空数组的最低成本，其中删除元素的成本为2 ^(removed_count)* arr [i]

给定一个数组arr []，任务是找到从数组中删除所有元素的最小开销，其中删除元素的开销为2 ^ j * arr [i]。在此，j是已经删除的元素数。

例子：

Input: arr[] = {3, 1, 3, 2}
Output: 25
Explanation: 
First remove 3. Cost = 2^(0)*3 = 3 
Then remove 3. Cost = 2^(1)*3 = 6 
Then remove 2. Cost = 2^(2)*2 = 8 
At last, remove 1. Cost = 2^(3)*1 = 8 
Total Cost = 3 + 6 + 8 + 8 = 25

Input: arr[] = {1, 2}
Output: 4
Explanation: 
First remove 2. Cost = 2^(0)*2 = 2 
Then remove 1. Cost = 2^(1)*1 = 2 
Total Cost = 2 + 2 = 4

方法：这个想法是使用贪婪的编程范例来解决这个问题。  
我们必须最小化表达式(2 ^ j * arr [i])。这可以通过以下方式完成：

• 以降序对数组进行排序。
• 将pow(2，i)乘以每个元素，i从0开始直到数组的大小。

因此，从数组中删除元素的总成本为：

$\text{Total Cost = }arr[0]*2^{0} + arr[1] * 2^{1} + .... arr[n]*2^{n}$

当数组降序排列时。

下面是上述方法的实现：

C++

// C++ implementation to find the
// minimum cost of removing all
// elements from the array
  
#include 
using namespace std;
  
#define ll long long int
// Function to find the minimum
// cost of removing elements from
// the array
int removeElements(ll arr[], int n)
{
  
    // Sorting in Increasing order
    sort(arr, arr + n, greater());
    ll ans = 0;
      
    // Loop to find the minimum
    // cost of removing elements
    for (int i = 0; i < n; i++) {
        ans += arr[i] * pow(2, i);
    }
  
    return ans;
}
  
// Driver Code
int main()
{
    int n = 4;
    ll arr[n] = { 3, 1, 2, 3 };
  
    // Function Call
    cout << removeElements(arr, n);
}

Java

// Java implementation to find the
// minimum cost of removing all
// elements from the array
import java.util.*;
  
class GFG{
  
// Reverse array in decreasing order
static long[] reverse(long a[])
{
    int i, n = a.length;
    long t;
      
    for(i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}
  
// Function to find the minimum
// cost of removing elements from
// the array
static long removeElements(long arr[],
                           int n)
{
      
    // Sorting in Increasing order
    Arrays.sort(arr);
    arr = reverse(arr);
  
    long ans = 0;
  
    // Loop to find the minimum
    // cost of removing elements
    for(int i = 0; i < n; i++)
    {
        ans += arr[i] * Math.pow(2, i);
    }
    return ans;
}
  
// Driver Code
public static void main(String[] args)
{
    int n = 4;
    long arr[] = { 3, 1, 2, 3 };
  
    // Function call
    System.out.print(removeElements(arr, n));
}
}
  
// This code is contributed by amal kumar choubey

Python3

# Python3 implementation to find the
# minimum cost of removing all
# elements from the array
  
# Function to find the minimum
# cost of removing elements from
# the array
def removeElements(arr, n):
  
    # Sorting in Increasing order
    arr.sort(reverse = True)
    ans = 0
  
    # Loop to find the minimum
    # cost of removing elements
    for i in range(n):
        ans += arr[i] * pow(2, i)
  
    return ans
  
# Driver Code
if __name__ == "__main__":
      
    n = 4
    arr = [ 3, 1, 2, 3 ]
  
    # Function call
    print(removeElements(arr, n))
      
# This code is contributed by chitranayal

C#

// C# implementation to find the
// minimum cost of removing all
// elements from the array
using System;
  
class GFG{
  
// Reverse array in decreasing order
static long[] reverse(long []a)
{
    int i, n = a.Length;
    long t;
      
    for(i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}
  
// Function to find the minimum
// cost of removing elements from
// the array
static long removeElements(long []arr,
                           int n)
{
      
    // Sorting in Increasing order
    Array.Sort(arr);
    arr = reverse(arr);
  
    long ans = 0;
  
    // Loop to find the minimum
    // cost of removing elements
    for(int i = 0; i < n; i++)
    {
        ans += (long)(arr[i] * Math.Pow(2, i));
    }
    return ans;
}
  
// Driver Code
public static void Main(String[] args)
{
    int n = 4;
    long []arr = { 3, 1, 2, 3 };
  
    // Function call
    Console.Write(removeElements(arr, n));
}
}
  
// This code is contributed by amal kumar choubey

输出

25

时间复杂度：O(N * log N)
辅助空间：O(1)

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。