README_EN.md

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

Python3

class Solution:
    def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int:
        res = 0
        x0, y0 = points[0][0], points[0][1]
        for x1, y1 in points[1:]:
            res += max(abs(x0 - x1), abs(y0 - y1))
            x0, y0 = x1, y1
        return res

Java

class Solution {
    public int minTimeToVisitAllPoints(int[][] points) {
        int res = 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];
            res += Math.max(Math.abs(x0 - x1), Math.abs(y0 - y1));
        }
        return res;
    }
}

TypeScript

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;
}

C++

class Solution {
public:
    int minTimeToVisitAllPoints(vector<vector<int>>& points) {
        int res = 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];
            res += max(abs(x0 - x1), abs(y0 - y1));
        }
        return res;
    }
};

Go

func minTimeToVisitAllPoints(points [][]int) int {
	res := 0
	for i := 1; i < len(points); i++ {
		x0, y0 := points[i-1][0], points[i-1][1]
		x1, y1 := points[i][0], points[i][1]
		res += max(abs(x0-x1), abs(y0-y1))
	}
	return res
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func abs(a int) int {
	if a > 0 {
		return a
	}
	return -a
}

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