Given that the heapq is the minimum stack in the dragon as specified in the dragon doctor, suppose i have elements There is a hap with, what is the timing complexity of call? I do not think that complexity is O (n * lg (m)) because only you get the best while sliding the root and again covering it in a heap?
You can see the code, assuming that you heapq Nlargest (n, this) , where is an iterable with the m elements; first of n elements < Creates a minimum stack with code> it . Then, the mn element for the rest, if they are larger than the root, then it takes the root out, holds the new element, and takes it down. Finally, the complexity O (log (n) * m) .
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