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Easy-Medium

# Pre-requisites:

Persistent Data Structures, Segment Tree, Merge Sort Tree

# Problem:

Find the number of comparisons that will be made during the sorting of the given permutation of integers with the provided QuickSort code.

# Explanation:

## Solution to the sub task 1:

It's enough just to translate the given code into your favorite programming language. Given implementation is naturally a QuickSort and it is widely known that it will not make more than N*N comparisons.

## Solution to the sub task 2:

Again, it's enough just to run the provided code. N is quite large here, but again, it's widely known that QuickSort makes about O(N*log(N)) comparisons on randomly generated arrays, so that's still enough.

## Solution to all the sub tasks:

The solutions to sub tasks 3 and 4 are very similar in fact. Your score depends on how efficiently you implement the key idea.

Observation: every time the list A[] is given to the sorting function, A[] always consists of the permutation of the numbers in some range [L; R]. Initially, it's [1; N], then it's [1; pivot-1] and [pivot+1; N] and so on.

Another observation: the relative order of numbers in every possible list A[] given to the sorting will not change.

So, we just have to calculate the middle number by position (i.e. (R-L+2)/2-th number) among all the numbers with values in the range [L; R] in order to get pivot.

For this, we generally use the concept of the "rank-query". The rank query basically asks, what is the k-th least element in a given set, for a queried k. If the set contained the positions of the elements in [L; R], then the (R-L+2)/2-th element is what we are looking for.

This is generally done by using a '1' at the value of the element in the set, and then the k-th element is got by just asking what is the first position to have k 1's to the left of it (inclusive). This data structure is generally implemented by a BIT or a segment tree, but in this problem we will see how using a segment tree is of use to us.

Lets define Pos(x) = position in original array where element x is,
Set(a, b) = the set {Pos(a), Pos(a+1), ..., Pos(b)}.
We are generally interested in Set(L, R) and finding the (R-L+2)/2-th element, but notice:
Set(1, a-1) is a subset of Set(1, b) and
Set(a, b) = Set(1, b) \ Set(1, a-1)

Thus, if just define S(x) = Set(1, x), we would get that given (L, R), we need to find the (R-L+2)/2-th element in S(b) \ S(a-1).

Using our concept of ones to denote positions, set difference where one set is subset of the other is now just finding the first position x where number of 1's to the left of x in S(R) - number of 1's to the left of x in S(L-1) = (R-L+2)/2 !!

Notice now, that once we have N+1 segment trees (each segment tree corresponds to a set S(x)), we would be done, because traversal for a given query (L, R, k) is merely : go to left child with (L, R, k) if number of ones in left child(R) - number of ones in left child(L) <= k, else go to right child with query (L, R, k - #ones).

But, we cannot have N+1 segment trees!! Thats too much memory!! and time to create !! But notice how S(x+1) = S(x) U {Pos(x+1)}. There is just one element being added to the set! Can we save memory and time now? The answer, is yes. The concept used is Persistence. You can read up about persistent data structures online through various sources : Article on Codeproject, Problem QUERY Editorial (the part on "Introducing Persistence"). For wikipedia lovers, what we are aiming to do in our persistent segment tree looks a lot like in the Example on Trees on the wiki page, whose diagrams are self-explanatory). One of the key things to note in persistence, and in persistent segment trees, is that there is no child-parent reference. All references and traversals are one-way: down the tree. Also, node numbers are dynamic and not the standard left-child = 2i, right-child = 2i+1.

Putting it all together:  Root[0] = Empty segment Tree for i in 1 to N Root[i] = Root[i-1].insert(Pos[i]) Answer = f(1, N)

 f(L, R) if(R <= L) return 0 x = getKth(Root[L-1], Root[R], (R-L+2)/2) pivot = A[x] return (R-L+1) + f(L, pivot-1) + f(pivot+1, R) 

getKth(Lnode, Rnode, k) if(Lnode is a single position) return Lnode's position ones = Rnode.leftchild.ones - Lnode.leftchild.ones if(ones <= k) return getKth(Lnode.leftchild, Rnode.leftchild, k) else return getKth(Lnode.rightchild, Rnode.rightchild, k - ones) 

Code for segment tree modification can be found for example in setter's code.

### Analysis of Time complexity and Memory Complexity:

Using persistence, each update operation take O(logN) time and memory.
We totally have an original segment tree taking about 2N memory, plus N more updates taking a total of 2N + NlogN memory.
Time complexity of an update is also O(logN).
Traversals (to get the K'th element) takes O(logN) time as well. And such queries come at most N times since there are at most N pivots.

