Minimum Time Visiting All Points

On a 2D plane, there are n points with integer coordinates points[i] = [xi, yi]. Return the minimum time in seconds to visit all the points in the order given by points.

You can move according to these rules:

• In 1 second, you can either:
  • move vertically by one unit,
  • move horizontally by one unit, or
  • move diagonally sqrt(2) units (in other words, move one unit vertically then one unit horizontally in 1 second).
• You have to visit the points in the same order as they appear in the array.
• You are allowed to pass through points that appear later in the order, but these do not count as visits.

Example 1:

Input: points = {{1,1},{3,4},{-1,0}}
Output: 7

Approach

Java

public class MinTimeVisitingAllPoints {
    public static void main(String[] args) {
        int points[][] = { { 1, 1 }, { 3, 4 }, { -1, 0 } };
        System.out.println(minTimeToVisitAllPoints(points));
    }
    static int minTimeToVisitAllPoints(int[][] points) {
        int cnt = 0;
        for (int i = 0; i < points.length - 1; i++) {
            int x1 = points[i][0];
            int y1 = points[i][1];
            int x2 = points[i + 1][0];
            int y2 = points[i + 1][1];
            cnt = cnt + Math.max(Math.abs(y2 - y1), Math.abs(x2 - x1));
        }
        return cnt;
    }
}

C++

#include <bits/stdc++.h>
using namespace std;

int minTimeToVisitAllPoints(vector<vector<int>>& points) 
{
    int cnt=0;
    for(int i=0;i<points.size()-1;i++) 
        {
             int x1=points[i][0];
             int y1=points[i][1];
             int x2=points[i+1][0];
             int y2=points[i+1][1];
             cnt=cnt+max(abs(y2-y1),abs(x2-x1));
        }
    return cnt;
}
int main()
{
    vector<vector<int>> points ={{1,1},{3,4},{-1,0}};
    cout<<minTimeToVisitAllPoints(points);
    return 0;
}