Sum of Array maximums after K operations by reducing max element to its half
Given an array arr[] of N integers and an integer K, the task is to find the sum of maximum of the array possible wherein each operation the current maximum of the array is replaced with its half.
Example:
Input: arr[] = {2, 4, 6, 8, 10}, K = 5
Output: 33
Explanation: In 1st operation, the maximum of the given array is 10. Hence, the value becomes 10 and the array after 1st operation becomes arr[] = {2, 4, 6, 8, 5}.
The value after 2nd operation = 18 and arr[] = {2, 4, 6, 4, 5}.
Similarly, proceeding forward, value after 5th operation will be 33.Input: arr[] = {6, 5}, K = 3
Output: 14
Approach: The given problem can be solved with the help of a greedy approach. The idea is to use a max heap data structure. Therefore, traverse the given array and insert all the elements in the array arr[] into a max priority queue. At each operation, remove the maximum from the heap using pop operation, add it to the value and reinsert the value after dividing it by two into the heap. The value after repeating the above operation K times is the required answer.
Below is the implementation of the above approach:
C++
// C++ program of the above approach#include <bits/stdc++.h>using namespace std;// Function to find maximum possible// value after K given operationsint maxValue(vector<int> arr, int K){ // Stores required value int val = 0; // Initializing priority queue // with all elements of array priority_queue<int> pq(arr.begin(), arr.end()); // Loop to iterate through // each operation while (K--) { // Current Maximum int max = pq.top(); pq.pop(); // Update value val += max; // Reinsert maximum pq.push(max / 2); } // Return Answer return val;}// Driver Callint main(){ vector<int> arr = { 2, 4, 6, 8, 10 }; int K = 5; cout << maxValue(arr, K);} |
Java
// Java code for the above approachimport java.util.Comparator;import java.util.PriorityQueue;class CustomComparator implements Comparator<Integer> { @Override public int compare(Integer number1, Integer number2) { int value = number1.compareTo(number2); // elements are sorted in reverse order if (value > 0) { return -1; } else if (value < 0) { return 1; } else { return 0; } }}class GFG { // Function to find maximum possible // value after K given operations static int maxValue(int[] arr, int K) { // Stores required value int val = 0; // Initializing priority queue // with all elements of array PriorityQueue<Integer> pq = new PriorityQueue<Integer>(new CustomComparator()); for (int i = 0; i < arr.length; i++) { pq.add(arr[i]); } // Loop to iterate through // each operation while (K != 0) { // Current Maximum int max = pq.poll(); // Update value val += max; // Reinsert maximum pq.add((int)Math.floor(max / 2)); K = K - 1; } // Return Answer return val; } // Driver Call public static void main(String[] args) { int[] arr = { 2, 4, 6, 8, 10 }; int K = 5; System.out.println(maxValue(arr, K)); }}// This code is conbtributed by Potta Lokesh |
Python3
# python3 program of the above approachfrom queue import PriorityQueue# Function to find maximum possible# value after K given operationsdef maxValue(arr, K): # Stores required value val = 0 # Initializing priority queue # with all elements of array pq = PriorityQueue() for dt in arr: pq.put(-1*dt) # Loop to iterate through # each operation while (True): # Current Maximum max = -1 * pq.get() # Update value val += max # Reinsert maximum pq.put(-1 * (max // 2)) K -= 1 if K == 0: break # Return Answer return val# Driver Callif __name__ == "__main__": arr = [2, 4, 6, 8, 10] K = 5 print(maxValue(arr, K)) # This code is contributed by rakeshsahni |
33
Time Complexity: O(K*log N)
Auxiliary Space: O(N)
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