Thus, overall time and memory complexity of the solutioN: O(N log N)

### Note on getting full points:

Finally, the above implementation will get you only 86 points. It is not because of efficiency (Subtask 3 and Subtask 4 are meant to differentiate time complexity of O(logN) with O(log^2N) per query), but due to the fact that with N as large as 5 * 10^5, the recursion stack would be too large resulting in a Runtime Exception/Segmentation Fault. The best way to overcome this issue, is to convert the implementation to a non-recursive manner.

In fact, since the [L; R] intervals being queried are independent of each other (subproblems don't even overlap), we need not worry about in which order we query the intervals, just that we need to query all the relevant ones. Thus, we can modify our DFS-like recursive implementation into a BFS-like version using queues. This will get you full 100.

There exists also an O(log^2)-per-query approach, however it's more complicated from the setter's point of view.

Problems to practice Persistence:
Codechef: Query
Codeforces: Noble Knight's Path
Hackerrank: Signal Tower
Spoj: DQUERY (online version of this can be solved using persistence)

# Common mistake

Bugs in subtask 1 and 2 can arise from incorrect implementation of Quicksort.  go(L, R): if(R - L <= 0) return 0 p = A[(L+R)/2] tmp1 = tmp2 = 0 for i from L to R: if(A[i] < pivot) less[tmp1++] = A[i] else if (A[i] > pivot) more[tmp2++] = A[i] for i from 0 to tmp1-1 A[L+i] = less[i] for i from 0 to tmp2-1 A[L+tmp1+i] = more[i] return go(L, tmp1-1) + go(tmp1+1, R) + (R-L+1) 

The bug is in the indices used in the last line of code.
In fact, we saw various versions of codes having bugs that were simply due to incorrect indices.

To avoid this kind of bug, it is best to have separate variables for different index-domains, and to completely avoid using "L+i" etc.
In the above example, you could have instead  int j1 = L; for(int i = 0; i < tmp1; i++, j1++) A[j1] = less[i] int j2 = L + tmp1; for(int i = 0; i < tmp2; i++, j2++) A[j2] = more[i] return go(L, j1-1) + go(j1+1, R) + (R-L+1); 

# Setter's Solution:

Solution to subtasks 1 and 2 can be found here
Solution to all the subtasks with O(N log^2 N) complexity can be found here
Solution to all the subtasks with O(N log N) complexity can be found here

# Tester's Solution:

Can be found here

asked 25 Aug '13, 14:31

719436377
accept rate: 0%

2561313

Tests are weak I think, for example this solution will get TL on test:

N

N-1 N-3 N-5 ... 1 2 4 6... N-2 N

where N = 500000.

