I know how to solve it for a problem which handles two queries 1)update an index to a new value 2)minimum element in a range.(using segment tree) But how to handle for the case like minimum in a range which is greater than K which is some number given in the query If anyone knows it using fenwick tree/segment tree please explain it clearly. For a segment tree we will build the tree first but here we dont know the value of k because k is given in the queries .So how to build and update in this case ? asked 06 Jan '18, 04:04

I think you can apply a Merge Sort Tree. I solved a similar problem a few days ago where I had to find the value closest to In your case you can do the following: Instead of storing a single value in any node, you can store a range. This range will be the sorted list of the values stored in its children. Then, while querying, you will query in this range. For a single query Edit: To handle point updates, we'll have to update the corresponding parent nodes too. We can update a node in $O(log(n))$ time by searching for the previous element and then inserting a new element, using something like Now, there can be at most $O(log(n))$ updates in a single update operation because the depth of the tree will be of the order of $log(n)$, making the complexity $O(log^2(n))$. answered 06 Jan '18, 13:54
@shubhambhattar but how to handle point updates?
(06 Jan '18, 19:25)
@beginner_1111 Updated my answer.
(06 Jan '18, 19:54)
@shubhambhattar sir i have gone through your AC code for LRQUER but why you have multiplied every array element by 2 and then found the answer accordingly i did it without multiplying by 2 but got WA on half of the subtasks
(07 Jan '18, 09:33)
@divik544 It was done so that I could avoid floating point calculations. If you remember, we had to divide the expression by $2$. Multiplying everything by $2$ helped me with that.
(07 Jan '18, 13:33)
Thanks sir i got it now :)
(07 Jan '18, 14:44)

how can i return index of minimum elemnet from segment tree? answered 09 Nov '18, 00:44
