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59 lines
1.9 KiB
Python
59 lines
1.9 KiB
Python
2 years ago
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"""
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Given a list of integers, made up of (hopefully) a small number of long runs
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of consecutive integers, compute a representation of the form
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((start1, end1), (start2, end2) ...). Then answer the question "was x present
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in the original list?" in time O(log(# runs)).
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"""
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import bisect
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from typing import List, Tuple
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def intranges_from_list(list_):
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# type: (List[int]) -> Tuple[int, ...]
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"""Represent a list of integers as a sequence of ranges:
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((start_0, end_0), (start_1, end_1), ...), such that the original
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integers are exactly those x such that start_i <= x < end_i for some i.
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Ranges are encoded as single integers (start << 32 | end), not as tuples.
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"""
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sorted_list = sorted(list_)
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ranges = []
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last_write = -1
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for i in range(len(sorted_list)):
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if i+1 < len(sorted_list):
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if sorted_list[i] == sorted_list[i+1]-1:
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continue
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current_range = sorted_list[last_write+1:i+1]
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ranges.append(_encode_range(current_range[0], current_range[-1] + 1))
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last_write = i
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return tuple(ranges)
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def _encode_range(start, end):
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# type: (int, int) -> int
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return (start << 32) | end
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def _decode_range(r):
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# type: (int) -> Tuple[int, int]
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return (r >> 32), (r & ((1 << 32) - 1))
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def intranges_contain(int_, ranges):
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# type: (int, Tuple[int, ...]) -> bool
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"""Determine if `int_` falls into one of the ranges in `ranges`."""
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tuple_ = _encode_range(int_, 0)
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pos = bisect.bisect_left(ranges, tuple_)
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# we could be immediately ahead of a tuple (start, end)
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# with start < int_ <= end
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if pos > 0:
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left, right = _decode_range(ranges[pos-1])
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if left <= int_ < right:
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return True
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# or we could be immediately behind a tuple (int_, end)
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if pos < len(ranges):
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left, _ = _decode_range(ranges[pos])
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if left == int_:
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return True
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return False
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