Code Implementation

class Solution:
    def minTimeToVisitAllPoints(self, points):
        total = 0
        for i in range(1, len(points)):
            x0, y0 = points[i-1]
            x1, y1 = points[i]
            total += max(abs(x1 - x0), abs(y1 - y0))
        return total
      

class Solution {
public:
    int minTimeToVisitAllPoints(vector<vector<int>>& points) {
        int total = 0;
        for (int i = 1; i < points.size(); ++i) {
            int x0 = points[i-1][0], y0 = points[i-1][1];
            int x1 = points[i][0], y1 = points[i][1];
            total += max(abs(x1 - x0), abs(y1 - y0));
        }
        return total;
    }
};
      

class Solution {
    public int minTimeToVisitAllPoints(int[][] points) {
        int total = 0;
        for (int i = 1; i < points.length; i++) {
            int x0 = points[i-1][0], y0 = points[i-1][1];
            int x1 = points[i][0], y1 = points[i][1];
            total += Math.max(Math.abs(x1 - x0), Math.abs(y1 - y0));
        }
        return total;
    }
}
      

var minTimeToVisitAllPoints = function(points) {
    let total = 0;
    for (let i = 1; i < points.length; i++) {
        let [x0, y0] = points[i-1];
        let [x1, y1] = points[i];
        total += Math.max(Math.abs(x1 - x0), Math.abs(y1 - y0));
    }
    return total;
};
      

Problem Description

You are given an array points where each element is a pair [x, y] representing coordinates on a 2D grid. Starting at the first point, you must visit each subsequent point in order. In one second, you can move from your current position to any of the 8 neighboring grid points (horizontal, vertical, or diagonal). The task is to calculate the minimum time required to visit all the points in the given order.

• You must visit the points in the order provided in points.
• Each move takes exactly 1 second, whether moving horizontally, vertically, or diagonally.
• There is only one valid solution for each input.
• You cannot skip or reorder points.

Thought Process

At first glance, it might seem like you need to simulate each movement step by step, moving horizontally, vertically, or diagonally until you reach the next point. However, this would be inefficient and unnecessary. Instead, we can optimize by realizing that diagonal movement allows us to cover both axes at once, so the fastest way to reach the next point is to move diagonally as much as possible, then finish with straight moves if needed.

For each pair of consecutive points, we can compute the horizontal and vertical distances. The minimum time to go from one point to another is the maximum of the absolute differences in x and y coordinates, because diagonal moves cover both axes at the same time. Summing this for all pairs gives the answer.

Solution Approach

Let's break down the solution step by step:

1. Iterate through each consecutive pair of points:
   • Start at the first point and for each subsequent point, calculate the difference in x and y coordinates.
2. Calculate the minimum time for each step:
   • Let dx = |x2 - x1| and dy = |y2 - y1|.
   • The minimum time to move from (x1, y1) to (x2, y2) is max(dx, dy) because diagonal moves can reduce both dx and dy by 1 at the same time.
3. Sum up the times:
   • Add up the minimum times for all consecutive pairs to get the total minimum time.
4. Return the total:
   • After processing all pairs, return the total time as the answer.

This approach is efficient and leverages the properties of diagonal movement to minimize the number of steps.

Example Walkthrough

Let's consider the input points = [[1,1],[3,4],[-1,0]].

1. From [1,1] to [3,4]:
   • dx = |3 - 1| = 2
   • dy = |4 - 1| = 3
   • Minimum time = max(2, 3) = 3
2. From [3,4] to [-1,0]:
   • dx = |-1 - 3| = 4
   • dy = |0 - 4| = 4
   • Minimum time = max(4, 4) = 4
3. Total minimum time:
   • 3 (first step) + 4 (second step) = 7

This matches the expected answer.

Time and Space Complexity

• Brute-force approach:
  • If we tried to simulate each step, the time complexity would depend on the distances between points and could be much higher, especially for large distances.
• Optimized approach:
  • Time Complexity: O(n), where n is the number of points. We process each pair of points exactly once.
  • Space Complexity: O(1), since we only use a fixed number of variables regardless of input size.

The optimized approach is efficient and suitable for large inputs.

Summary

The key insight is that diagonal movement allows us to minimize the time to traverse between points by always moving along the axis with the largest difference. By summing the maximum of horizontal and vertical distances for each consecutive pair, we efficiently compute the minimum time to visit all points in order. This solution is simple, elegant, and leverages the properties of grid movement to achieve optimal performance.