I think you have misunderstood the analysis. To keep a reasonable complexity,
you need to make sure that your set cannot contain several times the same
element.
See the comment saying "This is a set, not a multiset!". Therefore, heapq will
not be the right datastructure for this, as it does not allow to check if an
element is in the heap easily.
So basically, you need anything resembling a set, along with a way to count the
number of occurrences of a given element.
On Thursday, August 23, 2018 at 1:37:05 AM UTC+2, Eugene Yarmash wrote:
> Hello. I've tried to solve the problem using Python's heapq module, but
> that gave me 'Time limit exceeded' for the 2nd test case. I've checked
> the analysis and it claims that this approach should work for the 2nd
> test case. Am I missing something here? My code:
>
> from heapq import heappop, heappush
>
>
> def main():
> T = int(input()) # the number of test cases
>
> for case in range(1, T+1):
> N, K = map(int, input().split())
>
> h = [-N]
>
> for _ in range(K):
> chunk = -heappop(h)
> if chunk & 1:
> if chunk == 1:
> min_s = max_s = 0
> else:
> min_s = max_s = chunk >> 1
> heappush(h, -min_s)
> heappush(h, -max_s)
> else:
> max_s = chunk >> 1
> min_s = max_s - 1
> heappush(h, -min_s)
> heappush(h, -max_s)
>
> print('Case #{}: {} {}'.format(case, max_s, min_s))
>
>
> main()
>
>
>
>
>
> --
> Regards,
> Eugene
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