1266. Minimum Time Visiting All Points LeetCode Solution
In this guide, you will get 1266. Minimum Time Visiting All Points LeetCode Solution with the best time and space complexity. The solution to Minimum Time Visiting All Points problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Minimum Time Visiting All Points
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:
points.length == n
1 <= n <= 100
points[i].length == 2
-1000 <= points[i][0], points[i][1] <= 1000
Complexity Analysis
Time Complexity:
Space Complexity:
1266. Minimum Time Visiting All Points LeetCode Solution in C++
class Solution {
public:
int minTimeToVisitAllPoints(vector<vector<int>>& points) {
int ans = 0;
for (int i = 1; i < points.size(); ++i)
ans += max(abs(points[i][0] - points[i - 1][0]),
abs(points[i][1] - points[i - 1][1]));
return ans;
}
}; /* code provided by PROGIEZ */
1266. Minimum Time Visiting All Points LeetCode Solution in Java
class Solution {
public int minTimeToVisitAllPoints(int[][] points) {
int ans = 0;
for (int i = 1; i < points.length; ++i)
ans += Math.max(Math.abs(points[i][0] - points[i - 1][0]),
Math.abs(points[i][1] - points[i - 1][1]));
return ans;
}
} // code provided by PROGIEZ
1266. Minimum Time Visiting All Points LeetCode Solution in Python
class Solution:
def minTimeToVisitAllPoints(self, points: list[list[int]]) -> int:
ans = 0
for i in range(1, len(points)):
ans += max(abs(points[i][0] - points[i - 1][0]),
abs(points[i][1] - points[i - 1][1]))
return ans # code by PROGIEZ
