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:
\n\n1 second, you can either:\n\n\tsqrt(2) units (in other words, move one unit vertically then one unit horizontally in 1 second).\n
Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2019/11/14/1626_example_1.PNG\")
\n\nInput: points = [[1,1],[3,4],[-1,0]]\nOutput: 7\nExplanation: One optimal path is [1,1] -> [2,2] -> [3,3] -> [3,4] -> [2,3] -> [1,2] -> [0,1] -> [-1,0] \nTime from [1,1] to [3,4] = 3 seconds \nTime from [3,4] to [-1,0] = 4 seconds\nTotal time = 7 seconds\n\n
Example 2:
\n\n\nInput: points = [[3,2],[-2,2]]\nOutput: 5\n\n\n
\n
Constraints:
\n\npoints.length == n1 <= n <= 100points[i].length == 2-1000 <= points[i][0], points[i][1] <= 1000Compiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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