Question

1266. Minimum Time Visiting All Points

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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:
  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)
    )

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;
  }
};

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
  }
}

#define max(a, b) (((a) > (b)) ? (a) : (b))

int minTimeToVisitAllPoints(int** points, int pointsSize, int* pointsColSize) {
  int ans = 0;
  for (int i = 1; i < pointsSize; i++) {
    int x = abs(points[i - 1][0] - points[i][0]);
    int y = abs(points[i - 1][1] - points[i][1]);
    ans += max(x, y);
  }
  return ans;
}