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 ?
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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 k in a given range [L, R]. The problem was LRQUER.
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 [L, R], you can find the upper_bound() and then return the minimum of all the values found like this.
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 std::set (because we have to store the elements in sorted order) in C++. If duplicate elements are allowed, then we can use std::multiset in C++.
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)).
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how can i return index of minimum elemnet from segment tree?
@shubhambhattar but how to handle point updates?
@beginner_1111 Updated my answer.
@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
@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.
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when keepning minimum value in the node keep the index too
