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:
C++
#include <bits/stdc++.h> using namespace std;
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
priority_queue< int , vector< int >, greater< int > > pq;
// Sort the array in ascending order
sort(arr, arr + n);
// add max k elements to the heap
double sum = 0;
for ( int i = n - 1; i >= n - k; i--) {
pq.push(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.top()) {
int polled = pq.top();
pq.pop();
pq.push(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;
cout << max_avg << endl;
}
} int main()
{ int n = 4;
int k = 3;
int m = 4;
int arr[] = { 1, 2, 3, 4 };
int query[] = { 7, 2, 1, 5 };
max_average_k_numbers(n, k, m, arr, query);
return 0;
} |
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<Integer> pq = new PriorityQueue<Integer>();
// 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''' |
C#
using System;
using System.Collections.Generic;
using System.Linq;
public class MaxAverageKNumbers
{ public static void Calculate( int n, int k, int m, int [] arr, int [] query)
{
double maxAvg = 0.0;
// min-heap to maintain the max k elements at any point
// of time
var pq = new SortedSet< int >();
// Sort the array in ascending order
Array.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 += 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.Min)
{
int polled = pq.Min;
pq.Remove(polled);
pq.Add(query[i]);
// decrement the current sum by the polled
// element
sum -= polled;
// increment sum by the incoming element
sum += query[i];
}
// compute the average
maxAvg = sum / ( double )k;
Console.WriteLine(maxAvg);
}
}
public static void Main()
{
int n = 4;
int k = 3;
int m = 4;
int [] arr = { 1, 2, 3, 4 };
int [] query = { 7, 2, 1, 5 };
MaxAverageKNumbers.Calculate(n, k, m, arr, query);
}
} |
Javascript
function max_average_k_numbers(n, k, m, arr, query) {
let max_avg = 0.0;
// min-heap to maintain the max k elements at any point of time
const pq = new PriorityQueue();
// Sort the array in ascending order
arr.sort((a, b) => a - b);
// add max k elements to the heap
let sum = 0;
for (let i = n - 1; i >= n - k; i--) {
pq.push(arr[i]);
sum += arr[i];
}
// perform offline queries
for (let i = 0; i < m; i++) {
// if the minimum element in the heap is less than
// the incoming element
if (query[i] > pq.top()) {
const polled = pq.top();
pq.pop();
pq.push(query[i]);
// decrement the current sum by the polled element
sum -= polled;
// increment sum by the incoming element
sum += query[i];
}
// compute the average
max_avg = sum / k;
console.log(max_avg);
}
} class PriorityQueue { constructor() {
this .heap = [];
}
push(val) {
this .heap.push(val);
this .bubbleUp( this .heap.length - 1);
}
pop() {
const last = this .heap.pop();
const popped = this .heap[0];
if ( this .heap.length > 0) {
this .heap[0] = last;
this .bubbleDown(0);
}
return popped;
}
top() {
return this .heap[0];
}
bubbleUp(idx) {
const parent = Math.floor((idx - 1) / 2);
if (parent >= 0 && this .heap[idx] < this .heap[parent]) {
[ this .heap[idx], this .heap[parent]] = [ this .heap[parent], this .heap[idx]];
this .bubbleUp(parent);
}
}
bubbleDown(idx) {
const left = 2 * idx + 1;
const right = 2 * idx + 2;
let smallest = idx;
if (left < this .heap.length && this .heap[left] < this .heap[smallest]) {
smallest = left;
}
if (right < this .heap.length && this .heap[right] < this .heap[smallest]) {
smallest = right;
}
if (smallest !== idx) {
[ this .heap[idx], this .heap[smallest]] = [ this .heap[smallest], this .heap[idx]];
this .bubbleDown(smallest);
}
}
} // Driver Code const n = 4; const k = 3; const m = 4; const arr = [1, 2, 3, 4]; const query = [7, 2, 1, 5]; max_average_k_numbers(n, k, m, arr, query); // Contributed by sdeadityasharma |
Output
4.666666666666667 4.666666666666667 4.666666666666667 5.333333333333333
Complexity Analysis:
- Time Complexity: O(N(log(N))).
- Auxiliary Space: O(N) // N is the length of the array.