Schedule elevator to reduce the total time taken

Given an integer k and an array arr[] representing the destination floors for N people waiting currently at the ground floor and k is the capacity of the elevator i.e. maximum number of people it can hold at the same time. It takes 1 unit time for the elevator to reach any consecutive floor from the current floor. The task is to schedule the elevator in a way to minimize the total time taken to get all the people to their destination floor and then return back to the ground floor.

Examples:

Input: arr[] = {2, 3, 4}, k = 2
Output: 12
Second and the third persons (destination floors 3 and 4) shall go in the first turn taking 8 (4 + 4) unit time. The only person left will take 2 unit time to get to the destination
And then the elevator will take another 2 unit time to get back to the ground floor.
Total time taken = 8 + 2 + 2 = 12



Input: arr[] = {5, 5, 4}, k = 3
Output: 10
Every person can get on the elevator at the same time
Time required will be 10 (5 + 5).

Approach: Sort the given array in decreasing order of destination. Create groups of K (starting from the highest floor), the cost for each group will be 2 * (max(Elements in current group)). The summation across all groups will be the answer.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum time taken
// by the elevator when operating optimally
int minTime(int n, int k, int a[])
{
    // Sort in descending order
    sort(a, a + n, greater<int>());
    int minTime = 0;
  
    // Iterate through the groups
    for (int i = 0; i < n; i += k)
  
        // Update the time taken for each group
        minTime += (2 * a[i]);
  
    // Return the total time taken
    return minTime;
}
  
// Driver code
int main()
{
    int k = 2;
    int arr[] = { 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << minTime(n, k, arr);
  
    return 0;
}

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Java














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// Java implementation of the approach 


import java.util.*; 


  


class GFG 


{ 


  


// Function to return the minimum time taken 


// by the elevator when operating optimally 


static int minTime(int n, int k, int a[]) 


{ 


    // Sort in descending order 


    int temp; 


    for(int i = 0; i < n; i++) 


    {      


        for(int j = i + 1; j < n; j++) 


        { 


            if(a[i] < a[j]) 


            { 


                temp = a[i]; 


                a[i] = a[j]; 


                a[j] = temp; 


            } 


        } 


    } 


  


  


    int minTime = 0; 


  


    // Iterate through the groups 


    for (int i = 0; i < n; i += k) 


  


        // Update the time taken for each group 


        minTime += (2 * a[i]); 


  


    // Return the total time taken 


    return minTime; 


} 


  


// Driver code 


public static void main(String args[]) 


{ 


    int k = 2; 


    int arr[] = { 2, 3, 4 }; 


    int n = arr.length; 


    System.out.println(minTime(n, k, arr)); 


} 


} 


  


// This code is contributed by 


// Surendra_Gangwar 



















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Python3














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# Python3 implementation of the approach  


  


# Function to return the minimum time taken  


# by the elevator when operating optimally  


def minTime(n, k, a) : 


      


    # Sort in descending order  


    a.sort(reverse = True);  


      


    minTime = 0;  


  


    # Iterate through the groups  


    for i in range(0, n, k) : 


  


        # Update the time taken for  


        # each group  


        minTime += (2 * a[i]);  


  


    # Return the total time taken  


    return minTime;  


  


# Driver code  


if __name__ == "__main__" : 


      


    k = 2;  


    arr = [ 2, 3, 4 ];  


    n = len(arr) ; 


    print(minTime(n, k, arr));  


      


# This code is contributed by Ryuga 



















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C#














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// C# implementation of the approach  


using System; 


  


class GFG  


{  


  


// Function to return the minimum time taken  


// by the elevator when operating optimally  


static int minTime(int n, int k, int []a)  


{  


    // Sort in descending order  


    int temp;  


    for(int i = 0; i < n; i++)  


    {  


        for(int j = i + 1; j < n; j++)  


        {  


            if(a[i] < a[j])  


            {  


                temp = a[i];  


                a[i] = a[j];  


                a[j] = temp;  


            }  


        }  


    }  


  


  


    int minTime = 0;  


  


    // Iterate through the groups  


    for (int i = 0; i < n; i += k)  


  


        // Update the time taken for each group  


        minTime += (2 * a[i]);  


  


    // Return the total time taken  


    return minTime;  


}  


  


// Driver code  


public static void Main(String []args)  


{  


    int k = 2;  


    int []arr = { 2, 3, 4 };  


    int n = arr.Length;  


    Console.Write(minTime(n, k, arr));  


}  


}  


  


// This code is contributed by Arnab Kundu 



















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Output:


12





Time Complexity: O(N * log(N))



          
          
          

          

    

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