Codeforces and Polygon may be unavailable from May 23, 4:00 (UTC) to May 23, 8:00 (UTC) due to technical maintenance. ×

YMSeah's blog

By YMSeah, history, 3 years ago, In English

I have been doing competitive programming for some months now. While attempting problems above 1800 in difficulty, I noticed that some of them required the use of ordered sets/multisets that support lookup, addition and deletion of elements in O(log(n)) time. Their solutions usually involve the use of std::set or std::multiset in C++. However, I code primarily in Python and there are no built-in packages available on Codeforces. Normally outside of Codeforces I just import sortedcontainers.sortedlist which I use like an Order Statistic tree. Here I discuss some strategies which I have used to overcome these issues. I hope that it helps others who are stuck so that they do not give up on their favourite languages. For C++ users, this may not be as useful, but you can still see how alternative data structures may be used to solve problems.

The example I use here is CSES — Concert Tickets (for those with no CSES account, see here). This question requires the deletion of entries, and searching for the next largest available entry in sorted array arr smaller than a certain value x.

Method 1: Segment Trees

A segment tree is a very versatile data structure that work for cumulative searches for binary functions with associative properties. Go read up on it if you don't know what it is.

For Concert Tickets, we need to find the largest non-deleted value closest to x. To do this, initialize a maximum segment tree with its values corresponding to its index value. i.e. the segmentTreeArray=[0,1,2,3,...n-1], and queries between [l,r] will give the largest value within the range [l,r]. Everytime we delete a value, we set the index corresponding to that value to -inf (or -1e9, if you prefer). This way, the index will not "interfere" with future query ranges that includes it. To find a value, first perform a binary search on the original array arr to find where the search should end such that arr[i]<x. To find the next largest available index, query for the largest from 0 to i.

If you want to search for next smallest to the right, use a minimum segment tree, set value to inf (1e9) on delete, and search from i to n-1 instead.

Complexity: O(log(n)) for search, insert, or delete. Adding elements is only possible if you know beforehand what are the elements you can add (and thereby allocate a "space" in your segment tree).

What the code might look like:

# class SegmentTree...
# def getMaxSegTree(arr) ...
def main():
    
    n,m=readIntArr()
    
    actual=readIntArr()
    actual.sort()
    custMax=readIntArr()
    
    availIndexes=getMaxSegTree(list(range(n)))
    #if available, mark node as 1. Else, mark node as -inf
    ans=[]
    for cm in custMax:
        b=n
        i=-1
        while b>0: # binary search
            while i+b<n and actual[i+b]<=cm:
                i+=b
            b//=2
        if i==-1: # no smaller value
            ans.append(-1)
        else:
            idx=availIndexes.query(0,i)
            if idx==-inf: # no available smaller value
                ans.append(-1)
            else:
                ans.append(actual[idx])
                availIndexes.updateTreeNode(idx,-inf) # delete the node
    multiLineArrayPrint(ans)
        
    return

Note: If you want to query element counts (i.e how many elements between 0 and i has not been deleted), use a sum Segment Tree and set index to 1 if element is present, and 0 if element is deleted. Then query(0,i) gives you the number of elements present between [0,i].

Method 2: Linked List with Path Compression (like DSU)

Note: From my (rather shallow) experience, this only works with item deletion. Maintain an array parent in parallel to arr of same size. Originally parent looks like [0,1,2,...n-1]. Every time an item is deleted, replace parent[deleted_i] with deleted_i-1. When searching for a value, perform a binary search (like in segment tree example) for the index i such that arr[i]<x. Then, run a search for the parent of arr[i] with while(parent[i]!=i)i=parent[i]. Remember to perform a path compression after that, as this will make future searches faster.

Note that if the search queries are for next larger item (item on right that isn't deleted), then replace parent[deleted_i] with deleted_i+1 on deletion.

Complexity: O(1) for deletion. O(log(n)) for search, because of the binary search. I don't know the complexity of finding-of-parent-with-path-compression (see https://mirror.codeforces.com/blog/entry/2381). This linked-list with path compression runs quickly. For my submission for this question, it runs close to 2x faster than segment tree.

What the code might look like:

# no special classes needed :). only needs the parent array.
def main():
    
    n,m=readIntArr()
    h=readIntArr() # n prices
    parent=[i for i in range(n)] #price parent
    t=readIntArr() # m customers
    h.sort()
    
    ans=[]
    for x in t:
        b=n
        i=-1
        while b>0: # binary search
            while i+b<n and h[i+b]<=x:
                i+=b
            b//=2
        if i==-1:
            ans.append(-1)
        else:
            eventualParent=i
            while eventualParent!=-1 and parent[eventualParent]!=eventualParent:
                eventualParent=parent[eventualParent]
            if eventualParent==-1: # no smaller item found
                ans.append(-1)
                eventualParent+=1
            else:
                ans.append(h[eventualParent])
            eventualParent-=1 # eventualParent is deleted. to update its parent to eventualParent-1 later
            # path compression like DSU
            while i!=eventualParent:
                temp=parent[i]
                parent[i]=eventualParent
                i=temp
    multiLineArrayPrint(ans)
    
    return

Method 3: Actually using a self-implemented B-tree or Order Statistic Tree

This is straightforward. Just copy-paste already written code. For me, I copy-pasted the import library of sortedcontainers.sortedlist which is 800+ lines of code (eg. 111203644) after removing empty lines. I dislike this method because 1) it makes my code super long, 2) slows down my IDE which must parse all that code, 3) tends to have the largest overhead (usually slower than using segment tree). However, it doesn't require innovative thinking on how to manipulate a segment tree/linked list etc. to perform equivalent functions. Time complexity: O(log(n)) for search, insert and delete.

