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Formatted question description: https://leetcode.ca/all/1266.html

1266. Minimum Time Visiting All Points (Easy)

On a plane there are n points with integer coordinates points[i] = [xi, yi]. Your task is to find the minimum time in seconds to visit all points.

You can move according to the next rules:

  • In one second always you can either move vertically, horizontally by one unit or diagonally (it means to move one unit vertically and one unit horizontally in one second).
  • You have to visit the points in the same order as they appear in the array.

 

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

Related Topics:
Array, Geometry

Solution 1.

// OJ: https://leetcode.com/problems/minimum-time-visiting-all-points/
// Time: O(N)
// Space: O(1)
class Solution {
    int getDist(vector<int> &a, vector<int> &b) {
        return max(abs(a[0] - b[0]), abs(a[1] - b[1]));
    }
public:
    int minTimeToVisitAllPoints(vector<vector<int>>& points) {
        int ans = 0;
        for (int i = 1; i < points.size(); ++i) {
            ans += getDist(points[i - 1], points[i]);
        }
        return ans;
    }
};

  • class Solution {
    public int minTimeToVisitAllPoints(int[][] points) {
        int time = 0;
        int length = points.length;
        for (int i = 1; i < length; i++) {
            int[] prevPoint = points[i - 1];
            int[] curPoint = points[i];
            int curTime = Math.max(Math.abs(curPoint[0] - prevPoint[0]), Math.abs(curPoint[1] - prevPoint[1]));
            time += curTime;
        }
        return time;
    }
}

  • // OJ: https://leetcode.com/problems/minimum-time-visiting-all-points/
// Time: O(N)
// Space: O(1)
class Solution {
    int getDist(vector<int> &a, vector<int> &b) {
        return max(abs(a[0] - b[0]), abs(a[1] - b[1]));
    }
public:
    int minTimeToVisitAllPoints(vector<vector<int>>& points) {
        int ans = 0;
        for (int i = 1; i < points.size(); ++i) {
            ans += getDist(points[i - 1], points[i]);
        }
        return ans;
    }
};

  • 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

############

# 1266. Minimum Time Visiting All Points
# https://leetcode.com/problems/minimum-time-visiting-all-points/

class Solution:
    def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int:
        res = 0
        for (prevX,prevY), (currX,currY) in zip(points, points[1:]):
            res += max(abs(prevX-currX), abs(prevY-currY))
        
        return res

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