self.X is a sorted list of x-coordinates used by add/remove, where the tracking might start/stop. The corresponding
self.track values tell whether tracking is on at the coordinate and to its right. Removing is really just adding a range of False, so I reuse addRange for it.
class RangeModule(object): def __init__(self): self.X = [0, 10**9] self.track = [False] * 2 def addRange(self, left, right, track=True): def index(x): i = bisect.bisect_left(self.X, x) if self.X[i] != x: self.X.insert(i, x) self.track.insert(i, self.track[i-1]) return i i = index(left) j = index(right) self.X[i:j] = [left] self.track[i:j] = [track] def queryRange(self, left, right): i = bisect.bisect(self.X, left) - 1 j = bisect.bisect_left(self.X, right) return all(self.track[i:j]) def removeRange(self, left, right): self.addRange(left, right, False)
Share my python solution. Really the difficult part is to make sure I use
from bisect import bisect_left, bisect class RangeModule(object): def __init__(self): self.range = [-float('inf'), float('inf')] def addRange(self, left, right): self._updateRange(left, right, 0) def queryRange(self, left, right): li = bisect(self.range, left) ri = bisect_left(self.range, right) return li == ri and li % 2 == 0 def removeRange(self, left, right): self._updateRange(left, right, 1) def _updateRange(self, left, right, op): li = bisect_left(self.range, left) ri = bisect(self.range, right) if li % 2 == op: li = li - 1 left = self.range[li] if ri % 2 == op: right = self.range[ri] ri += 1 self.range[li:ri] = [left, right]
@StefanPochmann I always learn something new from your solutions
@derek3 Mind explaining your logic?
@derek3 Awesome. I had actually briefly considered that idea (i.e., a sorted list of alternating on and off coordinates) but didn't try to implement it, I think it seemed difficult. And reading and thinking through your code, I still think it's difficult :-P. But it's short and fast, I just submitted it three times and it took about 420 ms, beating 100%, 99% and 99%, losing only to an earlier version of itself. (Next fastest submission was my solution, at 476 ms).
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