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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 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 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 $O(1)$.

  • class Solution { public int minTimeToVisitAllPoints(int[][] points) { int ans = 0; for (int i = 1; i < points.length; ++i) { int dx = Math.abs(points[i][0] - points[i - 1][0]); int dy = Math.abs(points[i][1] - points[i - 1][1]); ans += Math.max(dx, dy); } return ans; } } 
  • class Solution { public: int minTimeToVisitAllPoints(vector<vector<int>>& points) { int ans = 0; for (int i = 1; i < points.size(); ++i) { int dx = abs(points[i][0] - points[i - 1][0]); int dy = abs(points[i][1] - points[i - 1][1]); ans += max(dx, dy); } return ans; } }; 
  • class Solution: def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int: return sum( max(abs(p1[0] - p2[0]), abs(p1[1] - p2[1])) for p1, p2 in pairwise(points) ) 
  • func minTimeToVisitAllPoints(points [][]int) (ans int) { for i, p := range points[1:] { dx := abs(p[0] - points[i][0]) dy := abs(p[1] - points[i][1]) ans += max(dx, dy) } return } func abs(x int) int { if x < 0 { return -x } return x } 
  • function minTimeToVisitAllPoints(points: number[][]): number { let ans = 0; for (let i = 1; i < points.length; i++) { let dx = Math.abs(points[i][0] - points[i - 1][0]), dy = Math.abs(points[i][1] - points[i - 1][1]); ans += Math.max(dx, dy); } return ans; } 
  • impl Solution { pub fn min_time_to_visit_all_points(points: Vec<Vec<i32>>) -> i32 { let n = points.len(); let mut ans = 0; for i in 1..n { let x = (points[i - 1][0] - points[i][0]).abs(); let y = (points[i - 1][1] - points[i][1]).abs(); ans += x.max(y); } ans } }

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