1266. Minimum Time Visiting All Points
Description
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 in1
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 Explanation: One optimal path is [1,1] -> [2,2] -> [3,3] -> [3,4] -> [2,3] -> [1,2] -> [0,1] -> [-1,0] Time from [1,1] to [3,4] = 3 seconds Time from [3,4] to [-1,0] = 4 seconds Total time = 7 seconds
Example 2:
Input: points = [[3,2],[-2,2]] Output: 5
Constraints:
points.length == n
1 <= n <= 100
points[i].length == 2
-1000 <= points[i][0], points[i][1] <= 1000
Solutions
Solution 1: Simulation
For two points $p1=(x_1, y_1)$ and $p2=(x_2, y_2)$, the distances moved in the x-axis and y-axis are $dx = |x_1 - x_2|$ and $dy = |y_1 - y_2|$ respectively.
If $dx \ge dy$, move along the diagonal for $dy$ steps, then move horizontally for $dx - dy$ steps. If $dx < dy$, move along the diagonal for $dx$ steps, then move vertically for $dy - dx$ steps. Therefore, the minimum distance between the two points is $max(dx, dy)$.
We can iterate through all pairs of points, calculate the minimum distance between each pair of points, and then sum them up.
The time complexity is $O(n)$, where $n$ is the number of points. The space complexity is $O(1)$.
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