Minimize cost to bring maximum element at Kth position by swapping

Given two integers N and K and an array of N positive integers (i.e., a₀, a₁, a₂…., a_n-1), the task is to find the minimum cost to bring maximum value to K^th position by swapping (1-based indexing). The cost of swapping the elements is defined as the sum of values of swapping elements.

Example:

Input: N = 6, K = 4, arr[] = {3, 7, 8, 7, 4, 5}
Output: 12
Explanation: Sum of arr[2] + arr[4]

Input: N = 8, K = 4, arr[] = {11, 31, 17, 18, 37, 14, 15, 11}
Output: 0
Explanation: Maximum is already at arr[4]

Approach:
The idea is to find the position of the maximum element if the maximum element is not at Kth position then swap it with the element which is at Kth position and the answer will be the cost of the swapping element which is the sum values of swapping elements. if the maximum element is at kth position then the answer will be 0.

Follow the steps below to implement the above idea:

• Find the position of the maximum element.
• Check if the maximum element is at Kth position 
  • If true, then return 0.
  • Otherwise, swap the maximum element with the element at Kth position and return the sum of values of swapping elements.

Below is the implementation of the above approach:

C++

// C++ program to implement the idea:
#include <bits/stdc++.h>
using namespace std;

// Function to minimize cost to bring max at Kth
// position by swapping
int minimizeCost(int arr[], int N, int K)
{

    // Store maximum element in array
    int max_element = INT_MIN;
    int idx = -1;

    for (int i = 0; i < N; i++) {
        if (max_element < arr[i]) {
            max_element = arr[i];
            idx = i;
        }
    }

    if (idx == K)
        return 0;

    swap(arr[K], arr[idx]);
    return arr[K] + arr[idx];
}

// Driver Code
int main()
{
    int N = 6;
    int k = 4;
    int arr[N] = { 3, 7, 8, 7, 4, 5 };

    cout << minimizeCost(arr, N, k);

    return 0;
}

Java

// Java program to implement the idea:

import java.io.*;
import java.util.*;

class GFG {

    // Function to minimize cost to bring max at Kth
    // position by swapping
    public static int minimizeCost(int[] arr, int N, int K)
    {

        // Store maximum element in array
        int max_element = Integer.MIN_VALUE;
        int idx = -1;

        for (int i = 0; i < N; i++) {
            if (max_element < arr[i]) {
                max_element = arr[i];
                idx = i;
            }
        }

        if (idx == K)
            return 0;

        // swap
        int temp = arr[K];
        arr[K] = arr[idx];
        arr[idx] = temp;

        return arr[K] + arr[idx];
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 6;
        int k = 4;
        int[] arr = { 3, 7, 8, 7, 4, 5 };

        System.out.println(minimizeCost(arr, N, k));
    }
}
// This code is contributed by Ishan Khandelwal

Python3

# Python3 program to implement the idea:

# Function to minimize cost to bring max at Kth
# position by swapping
import sys

def minimizeCost(arr, N, K):

    # Store maximum element in array
    max_element = -sys.maxsize -1
    idx = -1

    for i in range(N):
        if (max_element < arr[i]):
            max_element = arr[i]
            idx = i

    if (idx == K):
        return 0

    arr[K], arr[idx] = arr[idx],arr[K]
    return arr[K] + arr[idx]

# Driver Code
N = 6
k = 4
arr = [ 3, 7, 8, 7, 4, 5 ]

print(minimizeCost(arr, N, k))

# This code is contributed by shinjanpatra

C#

// C# program to implement the idea:

using System;

public class GFG {

  // Function to minimize cost to bring max at Kth
  // position by swapping
  public static int minimizeCost(int[] arr, int N, int K)
  {

    // Store maximum element in array
    int max_element = int.MinValue;
    int idx = -1;

    for (int i = 0; i < N; i++) {
      if (max_element < arr[i]) {
        max_element = arr[i];
        idx = i;
      }
    }

    if (idx == K)
      return 0;

    // swap
    int temp = arr[K];
    arr[K] = arr[idx];
    arr[idx] = temp;

    return arr[K] + arr[idx];
  }

  // Driver Code
  public static void Main(string[] args)
  {
    int N = 6;
    int k = 4;
    int[] arr = { 3, 7, 8, 7, 4, 5 };

    Console.WriteLine(minimizeCost(arr, N, k));
  }
}
// This code is contributed by AnkThon

Javascript

<script>
        // JavaScript code for the above approach

        // Function to minimize cost to bring max at Kth
        // position by swapping
        function minimizeCost(arr, N, K) {

            // Store maximum element in array
            let max_element = Number.MIN_VALUE;
            let idx = -1;

            for (let i = 0; i < N; i++) {
                if (max_element < arr[i]) {
                    max_element = arr[i];
                    idx = i;
                }
            }

            if (idx == K)
                return 0;

            let temp = arr[k];
            arr[k] = arr[idx];
            arr[idx] = temp;

            return arr[K] + arr[idx];
        }

        // Driver Code

        let N = 6;
        let k = 4;
        let arr = [3, 7, 8, 7, 4, 5];

        document.write(minimizeCost(arr, N, k));

    // This code is contributed by Potta Lokesh
    </script>

Output

12

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