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Maximize cost to empty given array by repetitively removing K array elements
  • Difficulty Level : Hard
  • Last Updated : 07 Dec, 2020

Given an array, arr[] of size N and an integer K ( N % K = 0), the task is find the maximum cost to remove all the array elements. In each operation, exactly K array elements can be removed and the cost of removal is equal to the second smallest element removed.

Examples:

Input: arr[] = { 1, 3, 4, 1, 5, 1, 5, 3 }, K = 4 
Output:
Explanation: 
Removing {arr[0], arr[3], arr[5], arr[7]} modifies arr[] to {3, 4, 5, 5}. Second smallest element removed = 1. Therefore, cost = 1 
Removing {arr[0], arr[1], arr[2], arr[3]} modifies arr[] to {}. Second smallest element removed = 4. Therefore, cost = 1 + 4 = 5 
Therefore, the required output is = 5

Input: arr[] = { 1, 2, 3, 4}, K = 4 
Output: 2

Approach: The problem can be solved using greedy technique. The idea is to sort the array and repetitively remove theK smallest array elements at each operation. Follow the steps below to solve the problem:



  • Initialize a variable, say maxCost, to store the maximum cost to remove all the array elements.
  • Sort the array in ascending order.
  • Traverse the array using variable i and update the value of maxCost = arr[i + 1] and i = i + K.
  • Finally, print the value of maxCost.

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximu cost to
// remove all array elements
int maxCostToRemove(int arr[], int N,
                             int K)
{
     
    // Stores maximum cost to
    // remove array elements
    int maxCost = 0;
     
     
    // Sort array in
    // ascending order   
    sort(arr, arr + N);
     
    // Traverse the array
    for (int i = 0; i < N;
                i += K) {
         
        // Update maxCost
        maxCost += arr[i + 1];
    }
     
    return maxCost;
}
 
// Driver Code
int main()
{
    int arr[]= { 1, 3, 4, 1, 5, 1, 5, 3};
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 4;
    cout<< maxCostToRemove(arr, N, K);
}

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Java

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// Java program for the above approach
import java.io.*;
import java.util.Arrays;
   
class GFG{
   
// Function to find the maximu cost to
// remove all array elements
static int maxCostToRemove(int arr[], int N,
                             int K)
{
      
    // Stores maximum cost to
    // remove array elements
    int maxCost = 0;
      
      
    // Sort array in
    // ascending order   
    Arrays.sort(arr);
      
    // Traverse the array
    for (int i = 0; i < N;
                i += K) {
          
        // Update maxCost
        maxCost += arr[i + 1];
    }
      
    return maxCost;
}
   
// Drive Code
public static void main(String[] args)
{
    int arr[]= { 1, 3, 4, 1, 5, 1, 5, 3};
    int N = arr.length;
    int K = 4;
    System.out.print(maxCostToRemove(arr, N, K));
}
}
 
// This code is contributed by sanjoy_62

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Python3

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# Python3 program to implement
# the above approach
 
# Function to find the maximu cost to
# remove all array elements
def maxCostToRemove(arr, N, K):
 
    # Stores maximum cost to
    # remove array elements
    maxCost = 0
 
    # Sort array in
    # ascending order
    arr = sorted(arr)
 
    # Traverse the array
    for i in range(0, N, K):
         
        # Update maxCost
        maxCost += arr[i + 1]
 
    return maxCost
 
# Driver Code
if __name__ == '__main__':
    arr= [1, 3, 4, 1, 5, 1, 5, 3]
    N = len(arr)
    K = 4
    print(maxCostToRemove(arr, N, K))
 
# This code is contributed by mohit kumar 29

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C#

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// C# program for the above approach
using System;
class GFG{
   
// Function to find the maximu cost to
// remove all array elements
static int maxCostToRemove(int []arr, int N,
                             int K)
{
      
    // Stores maximum cost to
    // remove array elements
    int maxCost = 0;
      
      
    // Sort array in
    // ascending order   
    Array.Sort(arr);
      
    // Traverse the array
    for (int i = 0; i < N;
                i += K) {
          
        // Update maxCost
        maxCost += arr[i + 1];
    }
      
    return maxCost;
}
   
// Drive Code
public static void Main(String[] args)
{
    int []arr= { 1, 3, 4, 1, 5, 1, 5, 3};
    int N = arr.Length;
    int K = 4;
    Console.Write(maxCostToRemove(arr, N, K));
}
}
 
// This code is contributed by shikhasingrajput

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Output: 

5

 

Time Complexity: O(N * log N) 
Auxiliary Space O(1)

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