Cost required to empty a given array by repeated removal of maximum obtained by given operations

Given an array arr[] consisting of N integers, the task is to find the cost to remove all array elements after performing the following operations in the specified order any number of times:

• Add the maximum element present in the given array to the cost.
• Remove the maximum element from the given array.
• Decrease all integers in the given array by 1.
• Remove the elements reduced to 0.

Examples:

Input: arr[] = {1, 6, 7, 4, 2, 5, 3}
Output: 16
Explanation:
Step 1: Current maximum element = 7. Removing 7 and reducing each element by 1 modifies the array to {0, 5, 3, 1, 4, 2}. Cost = 7. 
Step 2: Current maximum element = 5. Removing 5 and reducing each element by 1 modifies the array to {2, 0, 3, 1}. Cost = 7 + 5 = 12. 
Step 3: Current maximum = 3. Removing 3 and reducing each element by 1 modifies the array to {1, 1}. Cost = 12 + 3 = 15 
Step 4: Final cost obtained = 15 + 1 = 16

Input: arr[] = {6, 4, 6, 1}
Output: 13

Naive Approach: The simplest approach to solve the problem is as follows:

• Find the maximum from the given array
• Add it to the cost after removing it from the array.
• Once the maximum element is removed, reduce all remaining array elements by 1. Remove all elements reducing to 0.
• Repeat the above steps until all array elements are removed.
• Finally, print the value of cost.

C++

// C++ program to find the cost to remove all array elements
// after performing the specified operations.

#include <bits/stdc++.h>
using namespace std;

// Function to find the cost to remove all array elements
int findCost(vector<int>& arr, int n)
{
    // using long long to avoid overflow
    long long cost = 0;

    // loop until array arr is not empty
    while (!arr.empty()) {
        // find the maximum element
        int max_elem = *max_element(arr.begin(), arr.end());

        // add it to the cost
        cost += max_elem;
        // remove it from the array
        arr.erase(find(arr.begin(), arr.end(), max_elem));
        // reduce all remaining array elements by 1
        for (int i = 0; i < arr.size(); i++) {
            arr[i]--;
        }
        // remove all elements reducing to 0
        arr.erase(remove(arr.begin(), arr.end(), 0),
                  arr.end());
    }

    // Return the total cost
    return cost;
}

// Driver code
int main()
{
    // Input array
    vector<int> arr = { 1, 6, 7, 4, 2, 5, 3 };
    int n = arr.size();

    // Function Call
    int cost = findCost(arr, n);
    cout << cost << endl;

    return 0;
}

Java

// Java program to find the cost to remove all array
// elements after performing the specified operations.

import java.util.*;

public class GFG {
    // Function to find the cost to remove all array
    // elements
    static int findCost(ArrayList<Integer> arr, int n)
    {
        // using long to avoid overflow
        long cost = 0;
        // loop until array arr is not empty
        while (!arr.isEmpty()) {
            // find the maximum element
            int max_elem = Collections.max(arr);

            // add it to the cost
            cost += max_elem;
            // remove it from the array
            arr.remove(arr.indexOf(max_elem));
            // reduce all remaining array elements by 1
            for (int i = 0; i < arr.size(); i++) {
                arr.set(i, arr.get(i) - 1);
            }
            // remove all elements reducing to 0
            arr.removeIf(e -> e == 0);
        }

        // Return the total cost
        return (int)cost;
    }

    // Driver code
    public static void main(String[] args)
    {
        // Input array
        ArrayList<Integer> arr = new ArrayList<>(
            Arrays.asList(1, 6, 7, 4, 2, 5, 3));
        int n = arr.size();

        // Function Call
        int cost = findCost(arr, n);
        System.out.println(cost);
    }
}

Python3

# Python3 program to find the cost to remove all array elements
# after performing the specified operations.

# Function to find the cost to remove all array elements


def findCost(arr):
    # using long to avoid overflow
    cost = 0
    # loop until array arr is not empty
    while arr:
        # find the maximum element
        max_elem = max(arr)

        # add it to the cost
        cost += max_elem

        # remove it from the array
        arr.remove(max_elem)

        # reduce all remaining array elements by 1
        for i in range(len(arr)):
            arr[i] -= 1

        # remove all elements reducing to 0
        arr = [x for x in arr if x != 0]

    # Return the total cost
    return cost


# Driver code
if __name__ == '__main__':
    # Input array
    arr = [1, 6, 7, 4, 2, 5, 3]
    # Function Call
    cost = findCost(arr)
    print(cost)

C#

// C# program to find the cost to remove all array elements
// after performing the specified operations.

using System;
using System.Collections.Generic;
using System.Linq;

class GFG {
    // Function to find the cost to remove all array elements
    static int FindCost(List<int> arr, int n) {
        // using long to avoid overflow
        long cost = 0;

        // loop until array arr is not empty
        while (arr.Count > 0) {
            // find the maximum element
            int max_elem = arr.Max();

            // add it to the cost
            cost += max_elem;

            // remove it from the array
            arr.Remove(max_elem);

            // reduce all remaining array elements by 1
            for (int i = 0; i < arr.Count; i++) {
                arr[i]--;
            }

            // remove all elements reducing to 0
            arr.RemoveAll(elem => elem == 0);
        }

