1266. 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.

class Solution {
    int helper(vector<int>& p1, vector<int>& p2) {
        return max(abs(p1[0] - p2[0]), abs(p1[1] - p2[1]));
    }
public:
    int minTimeToVisitAllPoints(vector<vector<int>>& points) {
        int res = 0;
        for(int i = 1; i < points.size(); i++) {
            res += helper(points[i-1], points[i]);
        }
        return res;
    }
};