# 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
}
}