If an operation has an amortized time of O(1), can it ever, worst-case, take O(N^2) time?
Can an operation that takes O(1) amortized time have worst-case O(n^2) time?
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Yes, it can. Amortized complexity takes into account the frequency with which the worst case appears. Thus as soon as the worst case appears in about 1 in
N^2operations the amortized complexity will be constant.Let's take a simple example - the dynamically expanding array(I will call that
vectoras it is called in c++) in most languages has an amortized constant time for pushing an element to its back. Most of the time pushing an element is a simple assignment of a value, but once in a while all the elements allocated will be assigned and we need to re-allocate the vector. This would be the worst case of apush_backoperation and when that happens the operation is with linear complexity. Still the way vector grows makes sure that re-allocation is infrequent enough. Each time the vector is re-allocated it doubles its size. Thus before another re-allocation happens we will havensimple push_back operations(assumingnwas the capacity of the vector before re-allocation). As a result the worst case of linear complexity appears at most once in a linear number of operations.Analogously to the case above imagine a data structure that re-allocates in O(n^2), but makes sure that re-allocation is performed at most once in n^2 constant operations. This would be an example of an operation with amortized complexity of O(1) and worst-case complexity O(N^2).