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Maximum number of diamonds that can be gained in K minutes

  • Difficulty Level : Medium
  • Last Updated : 29 Jun, 2021
Geek Week

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:



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 minutes
void 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 Code
int 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 approach
import java.util.*;
 
class GFG{
 
// Function to find the maximum number
// of diamonds that can be gained in
// exactly K minutes
static 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 Code
public 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 minutes
def 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 Code
if __name__ == '__main__':
     
    A = [ 2, 1, 7, 4, 2 ]
    K = 3
    N = len(A)
     
    maxDiamonds(A, N, K)
 
# This code is contributed by SURENDRA_GANGWAR

Javascript




<script>
 
// JavaScript program for the above approach
 
 
// Function to find the maximum number
// of diamonds that can be gained in
// exactly K minutes
function 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 Code
 
let 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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