I am rated at 2100+ so I do agree that 1300 rating is low. But at the same time it solved https://codeforces.com/contest/1553/problem/D which is rated at 1500 which was actually non-trivial for me already. I had one wrong submit before getting that problem correct and I do estimate that 50% of the regular competitors (and probably the vast majority of the programmers commenting in this thread right now) should not be able to solve it within 2hrs.
On the AlphaCode Attention Visualization website [1], the Accepted code shown for 1553D is a O(n^2) Python one, which is supposed to be TLE. It correctly implements a two-pointer solution, but failed to "realize" that list.pop(0) is O(n) in Python. I'm not sure how it passed.
> AlphaCode's solution the "inner O(n) loop" is actually a memmove(), which is optimized to be insanely fast.
Again, it is not. CPython does not do these things.
The web page says, and this is corroborated in the paper,
> Solutions were selected randomly, keeping at most one correct (passes all test cases in our dataset) and one incorrect sample per problem and language. Note that since our dataset only has a limited number of test cases, passing all tests we have cannot completely rule out false positives (~4%), or solutions that are correct but inefficient (~46%).
The “54th percentile” measure did use estimated time penalties, which you can see discussed in Table 4 in the paper, but 1553D was not part of that.
Apologies, I thought you meant ‘optimized’ in a different sense, not in terms of how list.pop is implemented, as AlphaCode wasn't involved in that. You are entirely correct that list.pop uses memmove.
> Someone submitted this 1553D code to Codeforces and it passed
Ah, well that shows you have a 2 second time limit, which is quite a lot of time! Not quite enough to empty a 200k element list with list.pop(0)s, but not far off; a 140k element list squeaks in under the time limit for me.
The proposed O(N²) solution contains many unnecessary operations, e.g. the creation of list c or reversal of the input strings. Maybe it has been copied from a related problem? You can easily solve the task with half as many lines in O(N).
for _ in range(int(input())):
a = list(input())
b = list(input())
while a and b:
if a[-1] == b[-1]:
a.pop()
b.pop()
else:
a.pop()
if a: a.pop()
print("NO" if b else "YES")
I'm trying to solve this for fun, but I'm stuck! I've got a recursive definition that solves the problem by building a result string. I think it's a dynamic programming problem, but right now I can't see the shared sub-problems so :). Some real sour cherries being experienced from not getting this one!
from collections import defaultdict
def backspace(s1,s2):
h = defaultdict(lambda:0)
for x in s1:
h[x] = h[x] + 1
for x in s2:
h[x] = h[x] - 1
j = 0
maxj = len(s2) - 1
for x in s1:
if x != s2[j]:
h[x] -= 1
elif j < maxj:
j += 1
else:
break
return j == maxj and all(y >= 0 for y in h.values())
def random_backspace(s1):
res = []
for x in s1:
if randint(0,1) == 0:
res.append(x)
return "".join(res)
def backspaceTest(s1):
return all(backspace(s1,random_backspace(s1)) for _ in range(100))
I am rated at 2100+ so I do agree that 1300 rating is low. But at the same time it solved https://codeforces.com/contest/1553/problem/D which is rated at 1500 which was actually non-trivial for me already. I had one wrong submit before getting that problem correct and I do estimate that 50% of the regular competitors (and probably the vast majority of the programmers commenting in this thread right now) should not be able to solve it within 2hrs.