        // Return the total cost
        return (int) cost;
    }

      // Driver's code
    static void Main(string[] args) {
        // Input array
        List<int> arr = new List<int>() { 1, 6, 7, 4, 2, 5, 3 };
        int n = arr.Count;

        // Function Call
        int cost = FindCost(arr, n);
        Console.WriteLine(cost);
    }
}

Javascript

// Function to find the cost to remove all array elements
function findCost(arr) {
    // using long long to avoid overflow
    let cost = 0;

    // loop until array arr is not empty
    while (arr.length > 0) {
        // find the maximum element
        let max_elem = Math.max(...arr);

        // add it to the cost
        cost += max_elem;
        // remove it from the array
        arr.splice(arr.indexOf(max_elem), 1);
        // reduce all remaining array elements by 1
        for (let i = 0; i < arr.length; i++) {
            arr[i]--;
        }
        // remove all elements reducing to 0
        arr = arr.filter(item => item !== 0);
    }

    // Return the total cost
    return cost;
}

// Input array
let arr = [1, 6, 7, 4, 2, 5, 3];

// Function Call
let cost = findCost(arr);
console.log(cost);

Output

16

Time Complexity: O(N²), where N is the size of the given array.
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to observe that every i^th greatest element in the array is added arr[i] – i times to the cost where i is the index of each element arr[i]. Follow the below steps to solve the above problem:

1. Sort the given array in decreasing order.
2. Traverse the array and add the value arr[i] – i for each array element to cost if it is greater than 0.
3. Once the above steps are completed, print the value of cost.

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 total cost
// of removing all array elements
int findCost(int* a, int n)
{
    // Sort the array in descending order
    sort(a, a + n, greater<int>());

    // Stores the total cost
    int count = 0;

    for (int j = 0; j < n; j++) {

        // Contribution of i-th
        // greatest element to the cost
        int p = a[j] - j;

        // Remove the element
        a[j] = 0;

        // If negative
        if (p < 0) {
            p = 0;
            continue;
        }

        // Add to the final cost
        count += p;
    }

    // Return the cost
    return count;
}

// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 6, 7, 4, 2, 5, 3 };

    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    cout << findCost(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{

// Function to find the total cost
// of removing all array elements
static int findCost(Integer [] a, int n)
{
    // Sort the array in descending order
     Arrays.sort(a, Collections.reverseOrder());

    // Stores the total cost
    int count = 0;

    for (int j = 0; j < n; j++)
    {
        // Contribution of i-th
        // greatest element to the cost
        int p = a[j] - j;

        // Remove the element
        a[j] = 0;

        // If negative
        if (p < 0)
        {
            p = 0;
            continue;
        }

        // Add to the final cost
        count += p;
    }

    // Return the cost
    return count;
}

// Driver Code
public static void main(String[] args)
{
    // Given array arr[]
    Integer arr[] = {1, 6, 7,
                     4, 2, 5, 3};

    int N = arr.length;

    // Function Call
    System.out.print(findCost(arr, N));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach

# Function to find the total cost
# of removing all array elements
def findCost(a, n):

    # Sort the array in descending order
    a.sort(reverse = True)

    # Stores the total cost
    count = 0

    for j in range(n):

        # Contribution of i-th
        # greatest element to the cost
        p = a[j] - j

        # Remove the element
        a[j] = 0

        # If negative
        if(p < 0):
            p = 0
            continue

        # Add to the final cost
        count += p

    # Return the cost
    return count

# Driver Code

# Given array arr[]
arr = [ 1, 6, 7, 4, 2, 5, 3 ]

N = len(arr)

# Function call
print(findCost(arr, N))

# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
class GFG{

// Function to find the total cost
// of removing all array elements
static int findCost(int []a, int n)
{
    // Sort the array in descending order
    Array.Sort(a);
    Array.Reverse(a);

    // Stores the total cost
    int count = 0;

    for (int j = 0; j < n; j++)
    {
        // Contribution of i-th
        // greatest element to the cost
        int p = a[j] - j;

        // Remove the element
        a[j] = 0;

        // If negative
        if (p < 0)
        {
            p = 0;
            continue;
        }

        // Add to the readonly cost
        count += p;
    }

    // Return the cost
    return count;
}

// Driver Code
public static void Main(String[] args)
{
    // Given array []arr
    int []arr = {1, 6, 7,
                 4, 2, 5, 3};

    int N = arr.Length;

    // Function Call
    Console.Write(findCost(arr, N));
}
}

// This code is contributed by shikhasingrajput

Javascript

// JavaScript program for the above approach
// Function to find the total cost
// of removing all array elements
function findCost(a, n)
{

  // Sort the array in descending order
  a.sort((x, y) => y - x);

  // Stores the total cost
  var count = 0;

  for (var j = 0; j < n; j++)
  {

    // Contribution of i-th
    // greatest element to the cost
    var p = a[j] - j;

    // Remove the element
    a[j] = 0;

    // If negative
    if (p < 0) {
      p = 0;
      continue;
    }

    // Add to the final cost
    count += p;
  }

  // Return the cost
  return count;
}

// Driver Code
// Given array arr[]
var arr = [1, 6, 7, 4, 2, 5, 3];
var N = arr.length;

// Function Call
console.log(findCost(arr, N));

Output

16

Time Complexity: O(NlogN), where N is the size of the given array.
Auxiliary Space: O(1)