019f8fd211
git-subtree-dir: users/wpcarro git-subtree-mainline:464bbcb15cgit-subtree-split:24f5a642afChange-Id: I6105b3762b79126b3488359c95978cadb3efa789
57 lines
1.3 KiB
Python
57 lines
1.3 KiB
Python
def advance(position, xs):
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"""
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Return the next element in `xs` pointed to by the current `position`.
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"""
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return xs[position - 1]
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def find_duplicate(xs):
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"""
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Find the duplicate integer in the list, `xs`.
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"""
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beg = xs[-1]
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a = beg
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b = advance(a, xs)
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# Find the first element of the cycle
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cycle_beg = None
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while a != b:
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cycle_beg = a
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a = advance(a, xs)
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b = advance(b, xs)
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b = advance(b, xs)
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# The duplicate element is the element before the `cycle_beg`
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a = beg
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result = None
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while a != cycle_beg:
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result = a
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a = advance(a, xs)
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return result
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def find_duplicate(xs):
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"""
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This is the solution that InterviewCake.com suggests.
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"""
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# find length of the cycle
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beg = xs[-1]
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a = beg
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for _ in range(len(xs)):
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a = advance(a, xs)
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element = a
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a = advance(a, xs)
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n = 1
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while a != element:
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a = advance(a, xs)
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n += 1
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# find the first element in the cycle
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a, b = beg, beg
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for _ in range(n):
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b = advance(b, xs)
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while a != b:
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a = advance(a, xs)
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b = advance(b, xs)
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return a
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xs = [2, 3, 1, 3]
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result = find_duplicate(xs)
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print(result)
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assert result == 3
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print("Success!")
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