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 = 16Input: 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.
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 ith 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:
- Sort the given array in decreasing order.
- Traverse the array and add the value arr[i] – i for each array element to cost if it is greater than 0.
- 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 elementsint 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 Codeint 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 approachimport java.util.*;class GFG{// Function to find the total cost// of removing all array elementsstatic 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 Codepublic 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 elementsdef 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 callprint(findCost(arr, N))# This code is contributed by Shivam Singh |
C#
// C# program for the above approachusing System;class GFG{// Function to find the total cost// of removing all array elementsstatic 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 Codepublic 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 |
16
Time Complexity: O(NlogN), where N is the size of the given array.
Auxiliary Space: O(1)
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