57 lines
1.3 KiB
Python
57 lines
1.3 KiB
Python
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!")
|