Yep, that's a great elimination of the suspicious small overheads.
line_profiler is beautiful, I'll definitely be adding it to my
toolbox, thanks for that!
I tried a variant of accumulating the output and printing it all as
a single string, but of course this didn't help, printing is already
buffered.
Jonathan
On 6/8/2017 03:54, Stestagg wrote:
I honestly can't see a way to improve this in python. My best
solution is:
def main(lines):
stack = []
sa = stack.append
sp = stack.pop
si = stack.__getitem__
for line in lines:
meth = line[:3]
if meth == b'pus':
sa(int(line[5:]))
elif meth == b'pop':
sp()
else:
parts = line[15:].split()
end = len(stack)-1
amount = int(parts[1])
for x in range(int(parts[0])):
index = end - x
stack[index] += amount
print(stack[-1] if stack else None)
which comes out about 25% faster than your solution.
One tool that's interesting to use here is: line_profiler:
https://github.com/rkern/line_profiler
putting a @profile decorator on the above main() call, and running
with kernprof produces the following output:
Line # Hits Time Per Hit % Time Line Contents
==============================================================
12 @profile
13 def main(lines):
14 1 4 4.0 0.0 stack = []
15 2000001 949599 0.5 11.5 for line in
lines:
16 2000000 1126944 0.6 13.7 meth =
line[:3]
17 2000000 974635 0.5 11.8 if meth ==
b'pus':
18 1000000 1002733 1.0 12.2
stack.append(int(line[5:]))
19 1000000 478756 0.5 5.8 elif meth
== b'pop':
20 999999 597114 0.6 7.2
stack.pop()
21 else:
22 1 6 6.0 0.0 parts =
line[15:].split()
23 1 2 2.0 0.0 end =
len(stack)-1
24 1 1 1.0 0.0 amount
= int(parts[1])
25 500001 241227 0.5 2.9 for x
in range(int(parts[0])):
26 500000 273477 0.5 3.3
index = end - x
27 500000 309033 0.6 3.7
stack[index] += amount
28 2000000 2295803 1.1 27.8
print(stack[-1])
which shows that there's no obvious bottleneck (line by line) here
(for my sample data).
Note the print() overhead dominates the runtime, and that's with me
piping the output to /dev/null directly.
I had a go at using arrays, deques, and numpy arrays in various ways
without luck, but we're getting fairly close to the native python
statement execution overhead here (hence folding it all into one
function).
My only thoughts would be to see if there were some magic that could
be done by offloading the work onto a non-python library somehow.
Another thing that might help some situations (hence my previous
questions) would be to implement the add_to_first_n as a lazy
operator (i.e. have a stack of the add_to_first_n values and
dynamically add to the results of pop() but that would proabably be
much slow in the average case.
Steve
On Wed, Jun 7, 2017 at 7:34 PM Jonathan Hartley
<tart...@tartley.com> wrote:
Hey.
Thanks for engaging, but I can't help with the most important of
those questions - the large data sets on which my solution failed
due to timeout are hidden from candidates. Not unreasonable to
assume that they do exercise deep stacks, and large args to
add_to_first_n, etc.
Yes, the input looks exactly like your example. All args are
integers. The question asked for output corresponding to the top of
the stack after every operation. I omitted this print from inside
the 'for' loop in 'main', thinking it irrelevant.
I converted to integers inside 'dispatch'. 'args' must have actually
been created with:
args = [int(i) for i in tokens[1:]]
Where len(tokens) is never going to be bigger than 3.
Return values (from 'pop') were unused.
On 6/7/2017 13:25, Stestagg wrote:
Do you have any more context?
For example, is the add_to_first_n likely to be called with very
large numbers, or very often? Does the stack get very deep, or stay
shallow?
I'm assuming that lines look like this:
push 1
push 2
add_to_first_n 2 10
pop
pop
with all arguments as integers, and the final value being returned
from main()?
How did you convert from string inputs to numeric values?
How did you manage return values?
:D
On Wed, Jun 7, 2017 at 6:51 PM Jonathan Hartley
<tart...@tartley.com> wrote:
I recently submitted a solution to a coding challenge, in an
employment context. One of the questions was to model a simple
stack. I wrote a solution which appended and popped from the end of
a list. This worked, but failed with timeouts on their last few
automated tests with large (hidden) data sets.
From memory, I think I had something pretty standard:
class Stack:
def __init__(self):
self.storage = []
def push(arg):
self.storage.append(arg)
def pop():
return self.storage.pop() if self.storage else None
def add_to_first_n(n, amount):
for n in range(n):
self.storage[n] += amount
def dispatch(self, line)
tokens = line.split()
method = getattr(self, tokens[0])
args = tokens[1:]
method(*args)
def main(lines):
stack = Stack()
for line in lines:
stack.dispatch(line)
(will that formatting survive? Apologies if not)
Subsequent experiments have confirmed that appending to and popping
from the end of lists are O(1), amortized. So why is my solution too
slow?
This question was against the clock, 4th question of 4 in an hour.
So I wasn't expecting to produce Cython or C optimised code in that
timeframe (Besides, my submitted .py file runs on their servers, so
the environment is limited.)
So what am I missing, from a performance perspective? Are there
other data structures in stdlib which are also O(1) but with a
better constant?
Ah. In writing this out, I have begun to suspect that my slicing of
'tokens' to produce 'args' in the dispatch is needlessly wasting
time. Not much, but some.
Thoughts welcome,
Jonathan
--
Jonathan Hartley tart...@tartley.com http://tartley.com
Made out of meat. +1 507-513-1101 [1] twitter/skype:
tartley
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