Given an array arr[] consisting of N positive integers such that arr[i] represents that the ith bag contains arr[i] diamonds and a positive integer K, the task is to find the maximum number of diamonds that can be gained in exactly K minutes if dropping a bag takes 1 minute such that if a bag with P diamonds is dropped, then it changes to [P/2] diamonds, and P diamonds are gained.
Examples:
Input: arr[] = {2, 1, 7, 4, 2}, K = 3
Output: 14
Explanation:
The initial state of bags is {2, 1, 7, 4, 2}.
Operation 1: Take all diamonds from third bag i.e., arr[2](= 7), the state of bags becomes: {2, 1, 3, 4, 2}.
Operation 2: Take all diamonds from fourth bag i.e., arr[3](= 4), the state of bags becomes: {2, 1, 3, 2, 2}.
Operation 3: Take all diamonds from Third bag i.e., arr[2](= 3), the state of bags becomes{2, 1, 1, 2, 2}.
Therefore, the total diamonds gains is 7 + 4 + 3 = 14.Input: arr[] = {7, 1, 2}, K = 2
Output: 10
Approach: The given problem can be solved by using the Greedy Approach with the help of max-heap. Follow the steps below to solve the problem:
- Initialize a priority queue, say PQ, and insert all the elements of the given array into PQ.
- Initialize a variable, say ans as 0 to store the resultant maximum diamond gained.
- Iterate a loop until the priority queue PQ is not empty and the value of K > 0:
- Pop the top element of the priority queue and add the popped element to the variable ans.
- Divide the popped element by 2 and insert it into the priority queue PQ.
- Decrement the value of K by 1.
- After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum number// of diamonds that can be gained in// exactly K minutesvoid maxDiamonds(int A[], int N, int K){ // Stores all the array elements priority_queue<int> pq; // Push all the elements to the // priority queue for (int i = 0; i < N; i++) { pq.push(A[i]); } // Stores the required result int ans = 0; // Loop while the queue is not // empty and K is positive while (!pq.empty() && K--) { // Store the top element // from the pq int top = pq.top(); // Pop it from the pq pq.pop(); // Add it to the answer ans += top; // Divide it by 2 and push it // back to the pq top = top / 2; pq.push(top); } // Print the answer cout << ans;}// Driver Codeint main(){ int A[] = { 2, 1, 7, 4, 2 }; int K = 3; int N = sizeof(A) / sizeof(A[0]); maxDiamonds(A, N, K); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the maximum number// of diamonds that can be gained in// exactly K minutesstatic void maxDiamonds(int A[], int N, int K){ // Stores all the array elements PriorityQueue<Integer> pq = new PriorityQueue<>( (a, b) -> b - a); // Push all the elements to the // priority queue for(int i = 0; i < N; i++) { pq.add(A[i]); } // Stores the required result int ans = 0; // Loop while the queue is not // empty and K is positive while (!pq.isEmpty() && K-- > 0) { // Store the top element // from the pq int top = pq.peek(); // Pop it from the pq pq.remove(); // Add it to the answer ans += top; // Divide it by 2 and push it // back to the pq top = top / 2; pq.add(top); } // Print the answer System.out.print(ans);}// Driver Codepublic static void main(String[] args){ int A[] = { 2, 1, 7, 4, 2 }; int K = 3; int N = A.length; maxDiamonds(A, N, K);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approach# Function to find the maximum number# of diamonds that can be gained in# exactly K minutesdef maxDiamonds(A, N, K): # Stores all the array elements pq = [] # Push all the elements to the # priority queue for i in range(N): pq.append(A[i]) pq.sort() # Stores the required result ans = 0 # Loop while the queue is not # empty and K is positive while (len(pq) > 0 and K > 0): pq.sort() # Store the top element # from the pq top = pq[len(pq) - 1] # Pop it from the pq pq = pq[0:len(pq) - 1] # Add it to the answer ans += top # Divide it by 2 and push it # back to the pq top = top // 2; pq.append(top) K -= 1 # Print the answer print(ans)# Driver Codeif __name__ == '__main__': A = [ 2, 1, 7, 4, 2 ] K = 3 N = len(A) maxDiamonds(A, N, K)# This code is contributed by SURENDRA_GANGWAR |
C#
// C# program for the above approachusing System;using System.Collections;using System.Collections.Generic;class GFG{// Function to find the maximum number// of diamonds that can be gained in// exactly K minutesstatic void maxDiamonds(int []A, int N, int K){ // Stores all the array elements var pq = new List<int>(); // Push all the elements to the // priority queue for(int i = 0; i < N; i++) { pq.Add(A[i]); } // Stores the required result int ans = 0; // Loop while the queue is not // empty and K is positive while (pq.Count!=0 && K-- > 0) { pq.Sort(); // Store the top element // from the pq int top = pq[pq.Count-1]; // Pop it from the pq pq.RemoveAt(pq.Count-1); // Add it to the answer ans += top; // Divide it by 2 and push it // back to the pq top = top / 2; pq.Add(top); } // Print the answer Console.WriteLine(ans);}// Driver Codepublic static void Main(string[] args){ int []A= { 2, 1, 7, 4, 2 }; int K = 3; int N = A.Length; maxDiamonds(A, N, K);}}// This code is contributed by rrrtnx. |
Javascript
<script>// JavaScript program for the above approach// Function to find the maximum number// of diamonds that can be gained in// exactly K minutesfunction maxDiamonds(A, N, K) { // Stores all the array elements let pq = []; // Push all the elements to the // priority queue for (let i = 0; i < N; i++) { pq.push(A[i]); } // Stores the required result let ans = 0; // Loop while the queue is not // empty and K is positive pq.sort((a, b) => a - b) while (pq.length && K--) { pq.sort((a, b) => a - b) // Store the top element // from the pq let top = pq[pq.length - 1]; // Pop it from the pq pq.pop(); // Add it to the answer ans += top; // Divide it by 2 and push it // back to the pq top = Math.floor(top / 2); pq.push(top); } // Print the answer document.write(ans);}// Driver Codelet A = [2, 1, 7, 4, 2];let K = 3;let N = A.length;maxDiamonds(A, N, K);</script> |
Output:
14
Time Complexity: O((N + K)*log N)
Auxiliary Space: O(N)
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