Minimize Sum of an Array by at most K reductions
Given an array of integers arr[] consisting of N integers, the task is to minimize the sum of the given array by performing at most K operations, where each operation involves reducing an array element arr[i] to floor(arr[i]/2).
Examples :
Input: N = 4, a[] = {20, 7, 5, 4}, K = 3
Output: 17
Explanation:
Operation 1: {20, 7, 5, 4} -> {10, 7, 5, 4}
Operation 2: {10, 7, 5, 4} -> {5, 7, 5, 4}
Operation 3: {5, 7, 5, 4} -> {5, 3, 5, 4}
No further operation can be performed. Therefore, sum of the array = 17.Input: N = 4, a[] = {10, 4, 6, 16}, K = 2
Output: 23
Approach: To obtain the minimum possible sum, the main idea for every operation is to reduce the maximum element in the array before each operation. This can be implemented using MaxHeap. Follow the steps below to solve the problem:
- Insert all the array elements into MaxHeap.
- Pop the root of the MaxHeap and insert (popped element) / 2 into the MaxHeap
- After repeating the above step K times, pop the elements of the MaxHeap one by one and keep adding their values. Finally, print the sum.
Below is the implementation of above approach:
C++
// C++ program to implement the // above approach #include <bits/stdc++.h> using namespace std; // Function to obtain the minimum possible // sum from the array by K reductions int minSum( int a[], int n, int k) { priority_queue < int > q; // Insert elements into the MaxHeap for ( int i = 0; i < n; i++) { q.push(a[i]); } while (!q.empty() && k > 0) { int top = q.top() / 2; // Remove the maximum q.pop(); // Insert maximum / 2 q.push(top); k -= 1; } // Stores the sum of remaining elements int sum = 0; while (!q.empty()) { sum += q.top(); q.pop(); } return sum; } // Driver code int main() { int n = 4; int k = 3; int a[] = { 20, 7, 5, 4 }; cout << (minSum(a, n, k)); return 0; } // This code is contributed by jojo9911 |
Java
// Java Program to implement the // above approach import java.io.*; import java.util.*; class GFG { // Function to obtain the minimum possible // sum from the array by K reductions public static int minSum( int a[], int n, int k) { // Implements the MaxHeap PriorityQueue<Integer> maxheap = new PriorityQueue<>((one, two) -> two - one); // Insert elements into the MaxHeap for ( int i = 0 ; i < n; i++) maxheap.add(a[i]); while (maxheap.size() > 0 && k > 0 ) { // Remove the maximum int max_ele = maxheap.poll(); // Insert maximum / 2 maxheap.add(max_ele / 2 ); k -= 1 ; } // Stores the sum of remaining elements int sum = 0 ; while (maxheap.size() > 0 ) sum += maxheap.poll(); return sum; } // Driver Code public static void main(String[] args) { int n = 4 ; int k = 3 ; int a[] = { 20 , 7 , 5 , 4 }; System.out.println(minSum(a, n, k)); } } |
Python3
# Python3 program to implement the # above approach # Function to obtain the minimum possible # sum from the array by K reductions def minSum(a, n, k): q = [] # Insert elements into the MaxHeap for i in range (n): q.append(a[i]) q = sorted (q) while ( len (q) > 0 and k > 0 ): top = q[ - 1 ] / / 2 # Remove the maximum del q[ - 1 ] # Insert maximum / 2 q.append(top) k - = 1 q = sorted (q) # Stores the sum of remaining elements sum = 0 while ( len (q) > 0 ): sum + = q[ - 1 ] del q[ - 1 ] return sum # Driver code if __name__ = = '__main__' : n = 4 k = 3 a = [ 20 , 7 , 5 , 4 ] print (minSum(a, n, k)) # This code is contributed by mohit kumar 29 |
C#
// C# program to implement the // above approach using System; using System.Collections.Generic; class GFG{ // Function to obtain the minimum possible // sum from the array by K reductions static int minSum( int [] a, int n, int k) { // Implements the MaxHeap List< int > q = new List< int >(); for ( int i = 0; i < n; i++) { // Insert elements into the MaxHeap q.Add(a[i]); } q.Sort(); while (q.Count != 0 && k > 0) { int top = q[q.Count - 1] / 2; // Remove the maximum // Insert maximum / 2 q[q.Count - 1] = top; k--; q.Sort(); } // Stores the sum of remaining elements int sum = 0; while (q.Count != 0) { sum += q[0]; q.RemoveAt(0); } return sum; } // Driver Code static public void Main() { int n = 4; int k = 3; int [] a = { 20, 7, 5, 4 }; Console.WriteLine(minSum(a, n, k)); } } // This code is contributed by avanitrachhadiya2155 |
Javascript
<script> // Javascript Program to implement the // above approach // Function to obtain the minimum possible // sum from the array by K reductions function minSum(a,n,k) { // Implements the MaxHeap let maxheap = []; // Insert elements into the MaxHeap for (let i = 0; i < n; i++) maxheap.push(a[i]); maxheap.sort( function (a,b){ return a-b;}); while (maxheap.length > 0 && k > 0) { // Remove the maximum let max_ele = maxheap.pop(); // Insert maximum / 2 maxheap.push(Math.floor(max_ele / 2)); k -= 1; maxheap.sort( function (a,b){ return a-b;}); } // Stores the sum of remaining elements let sum = 0; while (maxheap.length > 0) sum += maxheap.shift(); return sum; } // Driver Code let n = 4; let k = 3; let a = [ 20, 7, 5, 4 ]; document.write(minSum(a, n, k)); // This code is contributed by unknown2108 </script> |
17
Time Complexity: O(Klog(N))
Auxiliary Space: O(N)
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