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 = 12Input: 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++
// 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; } |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
12
Time Complexity: O(N * log(N))
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