(25 Aug '13, 15:00) 7★

 6 I used an entirely different approach which may sound really interesting to the rest of you. As you may already know, the general idea of the task is to be able to quickly recover the element in the median position of some array A' - formed when only numbers in an interval [l,r] are taken from the initial array A. For example, if A=[4,3,5,1,2] and [l,r] = [2,4], then A'=[4,3,2] and the element in the median position is the second element, 3. or if [l,r] = [1,4] then we would have A'=[4,3,1,2]. Again, the element in the median position is 3. So, how do we do this? When I have an array A' described by [l,r], say [l,r] = [1,5] and A' = [4,5,3,2,1] I make it a double linked list, and I always keep a pointer to the median element. When I find the median, and now it's 3, i need to split the list into two - one described by [1,2] and another described by [4,5]. I take the shorter of these two lists, and ignore it for now. I make the longer of the two lists by starting from the original one (described by [1,5]) and removing elements. Element removal is O(1). What do we do with all the ignored shorter lists? We put them in a queue, every time building the linked list from scratch. Because we always build a new list from scratch from a shorter list the overall complexity ends up being O(n log n) - interestingly, without using complex data structures or even recursion. I am aware that my explanation is not very clear but as you will see, the code is quite simple. It solves the largest test case in 0.16 seconds. Edit: Just found out that using a different approach for sorting changes the recurrence relation, and then the solution ends up being O(n). Very good! We have an optimal solution! answered 26 Aug '13, 06:26 105●1●4 accept rate: 0% I got your explanation partially, can you tell me what do you mean by building from scratch and how much complexity would it take, see as far my understanding is concerned deletion can take O(n) in worst case and building from scratch can also take O(n), but surely that's not the case otherwise it wouldn't have got AC, so a little more explanation would be great. :) (31 Aug '13, 23:04) slash_it4★
 2 There exist an easier (in my opinion) solution without using a persistent data structures. We will maintain the same segment tree like in setters solution, but not persistent. When we will sort a range [L,R] we will have a segment tree for the Set(L,R). So, the pseudocode looks like this: sort(L, R, Root): if (R - L + 1 <= 1) return 0; ans = R - L + 1; pivot = getKth(Root, (R-L+2)/2); if (pivot - L < R - pivot) for i = L to pivot Root.remove(Pos[i]); ans += sort(pivot + 1, R, Root); Root = Empty Segment Tree for i = L to pivot - 1 Root.add(Pos[i]); ans += sort(L, pivot - 1, Root); else for i = pivot to R Root.remove(Pos[i]); ans += sort(L, pivot - 1, Root); Root = Empty Segment Tree for i = pivot + 1 to R Root.add(Pos[i]); ans += sort(pivot + 1, R, Root); return ans;  The worst case is when the pivot is (R-L+2)/2 all the time and the complexity is O(N log^2 N). answered 25 Aug '13, 15:15 384●1●6●11 accept rate: 28% How is this line Root = Empty Segment Tree implemented in your solution? It can be called O(n) times. (25 Aug '13, 15:30) mmaxio7★ Yes, here we need to implement dynamic Segment Tree (we will create nodes only when we need them), i.e. this line just creates a single node. (25 Aug '13, 15:42) And why is worst case O(N log^2 N) : why is worst case when pivot is in the middle? According to my analysis, if T(N) := worst case for a length N query, then, T(N) = max(over a: a <= N/2, and a is length of smaller breakup) (a logN + aloga + T(N-a) + T(a)). (25 Aug '13, 23:16) prad_adm0★ I can prove, that each position will be inserted and removed at most O(log N) times. At the very beginning all positions are in range of size N (Range [1,N]), if we remove some position - it means that at the next step it will be in part of size at most N/2, after second removing - N/4, after Kth - N/(2^K), so, K <= log N. When the pivot is in the middle all the time, this algorithm will perform exactly NlogN insertions and removing from the Segment Tree, so, it is the worst case. (26 Aug '13, 01:14)
 2 I used the range selection data structure mentioned in this paper: http://users-cs.au.dk/gerth/papers/isaac09median.pdf (subsection 2.1). Range selection was used in order to find the position with rank (R-L+2)/2 among the positions corresponding to the interval of numbers [L,R]. In essence the data structure is a segment tree with O(NlogN) memory. A standard implementation takes O(log^2(N)) per query (and, in fact, it only got me 86 points), but it can be improved to O(log(N)) by using fractional cascading (this way I got 100 points). answered 25 Aug '13, 16:21 10.0k●26●69●90 accept rate: 18% Some O(log^2 N)-per-query solutions for range selection can pass if the constant factor is low, you can take a look at one of my solutions. The key idea of it is very famous. However, there are many more solutions than I could imagine, that's why I like this problem :) (25 Aug '13, 17:09) @xcwgf666: Yes, indeed, it seems that there are many ways to successfully approach this problem, which is very nice. Which is your O(log^2(N)) solution which scores 100 points? Is it one of the solutions linked in the editorial or did you submit it in the practice section? (so I can find it there) (26 Aug '13, 02:08) @mugurelionut, there is a link in the editorial. (26 Aug '13, 03:06) Sir Mugurel i went through the paper but I am not able to understand it properly.....it will be very helpful if you will show the structure of the BST by taking a small example...I will be very thankfull to you.... (15 Oct '13, 12:15) hatim0093★
 1 Can someone please explain how to solve DQUERY using a persistent segment tree? I am unable to find any way to count the number of distinct integers in a particular range without going right down to the leaves and there's no point in doing that. answered 10 Feb '15, 00:06 31●1 accept rate: 0%
 0 how can we solve this by Persistence tree http://www.spoj.com/problems/DQUERY/link text answered 18 Feb '16, 16:34 1 accept rate: 0%
 0 saya sedang mencari Alat Pengecil Perut dan belajar Cara Membuat Website Sendiri di Indramayu Kota Wisata bersama Jolie Jogja Wirobrajan namun akhirnya ketemu Software ERP Indonesia yang mana ia Pilih Nahwa untuk Jasa Travel dan Rental Mobil di Malang dengan Sepatu Ardiles, Bonus Game Berhadiah oleh karena itu saya tidak jadi belajar malah ke Krakatau Steel Peduli Masyarakat bersama Cipto Junaedy answered 19 Feb '16, 13:51 1●1 accept rate: 0%
 0 can someone tell why my code is giving tle link answered 18 Jun '17, 16:38 1 accept rate: 0%
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question asked: 25 Aug '13, 14:31

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last updated: 18 Jun '17, 16:38