Given an integer N representing the number of cities near you along with two arrays pos[] representing the position of the city and time[] representing the time required to move 1 unit of distance for a particular city, the task is to find the minimum time required to visit any city where you are currently at cur position.
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:
Determine the lowest amount of time required by you to reach each city by first calculating the absolute distance between you and each city.
Below are the steps involved in the implementation of the code:
- Run a loop from 0 to N(number of taxis).
- Keep calculating the distance and then multiply it by the time taken by you to travel each city per unit distance.
- Now keep storing the minimum time.
- Return in the min time possible.
Below is the code implementation of the above code:
C++
#include <cmath>#include <iostream>#include <limits>using namespace std;// Minimum time required to// reach any cityint minimumTime(int N, int cur, int pos[], int time[]){ // Initialising with maximum // distance int mn = numeric_limits<int>::max(); // iterate over all positions for (int i = 0; i < N; i++) { // Calculate the distance from // the current position to the // current position in pos int dist = abs(pos[i] - cur); // Update mn if the time // taken to move to the // current position is // smaller than the // previous minimum time mn = min(mn, dist * time[i]); } // Return the minimum time taken // to move to any position return mn;}// Driver codeint main(){ int N = 3; int cur = 4; int pos[] = { 1, 5, 6 }; int time[] = { 2, 3, 1 }; // Function call int minTime = minimumTime(N, cur, pos, time); cout << minTime << endl; return 0;}// This code is contributed by Ayush pathak |
Java
import java.util.*;public class GFG { // Minimum time required to // reach any city public static int minimumTime(int N, int cur, int[] pos, int[] time) { // Initialising with maximum // distance int mn = (int)1e9; // iterate over all positions for (int i = 0; i < N; i++) { // Calculate the distance from // the current position to the // current position in pos int dist = Math.abs(pos[i] - cur); // Update mn if the time // taken to move to the // current position is // smaller than the // previous minimum time mn = Math.min(mn, dist * time[i]); } // Return the minimum time taken // to move to any position return mn; } // Driver code public static void main(String[] args) { int N = 3; int cur = 4; int[] pos = { 1, 5, 6 }; int[] time = { 2, 3, 1 }; // Function call int minTime = minimumTime(N, cur, pos, time); System.out.println(minTime); }} |
C#
using System;public class GFG{ // Minimum time required to // reach any city public static int minimumTime(int N, int cur, int[] pos, int[] time) { // Initialising with maximum // distance int mn = (int)1e9; // iterate over all positions for (int i = 0; i < N; i++) { // Calculate the distance from // the current position to the // current position in pos int dist = Math.Abs(pos[i] - cur); // Update mn if the time // taken to move to the // current position is // smaller than the // previous minimum time mn = Math.Min(mn, dist * time[i]); } // Return the minimum time taken // to move to any position return mn; } // Driver code public static void Main(string[] args) { int N = 3; int cur = 4; int[] pos = { 1, 5, 6 }; int[] time = { 2, 3, 1 }; // Function call int minTime = minimumTime(N, cur, pos, time); Console.WriteLine(minTime); }} |
Python3
import sys# Minimum time required to# reach any citydef minimumTime(N, cur, pos, time): # Initialising with maximum # distance mn = sys.maxsize # iterate over all positions for i in range(N): # Calculate the distance from # the current position to the # current position in pos dist = abs(pos[i] - cur) # Update mn if the time # taken to move to the # current position is # smaller than the # previous minimum time mn = min(mn, dist * time[i]) # Return the minimum time taken # to move to any position return mn# Driver codeif __name__ == "__main__": N = 3 cur = 4 pos = [1, 5, 6] time = [2, 3, 1] # Function call minTime = minimumTime(N, cur, pos, time) print(minTime) |
Javascript
function minimumTime(N, cur, pos, time) { // Initialising with maximum distance let mn = Number.MAX_SAFE_INTEGER; // iterate over all positions for (let i = 0; i < N; i++) { // Calculate the distance from // the current position to the // current position in pos let dist = Math.abs(pos[i] - cur); // Update mn if the time // taken to move to the // current position is // smaller than the // previous minimum time mn = Math.min(mn, dist * time[i]); } // Return the minimum time taken // to move to any position return mn;}// Driver codeconst N = 3;const cur = 4;const pos = [1, 5, 6];const time = [2, 3, 1];// Function callconst minTime = minimumTime(N, cur, pos, time);console.log(minTime); |
Output
2
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Share your thoughts in the comments
