vector
        top[k] = a[m]
   
        i = top.index(a[m])
        i -= 1
          
        while i >= 0:
  
            # compare the frequency and swap if higher
            # frequency element is stored next to it
            if (freq[top[i]] < freq[top[i + 1]]):
                t = top[i]
                top[i] = top[i + 1]
                top[i + 1] = t
              
            # if frequency is same compare the elements
            # and swap if next element is high
            elif ((freq[top[i]] == freq[top[i + 1]]) and (top[i] > top[i + 1])):
                t = top[i]
                top[i] = top[i + 1]
                top[i + 1] = t
            else:
                break
            i -= 1
          
        # print top k elements
        i = 0
        while i < k and top[i] != 0:
            print top[i],
            i += 1
    print
   
# Driver program to test above function
k = 4
arr = [ 5, 2, 1, 3, 2 ]
n = len(arr)
kTop(arr, n, k)
  
# This code is contributed by Sachin Bisht

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C#

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// C# program to find top k elements in a stream
using System;
using System.Collections.Generic;
  
class GFG {
    // function to search in top vector for element
    static int find(int[] arr, int ele)
    {
        for (int i = 0; i < arr.Length; i++)
            if (arr[i] == ele)
                return i;
        return -1;
    }
  
    // Function to print top k numbers
    static void kTop(int[] a, int n, int k)
    {
        // vector of size k+1 to store elements
        int[] top = new int[k + 1];
  
        // array to keep track of frequency
        Dictionary<int,
                   int>
            freq = new Dictionary<int,
                                  int>();
        for (int i = 0; i < k + 1; i++)
            freq.Add(i, 0);
  
        // iterate till the end of stream
        for (int m = 0; m < n; m++) {
            // increase the frequency
            if (freq.ContainsKey(a[m]))
                freq[a[m]]++;
            else
                freq.Add(a[m], 1);
  
            // store that element in top vector
            top[k] = a[m];
  
            // search in top vector for same element
            int i = find(top, a[m]);
            i--;
  
            // iterate from the position of element to zero
            while (i >= 0) {
                // compare the frequency and swap if higher
                // frequency element is stored next to it
                if (freq[top[i]] < freq[top[i + 1]]) {
                    int temp = top[i];
                    top[i] = top[i + 1];
                    top[i + 1] = temp;
                }
  
                // if frequency is same compare the elements
                // and swap if next element is high
                else if (freq[top[i]] == freq[top[i + 1]] && top[i] > top[i + 1]) {
                    int temp = top[i];
                    top[i] = top[i + 1];
                    top[i + 1] = temp;
                }
                else
                    break;
  
                i--;
            }
  
            // print top k elements
            for (int j = 0; j < k && top[j] != 0; ++j)
                Console.Write(top[j] + " ");
        }
        Console.WriteLine();
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        int k = 4;
        int[] arr = { 5, 2, 1, 3, 2 };
        int n = arr.Length;
        kTop(arr, n, k);
    }
}
  
// This code is contributed by
// sanjeev2552

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Output:

5 2 5 1 2 5 1 2 3 5 2 1 3 5

Complexity Analysis:

  • Time Complexity: O( n * k ).
    In each traversal the temp array of size k is traversed, So the time Complexity is O( n * k ).
  • Space Complexity:O(n).
    To store the elements in HashMap O(n) space is required.

This article is contributed by Niteesh Kumar. 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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