Minimize sum of minimum and second minimum elements from all possible triplets

Given an array arr[], the task is to minimize the sum of minimum and second minimum elements from all possible triplets. One element can be a part of exactly one triplet.

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

Approach: This problem can be solved by using the Greedy Approach. Follow the steps below to solve the given problem. 

• Sort the array arr[] in non-decreasing order, so that it gets easier to choose minimum elements in every triplet.
• Initialize a variable say ans = 0, to store the minimum possible answer.
• Traverse arr[] and make every triplet by taking two elements from the left side and one element from the right side.
• Return ans as the final answer.

C++14

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to minimize answer after choosing
// all the triplets from arr[]
int minTriplets(vector<int>& 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<int> 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

<script>
    // 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

</script>

Output

10

Time Complexity: O(N log N)
Auxiliary Space: O(1)