从所有可能的三元组中最小化最小和第二最小元素的总和

给定一个数组arr[] ，任务是最小化所有可能的三元组中最小元素和第二最小元素的总和。一个元素可以恰好是一个三元组的一部分。

Input: arr[] = {1, 2, 4, 6, 7, 8, 3}, N = 7
Output: 10
Explanation: Here two triplets are formed as the size of arr[] is 7 and 7/3 = 2.
Triplet 1 – {1, 6, 3} -> Two minimum elements are 1 and 3. 
Triplet 2 – {2, 4, 8} -> Two minimum elements are 2 and 4. 
Hence sum = 1 + 3 + 2 + 4 = 10

Input: arr[] = {5, 7, 3, 8, 9}
Output: 8

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。

• 以非递减顺序对数组arr[]进行排序，以便更容易选择每个三元组中的最小元素。
• 初始化一个变量ans = 0 ，以存储可能的最小答案。
• 遍历arr[]并通过从左侧取两个元素和从右侧取一个元素来制作每个三元组。
• 返回ans作为最终答案。

C++14

// C++ program for above approach
#include 
using namespace std;
 
// Function to minimize answer after choosing
// all the triplets from arr[]
int minTriplets(vector& arr, int N)
{
 
    // To store the final answer
    int ans = 0;
 
    // Sort the array
    sort(arr.begin(), arr.end());
 
    // Traverse the array
    for (int i = 0, j = N - 1;
         i + 1 < j;
         i += 2, j--) {
 
        // Add both the smallest numbers
        // of current triplet
        ans += arr[i];
        ans += arr[i + 1];
    }
 
    // Return the ans as the required answer
    return ans;
}
 
// Driver Code
int main()
{
    int N = 7;
 
    vector arr = { 1, 2, 4, 6, 7, 8, 3 };
 
    cout << minTriplets(arr, N);
}

Java

// Java program for above approach
import java.util.*;
public class GFG
{
 
  // Function to minimize answer after choosing
  // all the triplets from arr[]
  static int minTriplets(int []arr, int N)
{
 
  // To store the final answer
  int ans = 0;
 
  // Sort the array
  Arrays.sort(arr);
 
  // Traverse the array
  for (int i = 0, j = N - 1;
       i + 1 < j;
       i += 2, j--) {
 
    // Add both the smallest numbers
    // of current triplet
    ans += arr[i];
    ans += arr[i + 1];
  }
 
  // Return the ans as the required answer
  return ans;
}
 
// Driver Code
public static void main(String args[])
{
  int N = 7;
 
  int []arr = { 1, 2, 4, 6, 7, 8, 3 };
 
  System.out.print(minTriplets(arr, N));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python program for above approach
 
# Function to minimize answer after choosing
# all the triplets from arr[]
def minTriplets (arr, N) :
 
    # To store the final answer
    ans = 0
 
    # Sort the array
    arr.sort()
 
    i = 0
    j = N - 1
     
    # Traverse the array
    while( i + 1 < j):
       
        # Add both the smallest numbers
        # of current triplet
        ans += arr[i]
        ans += arr[i + 1]
        i += 2
        j -= 1
         
    # Return the ans as the required answer
    return ans
 
# Driver Code
N = 7
arr = [1, 2, 4, 6, 7, 8, 3]
print(minTriplets(arr, N))
 
# This code is contributed by gfgking

C#

// C# program for above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
   
// Function to minimize answer after choosing
// all the triplets from arr[]
static int minTriplets(int []arr, int N)
{
 
    // To store the final answer
    int ans = 0;
 
    // Sort the array
    Array.Sort(arr);
 
    // Traverse the array
    for (int i = 0, j = N - 1;
         i + 1 < j;
         i += 2, j--) {
 
        // Add both the smallest numbers
        // of current triplet
        ans += arr[i];
        ans += arr[i + 1];
    }
 
    // Return the ans as the required answer
    return ans;
}
 
// Driver Code
public static void Main()
{
    int N = 7;
    int []arr = { 1, 2, 4, 6, 7, 8, 3 };
    Console.Write(minTriplets(arr, N));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript program for above approach
 
    // Function to minimize answer after choosing
    // all the triplets from arr[]
    const minTriplets = (arr, N) => {
 
        // To store the final answer
        let ans = 0;
 
        // Sort the array
        arr.sort();
 
        // Traverse the array
        for (let i = 0, j = N - 1;
            i + 1 < j;
            i += 2, j--) {
 
            // Add both the smallest numbers
            // of current triplet
            ans += arr[i];
            ans += arr[i + 1];
        }
 
        // Return the ans as the required answer
        return ans;
    }
 
    // Driver Code
 
    let N = 7;
 
    let arr = [1, 2, 4, 6, 7, 8, 3];
 
    document.write(minTriplets(arr, N));
 
    // This code is contributed by rakeshsahni
 

输出

10

时间复杂度： O(N log N)
辅助空间： O(1)