>
>
> Min and max will be computationally expensive if re-computed on the entire 
> window each time. Fortunately, this is not necessary. As each new tick 
> arrives, it is compared to Min and to Max. If it is bigger / smaller than 
> Max/Min, it replaces them. This is O(1) operation as well.
>  
>

But in a moving window, the min/max may drop. I don't understand how you 
would maintain the min/max in O(1) time. 

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