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Given that heapq in python is min heap as specified in python doc, assume that I have a heapq with m elements, what is the time complexity of calling nlargest? I don't think the complexity is O(n*lg(m)) because simply popping the root and heapify again in a min heap only get you nsmallest?
How does heapq.nlargest work?
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Brute force approach: We can check all the nodes in the min-heap to get the maximum element. Note that this approach works on any binary tree and does not makes use of any property of the min-heap. It has a time and space complexity of O(n).
The time complexity to find the minimum element in a min-heap is O(1) , that is the primary purpose of such a container. It was literally made to find the smallest (or largest) element in constant time. The operation that is O(logn) is insertion.
(CLRS 6.1-4) Where in a min-heap might the largest element reside, assuming that all elements are distinct? Solution: Since the parent is greater or equal to its children, the smallest element must be a leaf node.
Time Complexity of Min/Max Heap The Time Complexity of this operation is O(1).
You can see the code here. Suppose you do a heapq.nlargest(n, it), where it is an iterable with m elements.
It first constructs a min heap with the first n elements of it. Then, for the rest m-n elements, if they are bigger than the root, it takes the root out, puts the new element, and shifts it down. At the end, the complexity is O(log(n) * m).
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