在 O(n) 中创建二叉树

Question

I have a sequence with n numbers and i want to create a data structure to answer the following question: sequence n = [5,7, 4, 24, 8, 3, 12, 34] I want the min(2,5) then the answer is 3 because a2=7, a3=4, a4=24, a5=8 . So the min(i,j) returns the position of minimum number between (i,j). I thought that a good data structure to save this sequence would be a complete binary tree to save the sequence numbers at leaves. But how can i implement this structure in O(n)?

Answer 1

All you need is a Segment Tree with range minimum query. Here is detailed explanation of it. Building time is O(n) , because there are in tree no more than 2 * n nodes, so final time complexity will be O(n) .

If you need to find not only the minimum value, but also the position, then inside the vertex you need to store not only the minimum, but also where it was reached. How to update such a structure seems clear: when you recalculate the minimum in the father, you need to see from which son it is received and take the corresponding position of the minimum from the son. For leaves, the positions are equal to the positions of the leaves themselves.