What the code might look like: (looks elegant, but imagine I have 800+ lines of code for the OrderedList data structure. By the way, the OrderedList is my implementation of the Order Statistic Tree)

EDIT: rishabnahar2025 has a good suggestion for an Order Statistic Tree template for Python users: here. This code is only 200+ lines long and runs faster than the OrderedList that I used previously. As pointed out by xuanji in the comments, the documentation for the original sortedcontainers.SortedList implementation (for Python) mentioned that its operations run in n^(1/3) to n^(1/2) with small overheads. I should point out that I have used PyRival's SortedList to solve many problems requiring nlogn time complexity.

# class OrderedList ... (800+ lines)
def main():
    
    n,m=readIntArr()
    h=readIntArr() # n ticket prices
    t=readIntArr() # m customers
    
    h.sort()
    ol=OrderedList(h)
    
    ans=[]
    for x in t:
        idx=ol.bisect_left(x)
        if idx==len(ol) or ol[idx]>x:
            idx-=1
        if idx==-1:
            ans.append(-1) #cant take anything
        else:
            ans.append(ol.pop(idx))
    multiLineArrayPrint(ans)
    
    return

Another Example

If you'd like to see these concepts used in another question, you may look at my solutions for 1642E - Anonymity Is Important where a part of each of them requires marking the earliest visited times for items in indexes 1 to n, and each time a range is given (ranges may overlap).

Method 1: Direct update in using Lazy-Propagated Segment Trees : 151336538 (Note that I subsequently transferred the final result to a sparse table for faster range querying of the final result. This is out of scope for thisdiscussion so ignore that code)

Method 2: Linked list with path compression : 151410511

Method 3: Using some sort of Order Statistic Tree (PyRival's SortedList) : 151340346

Alright that's it. Thanks for reading this far, and I hope that this has been helpful or informative. If there're mistakes in my post please let me know.

  • Vote: I like it
  • +36
  • Vote: I do not like it

| Write comment?
»
3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Auto comment: topic has been updated by YMSeah (previous revision, new revision, compare).

»
3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

You can include code for SortedList from here, This is really helpful, and you can include it in your template and forget. Using this sometimes leads to TLE(Ex. when problem authors allocate 1 sec for a problem, and you have an Nlog(N) solution with SortedList). But still, it is beneficial.

  • »
    »
    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    What is the complexity of inserting, deleting and finding the element for this ds?

  • »
    »
    3 years ago, # ^ |
    Rev. 2   Vote: I like it +5 Vote: I do not like it

    Wow thank you! I didn't know about this. I tried this SortedList and it ran faster than the one I ripped directly off the sortedcontainers library (probably due to more checks in the production-standard sortedcontainers library). I will be sure to update the Blog post to include this!

  • »
    »
    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I have read the implementation for a while, and i think the complexity for insertion/deletion operations is around $$$O(C) \sim O(\sqrt{N})$$$, and $$$O(\log N)$$$ for searching.

    • »
      »
      »
      3 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Ohh Are you sure...? Because I have used it at many occasions where O(sqrt(n)) will easily exceed the time limit. Maybe asymptotic complexity might be good enough.Tbh I was never able to understand the implementation.

      • »
        »
        »
        »
        2 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        xuanji shared a link discussing the implementation of SortedList and how it's really N^(1/3) to N^(1/2) with low overheads for some of the operations.

»
2 years ago, # |
Rev. 2   Vote: I like it +11 Vote: I do not like it

Actually the cost of inserting into sortedcontainers.SortedList not log N, but rather N^1/2 or N^1/3. You can see the analysis at https://grantjenks.com/docs/sortedcontainers/performance-scale.html.

According to them, even though the asymptotic complexity is higher, the smaller memory overhead (vs e.g. red-black tree like in stl::set) and less pointer-chasing means that the constant factor is much lower. The benchmark against std::map here https://grantjenks.com/docs/sortedcontainers/performance.html (but note that it's actually std::map with a wrapper to let python code call it),

And the code for sortedcontainers.SortedList and PyRival's SortedList is quite similar, I think the complexity is the same

»
2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Auto comment: topic has been updated by YMSeah (previous revision, new revision, compare).

»
2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Auto comment: topic has been updated by YMSeah (previous revision, new revision, compare).

»
5 months ago, # |
  Vote: I like it -8 Vote: I do not like it

Why not create a template for sortedList and use it whenever you want xD? I know it makes the code size very long but still works. Something like: https://mirror.codeforces.com/contest/1915/submission/239598054