 # REBXOR - Editorial

Author: Yurii Rebryk
Tester: Kevin Atienza
Editorialists: Pushkar Mishra and Suhash Venkatesh

Medium

### PREREQUISITES:

Tries, Properties of Bitwise XOR

### PROBLEM:

Given is an array A containing numbers in the range 0 to 10^9. We have to find l_1, r_1, l_2 and r_2 (l_1 \leq r_1 < l_2 \leq r_2) such that (\,\,A[l_1] ⊕ A[l_1 + 1] ⊕ \dots ⊕ A[r_1]\,\,) + (\,\,A[l_2] ⊕ A[l_2 + 1] ⊕ \dots ⊕ A[r_2]\,\,) is maximised. Here x ⊕ y refers to Bitwise XOR of x and y.

### EXPLANATION:

Let’s start off by thinking of the simplest algorithm that we can. We we will then optimise it step by step by making certain observations.
So, the simplest thing to do is to implement what the problem says. We can run four loops, one each for l_1, r_1, l_2 and r_2. This gives us an \mathcal{O}(N^4) algorithm which will exceed the time limit for all the subtasks.

How can we optimise this? Let’s make an observation. If we know for each i = 1 to N, the maximum value we can get for (\,\,A[l_1] ⊕ A[l_1 + 1] ⊕ \dots ⊕ A[r_1]\,\,) and (\,\,A[l_2] ⊕ A[l_2 + 1] ⊕ \dots ⊕ A[r_2]\,\,) such that r_1 \leq i and i < l_2, then we can tell the answer in \mathcal{O}(N) by simply taking the maximum of the sum of the two terms over all i.

What remains now is to calculate efficiently for each i the max (\,\,A[l_1] ⊕ A[l_1 + 1] ⊕ \dots ⊕ A[r_1]\,\,) and (\,\,A[l_2] ⊕ A[l_2 + 1] ⊕ \dots ⊕ A[r_2]\,\,) such that r_1 \leq i and i < l_2. Let’s call the two terms lbest* and rbest* respectively. For calculating lbest*, we iterate from j = i to 1 and find the j such that val = (\,\,A[j] ⊕ A[j + 1] ⊕ \dots ⊕ A*\,\,) is maximised. Then, lbest* = max(\,lbest[i-1],\, val\,)\,\,. Similarly, we can fill the array rbest.

Calculating the arrays lbest and rbest require \mathcal{O}(N^2) time. After that, the answer can be computed in \mathcal{O}(N). For this subtask, an \mathcal{O}(N^2) solution will pass.

We need to somehow speed up the calculation of arrays lbest and rbest. Up till now, we haven’t utilised any property of XOR operation in our solution. Let’s try to optimise our algorithm using the following property:
Let cumul* = (\,\,A ⊕ A ⊕ \dots ⊕ A*\,\,).
Then for some j\leq i, cumul[j-1] ⊕ cumul*\, = (\,\,A[j] ⊕ A[j + 1] ⊕ \dots ⊕ A*\,\,).
The proof behind this is left to the reader as a simple exercise.

We can now say that lbest* = max(\,lbest[i-1],\, val\,)\,\, where val = maximum of (cumul[j-1] ⊕ cumul*\,\,) for j = 1 to i.

Calculating lbest this way still takes \mathcal{O}(N^2). For i = 1 to N, we need to somehow optimise finding the index j such that (cumul[j-1] ⊕ cumul*\,\,) is maximum. If we can do it in \mathcal{O}(\log A*)\,, then we can compute the array lbest (and analogously, rbest too) in \mathcal{O}(N\log A_{max})\,.

We can use Tries to get the required complexity. We will keep a trie which holds the binary representation of elements in the array cumul. While processing for i from 1 to N, we will first use the trie to get the value which is maximum of (cumul[j-1] ⊕ cumul*\,\,) for 1 \leq j \leq i, and then insert cumul* into the trie. This will allow us to calculate lbest in the required complexity. Similarly, we can calculate rbest also.

Inserting into a trie is a standard operation. Let us look at how we do the other operation, i.e., for a value x, finding the value y in the trie such that x ⊕ y is maximised. Since, the trie stores binary representations of numbers, we first convert x to its binary representation. Then, we can go down the trie and for each bit in the representation of x, we try to find whether bit of opposite value exists or not. It is easy to see why we seek opposite; because 1 ⊕ 1 and 0 ⊕ 0 = 0 while 0 ⊕ 1 = 1.

### COMPLEXITY:

\mathcal{O}(N\log A_{max})

### SAMPLE SOLUTIONS:

11 Likes

Really nice problem. These are the steps I followed.

For first subtask, it was clear a brute force approach will work. That is exactly what I did. I preprocessed the array and got prefix-xor array. From that, an O(n^2) algorithm checking all subsets worked. But got TLE and SIGSEGV for all the test cases of subtask 2.

