Welcome to Subscribe On Youtube
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