# Average of max K numbers in a stream

• Difficulty Level : Easy
• Last Updated : 04 Aug, 2022

Given a list of N numbers, and an integer ‘K’. The task is to print the average of max ‘K’ numbers after each query where a query consists of an integer element that needs to be added to the list of elements.

Note: The queries are defined with an integer array ‘q’

Examples:

Input: N = 4, K = 3, arr = {1, 2, 3, 4}, q = {7, 2, 1, 5}

Output: 4.666666 4.666666 4.666666 5.333333
After query 1, arr = {1, 2, 3, 4, 7} and
the average of max K (i.e. {3, 4, 7}) elements is 4.666666.
After query 2, arr = {1, 2, 3, 4, 7, 2} and
the average is 4.666666 for {3, 4, 7}.
After query 3, arr = {1, 2, 3, 4, 7, 2, 1} and
the average is 4.666666 for {3, 4, 7}.
After query 4, arr = {1, 2, 3, 4, 7, 2, 5} and
the average is 5.333333 for {4, 5, 7}.
Input: N = 5, K = 4, arr = {1, 2, 2, 3, 3}, q = {2, 5, 1}
Output: 2.5 3.25 3.25

### Approach:

Heap (Min Heap) data structure can be used to solve problems like these where insertion and deletions of the elements can be performed in O(log n) time.

• Initially, store the max k elements from the given list of elements in the min heap.
• If the incoming element is less than or equal to the element currently at the root of the min heap then discard the element as it’ll have no effect on the average.
• Else if, if the number is greater than the root element then remove the root of the min heap followed by insertion of the new element, and then calculate the average of the elements currently in the heap.
• Print the average and repeat the above two steps for all incoming elements.

Below is the implementation of the above approach:

## Java

 // Java implementation of the approach import java.util.*;   class GFG {       // Function that returns the     // average of max k elements in     // the list after each query     static void max_average_k_numbers(int n,                                       int k,                                       int m,                                       int[] arr,                                       int[] query)     {         double max_avg = 0.0;           // min-heap to maintain         // the max k elements at         // any point of time;         PriorityQueue pq = new PriorityQueue();           // Sort the array         // in ascending order         Arrays.sort(arr);           // add max k elements         // to the heap         double sum = 0;         for (int i = n - 1; i >= n - k; i--) {             pq.add(arr[i]);             sum = sum + arr[i];         }           // perform offline queries         for (int i = 0; i < m; i++) {               // if the minimum element in             // the heap is less than             // the incoming element             if (query[i] > pq.peek()) {                 int polled = pq.poll();                 pq.add(query[i]);                   // decrement the current                 // sum by the polled element                 sum = sum - polled;                   // increment sum by the                 // incoming element                 sum = sum + query[i];             }               // compute the average             max_avg = sum / (double)k;             System.out.println(max_avg);         }     }       // Driver code     public static void main(String[] args)     {         int n = 4;         int k = 3;         int m = 4;         int[] arr = new int[] { 1, 2, 3, 4 };         int[] query = new int[] { 7, 2, 1, 5 };           max_average_k_numbers(n, k, m, arr, query);     } }

## Python3

 # implementation of the approach # importing heapq module import heapq   # Function that returns the # average of max k elements in # the list after each query def max_average_k_numbers(n, k, m, arr, query):     max_avg = 0.0           # min-heap to maintain     # the max k elements at     # any point of time     pq = []     Sum = 0           # Sort the array in ascending order     arr.sort()           # add max k elements to heap pq     for i in range(n - 1, n - k - 1, -1):         pq.append(arr[i])         Sum += arr[i]               # heapify the heap pq for maintaining the     # heap property     heapq.heapify(pq)           # perform offline queries     for i in range(m):                 # if the minimum element in         # the heap is less than          # the incoming element         if query[i] > pq[0]:             polled = pq[0]             pq[0] = pq[-1]             pq.pop()                           # heapq.heapify(pq)             pq.append(query[i])                           # decrement the current             # sum by the polled element             Sum -= polled                           # increment sum by the             # incoming element             Sum += query[i]                           # Again maintaining the heap property             heapq.heapify(pq)                       # compute the average         max_avg = Sum/float(k)         print(max_avg)     # Driver Code if __name__ == '__main__':     n = 4     k = 3     m = 4     arr = [1, 2, 3, 4]     query = [7, 2, 1, 5]     max_average_k_numbers(n, k, m, arr, query)   '''This Code is written By RAJAT KUMAR'''

Output:

4.666666666666667
4.666666666666667
4.666666666666667
5.333333333333333

Time Complexity: O(N(log(N)))

Auxiliary Space: O(N) // N is the length of the array

My Personal Notes arrow_drop_up
Recommended Articles
Page :