Then while surfing, I got this link . From this I got the idea. But I had never coded a trie before. So I googled and studied it. I then implemented it using a tree. But apparently this approach was slow. Got TLE in three test files in subtask 2. Here is my code

Can anyone tell me why it got TLE and how I could have improved it?

Then I studied how to implement it using three and two arrays from this link hoping it would be faster. But I could not understand them. Can anyone explain them please?

Then I read a codeforces blog, that explained how to implement trie using a 2D array. I understood that and implemented it quickly. But I didn’t know what the array size should be. I set it to 1000000 (10^6) and got SIGSEGV.
My code

Then I tried increasing the array size by 10 times and got AC with 100 points.
My final code

Can anyone plz explain what the array size should be and why? Also, shouldn’t array size 10^7 * 2 give SIGSEGV?

4 Likes

+1 to the problem setter for this problem and +5 for the awesome test data. Learned a new way of creating binary trees without using dynamic memory allocation.

Used dynamic memory allocation and this happened : https://www.codechef.com/viewsolution/8082903
Used a different approach and Voila! https://www.codechef.com/viewsolution/8145313

This is one of the best problems i solved till date. 2 Likes

Can anybody help me finding where am I going wrong ?
https://www.codechef.com/viewsolution/8164680

Providing failed test case will be much appreciated.

Hi. I implement a Trie using a struct node and traversed it recursively while inserting and querying. I passed 18/19 tests but got TLE in one. Can anyone explain how can I improve this method to pass all the tests? I think the reason I’m getting TLE is because complexity is O(120 * N) = 1.2 * 10^7 But can anyone explain how can I improve on this?
Here is my solution


 Why it is giving TLE as time complexity is same as editorial.

: https://www.codechef.com/viewsolution/8011026

@anishshah I faced exactly the same problem. Dynamic memory allocation is costly. There is a better way of doing it.
Struct node {int x,y};
Maintain a array of node(of size 410^530 say) and a array index initialized to 1. Now whenever u need a new node just increment the array index and store the updated value in x or y variable of the current node. This way you can avoid dynamic allocation. See this https://www.codechef.com/viewsolution/8145313
I hope the code is clear enough.

I researched / problem solve the approach described in the editorial. Still, TLE on task 2. Times for task 1 suggest I was WAY over time on task 2. I am using python. Is this an impossible task on python or is there a much faster way to implement tries in that language? Thanks so much for any input.

https://www.codechef.com/viewsolution/8161088

I used the same approach with dynamic memory allocation and got TLE in 4 tasks of subtask 2.
my solution is : https://www.codechef.com/viewsolution/8083878
After that I realised an optimisation that maximum posible XOR for any subarray till i 0<=i<n is the value when all the k bits are one, where k is the number of bits in the maximum number in the array.
So whenever this value is encountered, maximum xor subarray after that index will be that value and their is no need to play with the tries further, I used this fact and execution time was nearly halved.
My final AC’ed solution is : https://www.codechef.com/viewsolution/8089627

I used this reference to study the concept of tries : http://threads-iiith.quora.com/Tutorial-on-Trie-and-example-problems

3 Likes

my solution complexity : N+2Nlog Amax+2Nlog Amax ~ O(N log Amax)**

same approach I have I also implemented but I don’t know why test case : 18 gave TLE??

any genuine region??

i dont know why so many people got tle using dynamic memory allocation my solution (using dma) passed in time

https://www.codechef.com/viewsolution/8006674

Can anyone tell me why my solution passed, even after having the hieght of the tree 27 and not 30? also, do we have to make 2 tries, one for calculating lbest* and one for rbest*, or just one trie is sufficient(if yes then how would we calculate the other best of the other part using that tree)?

I kept getting TLE for task 18 under subtask 2 the whole time. I used the same approach as above. Can anybody tell me where I went wrong?
This is my solution: https://www.codechef.com/viewsolution/8165745

well,

i was thinking about any other method to solve this problem because i was getting TLE in 18 subtask using structure (link list) then i decided to use array because struct. is slow but problem was limited array size. then i use array as link list to minimize the no of node instead of storing address of child node i stored index of child node this was new thing for me using structure :-https://www.codechef.com/viewsolution/8162203

using array :-https://www.codechef.com/viewsolution/8166130

2 Likes

Could somebody explain the problem in my code ?
https://www.codechef.com/viewsolution/8098441

can anyone explain the logic how to use trie to solve this problem ??

1 Like

Check this video editorial for stepwise explaination of the solution and check the other videos also …

Happy Coding 1 Like

A suggestion: I think the most time consuming part of the program is converting the integer to its reverse in binary. Earlier, I was using a vector and doing it and it gave me a TLE for both, DMA and array method. So, I suggest you to use

bitset<32> vals = num;

This reduced the runtime to half.

Each insertion in Trie creates O(log(Max(A*))) New Nodes . So here N is 4 * 10 ^ 5 and log(Max[A*]) is 30 hence you need about 1.2 * 10 ^ 7 space

1 Like

@grebnesieh Thanks a lot for teaching me this wonderful way. I always used dynamic memory allocation. This approach is new for me.