[Persistent Segment Trees] 351 - D Jeff and Removing Periods

351 - D - Jeff and Removing Periods

Problem statement:

Problem can be found here.The gist of the  problem is:

We're given an array $a$ of length $n$. With that we're given $q$ queries too, which are of the form $l r$. We have to answer the total number of unique numbers in that range plus if any number present with a constant gap (AP of indices) in the range then add 0, else if no such element is present then add 1.

Notice that if an element is present two or one times, then it automatically satisfied the condition mentioned

Approach:

There are two approaches to solve this question:

1. Mo's algorithm for sorting the queries and finding the answer. I have done this question and here is the solution. We will not discuss it here, since it is already discussed in the editorials.
2. Persistent segment trees. 

I followed rlac's submission to understand  this method.

Description:

You need to know about persistent segment trees, in order to understand this solution. If you don't then you can find material regarding them at these places:

• GKCS-MKTHNUM (SPOJ)
• This codechef tutorial

In my opinion, the idea isn't that much hard to understand. It's just like version control. Once you update some value, which is a leaf, you create new parents till the root, for each update.

Now to solve this problem we need two persistent segment trees:

1. To find the total number of  unique numbers for a range $l,r$
2. To find whether any element's indices follow any AP for a range $l,r$

For the first tree, the logic can be like:

We go from 0, to n-1 in this case.

You create an array pos, which saves the element's last occurrence. So, if the current element has been seen before, then we update that position with a zero (this creates a new root). Then for the current number, we update the for the current index with a 1. We can save this root as the ith root.

When queried, we can query the segment tree root r. The operation performed is the sum operation. That is, the parent's value is calculated by adding the values of the left child and the right child.

Initially, I thought why not just use a regular segment tree? But actually changing the one to zero, might affect some answers, so a regular segment tree might have not worked in that case.

Now, for the second tree, we can again use pos to save last occurrences, next_pos, to save the next position of an element in the array, and last to save the element after the last element of the AP.

Note that we go from n-1 to 0, in this case.

So, if the current element was present before, then it's index' value is set to zero for this segtree. Now if we have a next->next element, then we check if an AP is being formed. If not, then we know that two elements always form an AP, so the last value of the current element is the next->next value.

If next->next indeed follows an AP, then we have atleast three elements in AP, maybe more, maybe less. A check must have also been done for the next element which was saved in pos[a[i]]. We can save it's last value as last[i].

Now when queried, we go for the lth 2nd segment tree, and check for (l,r). Here the operator is the max. That is, the value for the parent node is calculated by taking the max of the left and right child.

If the queried result is less that r, then we know the AP got fucked up before the query r. So we know no AP is present in the range.

If the queries results is more than r, then we know that there is an AP in the given range, as the last element after the AP is more than r.

I hope this explanation helps

Here is my submission which I absolutely stole from rlac.

[Trees][Binary Lifting] 501 - D Misha And Permutations

Misha and Permutations

Problem statement:

The problem can be found here. The gist of the problem is:

You're given two permutations of $n$. Their order might be $a$, and $b$. By "order" I mean, the index of the permutation. That is, for $n=3$, $0,1,2$ is the first permutation, whereas, $2,1,0$ is the last (=6th). 

So, we have to find $(a+b)%(n!)$th permutation. 

This is it. Simple. Isn't is? 

No. Nope. Nada. Wtf is this question man?

Okay, rants apart.

Ideas:

The major point here is the conversion of order, and the permutation. We convert the initial permutations into the order. Add them. Mod by n!, and then find the resulting permutation.

There are many problems with this approach. You'll have to find ways to interconvert order and permutation, plus the mod by n!. Since $1 \leq n \leq 200000$, thus $n!$ can be a very large number. You don't want to solve, this way.

Enter the [Factorial Number System](https://en.wikipedia.org/wiki/Factorial_number_system).

As the name suggests, it's the factorial number system. Instead of bases, you have factorials! An example would be:

$$ 4\times 4! + 3\times 3!+2\times 2!+1\times 1!+0\times 0!$$.

This number in factorial representation (or factoradic) can be written as $43210_{!}$, which in decimal representation can be written as $119_{10}$. 

A good property is that, we can represent permutations using factoradics. This way, we can ignore the decimal representation altogether. But you'll say what about the n! mod? We'll come back to that later. I promise.

Pemutation to factoradic conversions:

Given a permuation $3,0,2,1$. It's factoradic representation is not simply $4132_{!}$. We have to put some mind to it. 

