PUSSY_LICKING_LOLI_69's blog

By PUSSY_LICKING_LOLI_69, history, 3 months ago, In English

Given an array A of size n. Do these q queries:

1 pos x: assign A[pos] = x;

2 l r k: find k-th smallest element in range [l,r]

Constraints:

n,q <= 1e5

A[i], x <= 1e9

My first aproach was to use a Wavelet tree with a BST(specifically a Treap) on each of its nodes. This gives a time complexity of O(n*log^2) and can do online queries(I still did it offline, i had to read all the input and compress the values because i don't want to make the Wavelet tree dynamic). However it has a really high constant factor and it takes 7-8 seconds to run with the full constraints.

Here's my code if you're interested (sorry if it's poorly written):

My Code

It is allowed to do queries offline, and the problem also has a tag of paralel binary search.

Can someone please show me the intended solution with paralel binary search (or even optimize my current code) ?

UPDATE: Solution founded :D https://robert1003.github.io/2020/02/05/parallel-binary-search.html#a-hrefhttpacmhdueducnshowproblemphppid5412hdu5412---crb-and-queriesa

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3 months ago, # |
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You can solve it using persistent segment tree

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    3 months ago, # ^ |
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    How would you handle the update queries? I can do it with persistent segment tree if there are no updates

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3 months ago, # |
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Here’s a simpler way to tackle the problem:

Use a Segment Tree: Think of it like a big tree where each node covers a part of the array and keeps a sorted list of the numbers in that part.

For Updates: When you change a number in the array, you update the sorted lists in the parts of the tree that cover that number.

For Queries: To find the k-th smallest number in a range, you collect and merge the sorted lists from the relevant parts of the tree, then pick the k-th smallest from the merged list.

This method is pretty efficient and manages both updates and queries well.

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    3 months ago, # ^ |
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    I remember doing this for a similar problem without updates and small k constraint, but this problem has larger constraints for k, which is up to n. I can be mistaken but it seems your solution has a time complexity of O(n^2*log^2)

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      3 months ago, # ^ |
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      It's probably chatgpt solution

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    3 months ago, # ^ |
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    This approach is O(nq), since merging the lists take O(n).

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    3 months ago, # ^ |
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    How would you merge the lists to find the kth quickly?

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3 months ago, # |
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Could you give the problem link?

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3 months ago, # |
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You can use segment tree or Fenwick tree where $$$t[i]$$$ is ordered_set of indices $$$j$$$ such that value $$$a[j]$$$ is in the $$$i$$$-th segment of segment/Fenwick tree.

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    3 months ago, # ^ |
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    I think that it is similar to my solution, but i used treaps instead of ordered sets. This is the first time I've heard about it and it seems pretty powerful. I wonder if it is allowed to be used in competitions, as it is pretty wild for a pre implemented data structure.

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3 months ago, # |
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Use a merge sort tree and on each node use an ordered set for updates

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    3 months ago, # ^ |
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    That was also my first though but how are you getting kth smallest number in O(logn^2).

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      3 months ago, # ^ |
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      My bad yeah the query will take O(logn ^ 3)

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3 months ago, # |
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You can use MergeSort Tree and then binary search or in this case use a ordered_multiset (since it also suports order_of_key() operation) for update queries. But since, binary search takes up extra $$$O(logN)$$$ time we can use fractional cascading to precompute this at the cost of extra memory by storing the iterators of lower_bound in the right and left child of current node in the Segment Tree, along with the element values.

Thus, each time we query we get a $$$O(logN)$$$ complexity. However, building the update function will be trickier here.

I think combining this with this should work.

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3 months ago, # |
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Holy username. I couldn't help but laugh at it. Definitely a normal and not sussy at all name.

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3 months ago, # |
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3 months ago, # |
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Auto comment: topic has been updated by PUSSY_LICKING_LOLI_69 (previous revision, new revision, compare).

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2 months ago, # |
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Alternative: we can try using a Splay tree. Updates is just updating the value of the node, and for querying, we first splay(l-1), then splay(r+1). Then, binary searching on $$$[l, r]$$$ gives us the answer in $$$O(N\log N)$$$, albeit with a pretty high constant.