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75 lines
2.1 KiB
Python
75 lines
2.1 KiB
Python
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from __future__ import annotations
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def largest_divisible_subset(items: list[int]) -> list[int]:
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"""
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Algorithm to find the biggest subset in the given array such that for any 2 elements
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x and y in the subset, either x divides y or y divides x.
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>>> largest_divisible_subset([1, 16, 7, 8, 4])
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[16, 8, 4, 1]
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>>> largest_divisible_subset([1, 2, 3])
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[2, 1]
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>>> largest_divisible_subset([-1, -2, -3])
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[-3]
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>>> largest_divisible_subset([1, 2, 4, 8])
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[8, 4, 2, 1]
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>>> largest_divisible_subset((1, 2, 4, 8))
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[8, 4, 2, 1]
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>>> largest_divisible_subset([1, 1, 1])
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[1, 1, 1]
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>>> largest_divisible_subset([0, 0, 0])
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[0, 0, 0]
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>>> largest_divisible_subset([-1, -1, -1])
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[-1, -1, -1]
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>>> largest_divisible_subset([])
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[]
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"""
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# Sort the array in ascending order as the sequence does not matter we only have to
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# pick up a subset.
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items = sorted(items)
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number_of_items = len(items)
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# Initialize memo with 1s and hash with increasing numbers
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memo = [1] * number_of_items
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hash_array = list(range(number_of_items))
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# Iterate through the array
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for i, item in enumerate(items):
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for prev_index in range(i):
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if ((items[prev_index] != 0 and item % items[prev_index]) == 0) and (
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(1 + memo[prev_index]) > memo[i]
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):
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memo[i] = 1 + memo[prev_index]
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hash_array[i] = prev_index
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ans = -1
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last_index = -1
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# Find the maximum length and its corresponding index
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for i, memo_item in enumerate(memo):
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if memo_item > ans:
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ans = memo_item
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last_index = i
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# Reconstruct the divisible subset
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if last_index == -1:
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return []
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result = [items[last_index]]
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while hash_array[last_index] != last_index:
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last_index = hash_array[last_index]
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result.append(items[last_index])
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return result
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if __name__ == "__main__":
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from doctest import testmod
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testmod()
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items = [1, 16, 7, 8, 4]
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print(
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f"The longest divisible subset of {items} is {largest_divisible_subset(items)}."
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)
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