We have $n=4$. So we have $A\times 3! + B\times 2! + C\times 1! + D\times 0!$. We have to find the values of $A,B,C$ and $D$. 

A = We can simply do that by finding number of elements smaller than 3. That is 3. 

Now we remove 3 from consideration, and have $\{0,1,2\}$. 

B = Number of elements smaller than 0? None. So 0. Remove 0. We're left with $\{1,2\}$.

C = Number of elements smaller than 2? 1. So 1. We have $\{1\}$ left.

D = 0

So, we have $3010_{!}$, representing $3,0,2,1$. To double check, you can find the decimal value, which comes out to be: $3\times 3!+0\times 2!+1\times 1!+ 0 \times 0! = 18+0+1+0 = 19$

As you can see here, 3021, is indeed the 19th permutation for n=4. 

The basic inference you can get from here is that factoradic representation gives the rank of the number, for the remaining numbers. 

This can be done by fenwick trees / Binary indexed trees, easily.

Modulus n!: 

Thanks to Ecenerwala's solution.

So, until now, we have converted the permutations into their factoradic representations, and added them. How do we do the mod n!?

A factoradic representation is of the form (for n):

$$X_{(n-1)!}\times (n-1)!+X_{(n-2)!}\times(n-2)!+...+X_{2!}\times 2!+X_{1!}\times 1!$$.

Thus if $X_{(i-1)!}>=i$, then it can be written in the form: $X_{(i-1)!} = i+(X_{(i-1)!}-i)$. Thus, the term $$X_{(i-1)!}\times (i-1)! = i \times (i-1)! + (X_{(i-1)!}-i\times (i-1)!) \\= i!+(X_{(i-1)!}-i\times (i-1)!$$

This extra $i!$ can be added to the next greater element. So, if we percolate up to $(n-1)$, then we might get an extra $n!$ element. That's it!

Isn't that what we want? We know that there can be at most one n! extra. You can check that. (Kuchh to krlo khud se)

Code for that:

Good. Beautiful.

Factoradic to permutation conversion:

So, as I mentioned earlier, factoradic-> Rank (well actually, number of elements smaller). So you have ranks, and  you want the permutation now. Consider $3010_{!}$ again. You want the (3+1)th element, amongst $0,1,2,3$ = 3. Remove 3. $\{0,1,2\}$ left. You want the (0+1)st element. You get 0. Remaining $\{1,2\}$. You want the (1+1)nd element. You get 2. Remaining $\{1\}$. You want the (0+1)st element. You get 1. So $3,0,2,1$ formed.

This can be done by binary searching on he fenwick tree, or doing binary lifting on it. 

My submission can be found here.

[Math][Codeforces] 1247C P Binary

P Binary:

My solution:

I was giving a virtual contest for this. The question before this one took 1 hour (ughh), and thus I couldn't solve this in time. So I solved this later.

So you're given a number $n$, which we need to check how many p-binary numbers are needed to make this $n$ number. 

What are you gonna do?

So basically let's say we have $n$ which we describe using $m$ p-binary numbers. That will look like:

$$ n = (2^{x_1}+p)+(2^{x_2}+p)+...+(2^{x_m}+p)$$

That can be made to look like:

$$ n = (\sum_{i=1}^{m} 2^{x_i})+m*p$$

That is equal to:

$$ n - m*p = (\sum_{i=1}^{m} 2^{x_i}) $$

We do not know the value of $m$ here. So,we can create an array with:

$$ n-i*p$$ where i goes from ${1,...,100}$, because you don't need more than that.

Now, check the number of set bits for each number in the array, starting from the left. Once you reach a point, where $\text{number of set bits} <= i$, you stop. As you can create $i$ such elements in the expansion given above, from a lesser number of bits. Just remember that you can create 2 elements from one set bit if the number is not 1. 

For example if you have a 2 (10), you can have two 1s (01). Similarly, you can see that you can the same for 4,8, et cetera. But you can't divide 1. So remember to check that $n-i*p>=i$, which means that the number can create $i$ elements. 

The logic here is that a number $j$ can, at max, be divided into $j$ elements.

It's a good question. 

