I read the approach for finding Range minimum value using 2 Binary Indexed Trees. Is it possible to do this problem using only one BIT when updates are also there?
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No, a normal BIT is only able to answer prefix RMQs. A BIT doesn't have enough information to answer a general RMQ (as it doesn't even store the value of each element).
You sure?
I have read this paper previously .In this paper two BITs have been used . I want to know if we can achieve the same result using a single BIT.
Found one approach for solving RMQ using BIT(without updates). Suppose our query range is [L,R]. We know that BIT[i] stores the result from index [i-(i & (-i))+1] to index i.
So for given query we start from R. Now first we check if index [R-(R & (-R))+1]>=L or not. if yes we just update the minimum value using the value in BIT[R] and update R by subtracting (R & (-R)) from R . if not then we just take minimum of A[R] and current answer and update R to R-1.
We keep on doing this while(R>=L).
This algorithm will return the correct result but cost
time complexity if you query the range [2, n].
Can you provide some explanation for this complexity. Not able to get it.
Firstly R will go to the biggest 2k satisfying 2k ≤ n, then the progress will be
, between the progress 
it would take k times. So the total time complexity is Θ(k2).