LeetCode Entry
1266. Minimum Time Visiting All Points
Path coordinates distance in XY plane
1266. Minimum Time Visiting All Points easy
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Problem TLDR
Path coordinates distance in XY plane
Intuition
For each pair of points lets compute diagonal distance and the remainder: time = diag + remainder. Given that remainder = max(dx, dy) - diag, we derive the formula.
Approach
Let’s use some Kotlin’s API:
Complexity
-
Time complexity: \(O(n)\)
-
Space complexity: \(O(1)\)
Code
fun minTimeToVisitAllPoints(points: Array<IntArray>): Int =
points.asSequence().windowed(2).sumBy { (from, to) ->
max(abs(to[0] - from[0]), abs(to[1] - from[1]))
}