Here's the code:

[SegmentTree][SPOJ] ORDERS

ORDERS

Problem statement:

As you are probably well aware, in Byteland it is always the military officer's main worry to order his soldiers on parade correctly. In Bitland ordering soldiers is not really such a problem. If a platoon consists of n men, all of them have different rank (from 1 - lowest to n - highest) and on parade they should be lined up from left to right in increasing order of rank.
Sounds simple, doesn't it? Well, Msgt Johnny thought the same, until one day he was faced with a new command. He soon discovered that his elite commandos preferred to do the fighting, and leave the thinking to their superiors. So, when at the first rollcall the soldiers lined up in fairly random order it was not because of their lack of discipline, but simply because they couldn't work out how to form a line in correct order of ranks. Msgt Johnny was not at all amused, particularly as he soon found that none of the soldiers even remembered his own rank. Over the years of service every soldier had only learned which of the other soldiers were his superiors. But Msgt Johnny was not a man to give up easily when faced with a true military challenge. After a moment's thought a solution of brilliant simplicity struck him and he issued the following order: "men, starting from the left, one by one, do: (step forward; go left until there is no superior to the left of you; get back in line).". This did indeed get the men sorted in a few minutes. The problem was solved... for the time being.
The next day, the soldiers came in exactly the same order as the day before, and had to be rearranged using the same method. History repeated. After some weeks, Msgt Johnny managed to force each of his soldiers to remember how many men he passed when going left, and thus make the sorting process even faster.
If you know how many positions each man has to walk to the left, can you try to find out what order of ranks the soldiers initially line up in?

Input

The first line of input contains an integer t<=50, the number of test cases. It is followed by t test cases, each consisting of 2 lines. The first line contains a single integer n (1<=n<=200000). The second line contains n space separated integers w_i, denoting how far the i-th soldier in line must walk to the left when applying Msgt Johnny's algorithm.

Output

For each test case, output a single line consisting of n space separated integers - the ranks of the soldiers, given from left to right in their initial arrangement.

Example

Input:
2
3
0 1 0
5
0 1 2 0 1

Output:
2 1 3
3 2 1 5 4 

My approach:

There was an approach of mine that worked in $O(n^2 log n)$ time. So bravo I'm a genius. So I read solutions for this question everywhere, then I understood the answer, and eventually it was pretty easy.

So you're at the right most element. It says "1". That means it has one element that is greater than it on the left of it. We have 5 numbers available, 1,2,3,4 and 5. Now, if it were 3, then we would have a 2, as 4 and 5 are definitely in the array. So, no, it must be a 4. As 4 has one element that is greater than it. Cool. Now we can move forward without 4. So we have 1,2,3, and 5.

We are at the 4th position now, which says 0. So, it has no element on the left of it that is greater than it. What such element can be possible? The greatest of those left. Right? That is $(\text{total elements left} - \text{number of elements greater than it})^{th}$ element. That's $4-0 = 4$, that is, the 4th element in the numbers that are left out. That seems to be 5. So we remove it, and we're left with 1,2,and 3.

Again, we are at the 3rd position, which says  2. So, we find the element which has 2 elements greater than in in the remaining elements. That is, $3-2=1$. So we're talking about 1. So, let's put 1 at this position and move forward. So, now we have only 2 and 3 remaining.

You get my point. Right?

So how will you do it optimally? We have to care for the removed elements. We should know that the 4th element after the removal of 2 is 5. So, we can create a segment tree for that, where each number has a 1 assigned to it.

Once we use that number, it's 1 becomes a 0. The parent nodes have the sum of the child nodes. Updation is easy peasy too.

Here is the code:

[SegmentTree][SPOJ] BRCKTS

BRCKTS

Question statement:

We will call a bracket word any word constructed out of two sorts of characters: the opening bracket "(" and the closing bracket ")". Among these words we will distinguish correct bracket expressions. These are such bracket words in which the brackets can be matched into pairs such that

• every pair consists of an opening bracket and a closing bracket appearing further in the bracket word
• for every pair the part of the word between the brackets of this pair has equal number of opening and closing brackets

On a bracket word one can do the following operations:

• replacement -- changes the i-th bracket into the opposite one
• check -- if the word is a correct bracket expression

Task

Write a program which

• reads (from standard input) the bracket word and the sequence of operations performed,
• for every check operation determines if the current bracket word is a correct bracket expression,
• writes out the outcome (to standard output). 

My approach:

Well well well. This one took a lot of time for me to understand how exactly I would get the answer for the current node. Initially I thought in the direction of having a single value for each node, which is a total for the range, in which a '(' is counted as a 1 and ')' as a -1.

Then I thought of having valid and invalid nodes, depending on the values of their children. But no point into going into the wrong approach.

Well what struck me was the point that all we needed to see the brackets which were imbalanced. That is, the ones that are not matched. So, I could keep the track of left unmatched brackets and right unmatched brackets, for a node, for the range to which it corresponds. Cool.

