Kth minimum in a Range

Question

Given an array of integers and some query operations.
The query operations are of 2 types
1.Update the value of the ith index to x.
2.Given 2 integers find the kth m

Answer 1

Here is a O(polylog n) per query solution that does actually not assume a constant k, so the k can vary between queries. The main idea is to use a segment tree, where every node represents an interval of array indices and contains a multiset (balanced binary search tree) of the values in the represened array segment. The update operation is pretty straightforward:

1. Walk up the segment tree from the leaf (the array index you're updating). You will encounter all nodes that represent an interval of array indices that contain the updated index. At every node, remove the old value from the multiset and insert the new value into the multiset. Complexity: O(log^2 n)
2. Update the array itself.

We notice that every array element will be in O(log n) multisets, so the total space usage is O(n log n). With linear-time merging of multisets we can build the initial segment tree in O(n log n) as well (there's O(n) work per level).

What about queries? We are given a range [i, j] and a rank k and want to find the k-th smallest element in a[i..j]. How do we do that?

1. Find a disjoint coverage of the query range using the standard segment tree query procedure. We get O(log n) disjoint nodes, the union of whose multisets is exactly the multiset of values in the query range. Let's call those multisets s_1, ..., s_m (with m <= ceil(log_2 n)). Finding the s_i takes O(log n) time.
2. Do a select(k) query on the union of s_1, ..., s_m. See below.

So how does the selection algorithm work? There is one really simple algorithm to do this.

We have s_1, ..., s_n and k given and want to find the smallest x in a, such that s_1.rank(x) + ... + s_m.rank(x) >= k - 1, where rank returns the number of elements smaller than x in the respective BBST (this can be implemented in O(log n) if we store subtree sizes). Let's just use binary search to find x! We walk through the BBST of the root, do a couple of rank queries and check whether their sum is larger than or equal to k. It's a predicate monotone in x, so binary search works. The answer is then the minimum of the successors of x in any of the s_i.

Complexity: O(n log n) preprocessing and O(log^3 n) per query.

So in total we get a runtime of O(n log n + q log^3 n) for q queries. I'm sure we could get it down to O(q log^2 n) with a cleverer selection algorithm.

UPDATE: If we are looking for an offline algorithm that can process all queries at once, we can get O((n + q) * log n * log (q + n)) using the following algorithm:

• Preprocess all queries, create a set of all values that ever occured in the array. The number of those will be at most q + n.
• Build a segment tree, but this time not on the array, but on the set of possible values.
• Every node in the segment tree represents an interval of values and maintains a set of positions where these values occurs.
• To answer a query, start at the root of the segment tree. Check how many positions in the left child of the root lie in the query interval (we can do that by doing two searches in the BBST of positions). Let that number be m. If k <= m, recurse into the left child. Otherwise recurse into the right child, with k decremented by m.
• For updates, remove the position from the O(log (q + n)) nodes that cover the old value and insert it into the nodes that cover the new value.

The advantage of this approach is that we don't need subtree sizes, so we can implement this with most standard library implementations of balanced binary search trees (e.g. set in C++).

We can turn this into an online algorithm by changing the segment tree out for a weight-balanced tree such as a BB[α] tree. It has logarithmic operations like other balanced binary search trees, but allows us to rebuild an entire subtree from scratch when it becomes unbalanced by charging the rebuilding cost to the operations that must have caused the imbalance.