diff options
Diffstat (limited to 'scratch/facebook/find-duplicate-beast-mode.py')
-rw-r--r-- | scratch/facebook/find-duplicate-beast-mode.py | 57 |
1 files changed, 57 insertions, 0 deletions
diff --git a/scratch/facebook/find-duplicate-beast-mode.py b/scratch/facebook/find-duplicate-beast-mode.py new file mode 100644 index 000000000000..e246415efd1f --- /dev/null +++ b/scratch/facebook/find-duplicate-beast-mode.py @@ -0,0 +1,57 @@ +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!") |