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++ 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 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 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 |
// C# program for the above approach using 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 minutes static 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 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 rrrtnx. |
<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> |
14
Time Complexity: O((N + K)*log N)
Auxiliary Space: O(N)