如果选择每 2 个元素，则免费选择第 3 个元素，以最小化减少 Array 的成本

给定一个数组arr[] 。任务是最小化使用给定操作减少数组的成本。在一个操作中，选择两个元素将它们的值添加到总成本中并删除它们，并删除任何其他元素的值与两个选择的元素几乎相同。

例子：

Input: arr[] = {1, 2, 3}
Output: 5
Explanation: Choose 2 and 3, cost = 5. 1 is reduced from array for free.

Input: arr[] = {6, 5, 7, 9, 2, 2}
Output: 23

方法：这个问题可以通过使用贪心方法来解决。请按照以下步骤解决给定的问题。

• 对给定的数组进行排序。
• 从末尾遍历排序数组。
• 将 2 个元素添加到最终成本并跳过第 3 个元素。
• 返回最终成本。

下面是上述方法的实现。

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function to find minimum cost
// to get the desired array
int minCost(vector& arr)
{
    // Sorting the array
    sort(arr.begin(), arr.end());
    int ans = 0;
    for (int i = arr.size() - 1, k = 1;
         i >= 0;
         i--, k++)
        if (k == 3)
            k = 0;
        else
            ans += arr[i];
    return ans;
}
 
// Driver Code
int main()
{
    vector arr = { 6, 5, 7, 9, 2, 2 };
 
    // Function Call
    cout << minCost(arr);
    return 0;
}

Java

// JAVA program for above approach
import java.util.*;
class GFG
{
 
  // Function to find minimum cost
  // to get the desired array
  public static int minCost(ArrayList arr)
  {
    // Sorting the array
    Collections.sort(arr);
    int ans = 0;
    for (int i = arr.size() - 1, k = 1; i >= 0;
         i--, k++)
      if (k == 3)
        k = 0;
    else
      ans += arr.get(i);
    return ans;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    ArrayList arr = new ArrayList<>(
      Arrays.asList(6, 5, 7, 9, 2, 2));
 
    // Function Call
    System.out.print(minCost(arr));
  }
}
 
// This code is contributed by Taranpreet

Python3

# Python code for the above approach
 
# Function to find minimum cost
# to get the desired array
def minCost(arr):
 
    # Sorting the array
    arr.sort()
    ans = 0
    k = 1
    for i in range(len(arr) - 1, 0, -1):
        if k == 3:
            k = 0
        else:
            ans = ans + arr[i]
        k = k + 1
    return ans
 
# Driver Code
arr = [6, 5, 7, 9, 2, 2]
 
# Function Call
print(minCost(arr))
 
# This code is contributed by Potta Lokesh

C#

// C# program for above approach
using System;
class GFG
{
 
  // Function to find minimum cost
  // to get the desired array
  static int minCost(int[] arr)
  {
 
    // Sorting the array
    Array.Sort(arr);
    int ans = 0;
    for (int i = arr.Length - 1, k = 1; i >= 0;
         i--, k++)
      if (k == 3)
        k = 0;
    else
      ans += arr[i];
    return ans;
  }
 
  // Driver Code
  public static void Main()
  {
    int[] arr = { 6, 5, 7, 9, 2, 2 };
 
    // Function Call
    Console.Write(minCost(arr));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript program for above approach
 
    // Function to find minimum cost
    // to get the desired array
    const minCost = (arr) => {
     
        // Sorting the array
        arr.sort();
        let ans = 0;
        for (let i = arr.length - 1, k = 1;
            i >= 0;
            i--, k++)
            if (k == 3)
                k = 0;
            else
                ans += arr[i];
        return ans;
    }
 
    // Driver Code
    let arr = [6, 5, 7, 9, 2, 2];
 
    // Function Call
    document.write(minCost(arr));
 
// This code is contributed by rakeshsahni
 

输出

23

时间复杂度： O(N*log N)
辅助空间： O(1)