Find minimum time to board taxi from current position

Given that there is an infinite number of points 1, 2, 3… on the X axis and your current position is cur. There are N taxis near you, and the position of those taxis is given as an array pos[]. Where pos[i] denotes the position of the ith taxi. You are also given an array of time[], where time[i] denotes the time taken by the i^th taxi to cover 1 unit of distance. Your task is to find the minimum time to board a taxi.

Examples:

Input: N = 3, cur = 4, pos = [1, 5, 6], time = [2, 3, 1]
Output: 2
Explanation: 

• Total time taken by the 1st taxi will be: (4-1)*2 = 6
• Total time taken by the 2nd taxi will be: (5-4)*3 = 3
• Total time taken by the 3rd taxi will be: (6-4)*1 = 2

So, the minimum time will be 2 sec.

Input:N = 2, cur = 1, pos = [1, 6], time = [10, 3]
Output: 0
Explanation:

• Total time taken by the 1st taxi will be: (1-1)*10 = 0
• Total time taken by the 2nd taxi will be: (6-1)*3 = 15

So, the minimum time will be 0 sec.

Approach: This can be solved with the following idea:

Iterate over the array and count the minimum time for each.

Below are the steps involved in the implementation of the code:

• Initialize minn as the maximum integer possible
• Iterate over the array.
• Find the distance between the passenger and each taxi position.
• Find the minimum distance by multiplying it by the time[i] required.
• Update the minn.

Below is the implementation of the code:

C++

// C++ code for the above approach:
#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// Function to find minimum time required
int minimumTime(int N, int cur, vector<int>& pos,
                vector<int>& time)
{
    int minn = INT_MAX;
    int i = 0;

    // Iterate over the array
    while (i < N) {
        minn = min(minn, abs(pos[i] - cur) * time[i]);
        i++;
    }
    return minn;
}

// Driver code
int main()
{

    int N = 3;
    int cur = 4;
    vector<int> pos = { 1, 5, 6 };
    vector<int> time = { 2, 3, 1 };

    // Function call
    cout << minimumTime(N, cur, pos, time);

    return 0;
}

Java

// JAVA code for the above approach:

import java.util.*;

public class GFG {
    // Function to find minimum time required
    public static int minimumTime(int N, int cur,
                                  List<Integer> pos,
                                  List<Integer> time)
    {
        int minn = Integer.MAX_VALUE;
        int i = 0;

        // Iterate over the array
        while (i < N) {
            minn = Math.min(minn, Math.abs(pos.get(i) - cur)
                                      * time.get(i));
            i++;
        }
        return minn;
    }

    // Driver code
    public static void main(String[] args)
    {
        int N = 3;
        int cur = 4;
        List<Integer> pos
            = new ArrayList<>(Arrays.asList(1, 5, 6));
        List<Integer> time
            = new ArrayList<>(Arrays.asList(2, 3, 1));

        // Function call
        System.out.println(minimumTime(N, cur, pos, time));
    }
}

// This code is contributed by rambabuguphka

Python3

# Python3 code for the above approach:
# Function to find minimum time required


def minimumTime(N, cur, pos, time):
    minn = float('inf')
    i = 0

    # Iterate over the array
    while i < N:
        minn = min(minn, abs(pos[i] - cur) * time[i])
        i += 1
    return minn


# Driver code
if __name__ == "__main__":
    N = 3
    cur = 4
    pos = [1, 5, 6]
    time = [2, 3, 1]

    # Function call
    print(minimumTime(N, cur, pos, time))

# This code is contributed by rambabuguphka

C#

// C# code for the above approach:

using System;
using System.Collections.Generic;

class GFG {
    // Function to find minimum time required
    static int MinimumTime(int N, int cur, List<int> pos,
                           List<int> time)
    {
        int minn = int.MaxValue;
        int i = 0;

        // Iterate over the list
        while (i < N) {
            minn = Math.Min(minn, Math.Abs(pos[i] - cur)
                                      * time[i]);
            i++;
        }
        return minn;
    }

    static void Main(string[] args)
    {
        int N = 3;
        int cur = 4;
        List<int> pos = new List<int>{ 1, 5, 6 };
        List<int> time = new List<int>{ 2, 3, 1 };

        // Function call
        Console.WriteLine(MinimumTime(N, cur, pos, time));
    }
}

// This code is contributed by shivamgupta310570

Javascript

// Function to find minimum time required
function minimumTime(N, cur, pos, time) {
    let minn = Infinity;
    let i = 0;

    // Iterate over the array
    while (i < N) {
        minn = Math.min(minn, Math.abs(pos[i] - cur) * time[i]);
        i++;
    }
    return minn;
}

// Driver code
    const N = 3;
    const cur = 4;
    const pos = [1, 5, 6];
    const time = [2, 3, 1];

    // Function call
    console.log(minimumTime(N, cur, pos, time));

// This code is contributed by shivamgupta0987654321

Output

2

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