How to maintain this property?

That is, given that we have these values for the left and right children of a node, can we get the values for the parent node?

Let's say that the left child has $L_{l}$ left unmatched brackets, and $L_{R}$ right unmatched brackets. Similarly, the right child has $R_{l}$ left unmatched brackets, and $R_{r}$ right unmatched brackets.

Now when we're merging the ranges of the left and the right children, we can see that the left unmatched brackets of the left child and the right unmatched brackets of the right child cancel out!

Example: L: )((()) has one left and one right unmatched bracket
                R: )(      has one left and one right unmatched bracket
So, while merging, we can do $cancelledBrackets = min(L_{l},R_{r})$, and for the parent node:
$$P_{l} = L_{l}- cancelledBrackets+R_{l}$$
And:
$$P_{r} = L_{r}+R_{r}-cancelledBrackets$$

So, this is how we can build our segment tree.

For updation, we can simply change the 1 to 0 and a 0 to 1, for both the left and right brackets, at the leaf. And then we can update the parents using the method as described above.

Here is my elegant code minions:

[Segment Tree][SPOJ] FREQUENT

FREQUENT 

Here is the problem statement:

You are given a sequence of n integers a₁ , a₂ , ... , a_n in non-decreasing order. In addition to that, you are given several queries consisting of indices i and j (1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integers a_i , ... , a_j. 

Input Specification

The input consists of several test cases. Each test case starts with a line containing two integers n and q (1 ≤ n, q ≤ 100000). The next line contains n integers a₁ , ... , a_n (-100000 ≤ a_i ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: a_i ≤ a_i+1. The following q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), which indicate the boundary indices for the query.
The last test case is followed by a line containing a single 0.

Output Specification

For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

My Solution:

This is a segment tree question. Not because it can only be done via a segment tree, but because I am trying to learn segment trees, and it's amongst the questions I'm practicing. 

So, IMO there are two things one should think about when using segment trees:

1. Merge

2. Update

Merge  here means that assuming you've your answer for the left and the right child of a node, how will you percolate the result up? What intermediate values must you store in the child nodes such that the current node's solution can be reached efficiently?

Update depends upon the context. Do you have a point update or a range update? How will you manage to update the values of the node efficiently for whole of the segtree?

So these are the things I thought.

So, let's say a node contains the maximum occurring number and it's count. Cool cool. Now how does it's parent benefit from this information? It'll have the max count of it's left and right children. Good. But is there any merging required here? That can only be possible if there is an element that is in both the left and the right child. That means that the element is at the right most part of the left child and the leftmost part of the right child. 

But we don't save this information! So let's save that too. The count of the leftmost and the rightmost element of the node. When we merge, we check these too, and update the max element and it's count accordingly.

Now just go till the top and enjoy.

There's no update so no worries. :)

Some elegant code

My code :/

[Arrays][InterviewBit] Find the duplicate

Find the duplicate

Question:

Given an array of n elements which contains elements from 0 to n-1, with any of these numbers appearing any number of times. Find these repeating numbers in O(n) time and using only constant memory space. 

Here  is the link

 Solution:

So basically you can't use a hashSet initially. So we have to find some way so that the number of checked elements are decreased.

Buckets! These days they are everywhere. We can create equal

  sized buckets here too, and limit the search to a single bucket. What should be the size of such a bucket? If $ \delta = (MAX-MIN)$, and there are $n$ elements, then we can create each bucket of size $$\frac{n}{\delta}$$. But this too will be a $O(n)$ search, so let's do something better.

We can decide to use buckets of size $\sqrt{n}$. 

Now the plan.

We can have $\sqrt{n}$ buckets. While going through all the elements we are given, we can check which bucket it belongs, and then increase the size  of that  bucket. Now the bucket that overlaps will be the one which has the duplicate element. Now, use the hash set. Go again through all the elements and then put the ones in the range in the hash set. Boom. You're done. The element that was already in the hash set will be the duplicate element.

Corner case.

Now this is magical fucking information. When you'll implement this stuff,  you'll realize $\sqrt{n}$ is like never an integer. So you'll have to make an array of buckets of size $\lfloor \sqrt{n} \rfloor+1$ and you'll check if any bucket reaches $\lfloor \sqrt{n} \rfloor$. But there'll be a point when no such bucket will be found because the duplicate is in the last bucket, which doesn't even have a size near to $\sqrt{n}$, so boom you're done. Take care of this case and you'll be okay.

Here is the code: