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def advance(position, xs):
"""
Return the next element in `xs` pointed to by the current `position`.
"""
return xs[position - 1]
def find_duplicate(xs):
"""
Find the duplicate integer in the list, `xs`.
"""
beg = xs[-1]
a = beg
b = advance(a, xs)
# Find the first element of the cycle
cycle_beg = None
while a != b:
cycle_beg = a
a = advance(a, xs)
b = advance(b, xs)
b = advance(b, xs)
# The duplicate element is the element before the `cycle_beg`
a = beg
result = None
while a != cycle_beg:
result = a
a = advance(a, xs)
return result
def find_duplicate(xs):
"""
This is the solution that InterviewCake.com suggests.
"""
# find length of the cycle
beg = xs[-1]
a = beg
for _ in range(len(xs)):
a = advance(a, xs)
element = a
a = advance(a, xs)
n = 1
while a != element:
a = advance(a, xs)
n += 1
# find the first element in the cycle
a, b = beg, beg
for _ in range(n):
b = advance(b, xs)
while a != b:
a = advance(a, xs)
b = advance(b, xs)
return a
xs = [2, 3, 1, 3]
result = find_duplicate(xs)
print(result)
assert result == 3
print("Success!")
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