Unofficial editorials December Long challengr Part 2

Hello Guys, I’m back once again with editorials, hoping you would like them. :slight_smile:

This is the part 2 of three posts of Unofficial editorials for December Long Challenge.

For editorials of problems GIT01, CPLAY, and VK18, click here.

For editorials of problem REDBLUE, click here.

This post has editorials for problems CHEFHAM and CHEFEXQ.


Problem Difficulty : Easy-medium

Problem Explanation

Given an array, rearrange it in order to maximize the hamming distance between original array and the rearranged array. Hamming distance is defined as the number of indices i for which A[i] != B[i].

Important thing to note: Maximum frequency of a value is 2.


Well, I’m gonna share the most basic solution that comes in your mind naturally, with O(n^2) time Complexity.

Loop for every position and run a nested loop swapping elements if elements at both positions can be swapped or not.

Two elements at positions i and j can be swapped if A[i] != B[j] && A[j] != B[i]. (B array is another array with same elements in same order). (Swaps should be made only in B array)

After that, calculate hamming distance using a single loop and print the Hamming distance and the B array.

Simple, right. Well, this is my solution, except one observation.

Observation: Above solution will easily fit the time limit. (My solution ran within 0.47 seconds out of 6 second time limit.)

We run the internal loop only at that position i for which A[i] == B[j].

As we know that the maximum frequency of any element is 2, there can be at most 3 iterations (one for each position i and j, and last one is successful one) in the inner loop, making the solution complexity O(3*N), that is same as O(N).

That’s all.

Here’s a link to my


### Problem [CHEFEXQ][5]
### Problem difficulty: Medium

### Prerequisites: Square Root Decomposition

#### Problem Explanation

Given an array with N elements, you have two perform two type of queries:

 1. Update ith value to x
 2. Count Number of magical sub-arrays ending before i whose prefix xor is k.

Magical sub-array is defined as sub-array starting with index 0 of original array. (0 based indexing in my solution).

#### Solution

**Naive Solution**:

For every update query, update A[i] to x and for every query of second type, run a loop from 0 to i, taking prefix xor. whenever prefix xor == k, increment count by 1 and then print count.

Time complexity of update is O(1) and count operation is O(N), making overall complexity O(N*Q), which will give TLE.

Surprisingly, this solution gave 50 points instead of 20 points as above solution would give TLE for N and Q upto 1e4, probably weak test cases for those test files.

Anyways, we now are aiming for 100 points and that's where square root decomposition comes into action.

**Square root decomposition, or the buckets method**, means to divide the given range into blocks of any size (usually of sqrt(N) size), pre-compute answers for every block as a whole, and combine results of different blocks to answer count queries of given range. In case of query having partial start and end block, we will compute answer using naive method till start of next block and also from end of last complete block in range to r. Time complexity per count => Q(sqrt(N)) (Assuming we have sqrt(N) blocks).

For update operation, We just need to update one block, which can be done in O(sqrt(N)) time, assuming block size is sqrt(N).

Well, this was general overview of square root decomposition technique. Now, we need to decide what to store in each block such that we can answer queries for whole block in O(1).

I creates an array xor, which holds the xor of all values present in each block, and a map
(unordered in c++/HashMap in java) for every block, holding the the key value pairs where key is the prefix xor of block up-to any position and value is the frequency of key being the prefix xor of block.

Take example of one block of an array 1 3 5 4 1

prefix xor would be 1 2 7 3 2

Now, map for this block will contain following four key => value pairs:

 1. 1 => 1
 2. 2 => 2
 3. 3 => 1
 4. 7 => 1

Coming to update operation, for each update, we will rebuild whole map again the same way we initially built above, and also update xor[b] with new block xor, where b is block index.

Query operation is simple.

prev = 0 and count = 0

if i is the given index and k is given value, then for every block ending before i,

if map[b] contains key prev^k, count += map[b].get(prev^k)

prev = prev^xor[b].

For remaining portion of count operation, use the simple solution mentioned above.

The variable prev is holding the prefix xor for first b-1 blocks. Querying for prev^k works for following reason.

**Reason:** we need to find number of positions for which **prefix xor of whole array** == k. But every block stores prefix xor of their own block only. We know that if xor(i, j) denote prefix xor of range i to j, then xor(i, j) = xor(i, k) ^ xor(k+1, j) holds true.

xor(i, j) = xor(i, k) ^ xor(k+1, j)

If k = index of last element of last completed block i-1, Then prev = xor(0,k) for ith block.

We need the prefix xor to be k and we have prefix xor up-to last position is prev, So entry in map of ith block, whose key is prev^k, will have prefix xor of whole array as (prev^k) ^ prev = k, which is what we want.

Link to my 


Just like editorial part 1, I again hope you enjoy my editorials as much as you all do like coffee in winters (though i don’t). :slight_smile:

Enjoy coding.


Hi taran,

I have solved CHEFHAM using sorting. Make one copy of given array and sort it. Iterate the sorted and array put the element x at index having value ceiling(x) in original array. if ceiling(x) is null put it at the at the index having vacant space as well as lowest value.

my solution

I think my solution is the same as what chef_keshav is saying (in C++ though). I believe I had nlogn complexity and my solution ran in 0.33 seconds.


How even in the world can this solution pass all the test cases? Or am in not seeing something? :frowning:

CHEFEXQ problem had weak test cases, a lot of O(NQ) solutions managed to pass all test cases and get AC which is really disappointing.

Is there no other method of solving CHEFEXQ problem? If there is can somebody please explain it to me. Thank You

@taran_1407 I attempted the problem CHEFEXQ as you explained above

My Code:- Link to my code

Where am I getting it wrong? Please have a look at my code.

CHEFHAM can be solved in so many ways . The test cases were weak enough for hit and trial for obtaining the solutuion .
CHEFEXQ scoring was bit weird too . The most basic brute force solution of CHEFEXQ was able to fetch 50 points. THere was no need for three different scoring brackets .

Anyways thanks for the editorials . Keep up the good work Taran :slight_smile:

Thanks for the editorial.
Can we do CHEFEXQ with segment trees in any way?

Hey @taran_1407, used same method for CHEFEXQ but getting TLE. Doing Square_Root_Decomposition, storing frequencies in unordered_map, used lazy technique for update task, still getting TLE.

Here is MY CODE .

For the CHEFHAM I used a slightly different approach but got only 60pts because of one test case going wrong. Can somebody figure out where might I go wrong?

Here is my


I handled palindromes and non-palindromes separately.


For CHEFHAM I used the following approach. If n > 3 break the array into two subsequences of length greater than 1, each consisting of unique elements, and cyclically shift each subsequence. Otherwise cyclically shift the longest subsequence consisting of unique elements. All of this can be done in a single pass.


I am also getting TLE with the square root decomposition technique in c++. In Subtask 3, just one test case(the first one) is getting passed and others are TLE. Please Help!

hi Taran,
I have done this in O(n) and my code run in less than (0.1)sec for all test cases.
Just see my code.I think my solution is most optimised than the above posts.

My solution for CHEFHAM:

I LOL’ed hard after this passed.


Those who are getting TLE for CHEFEXQ, use normal array instead of an map/Hashmap. Map increases the time complexity by additional logn factor per query.

In order to set all index of certain block to zero during an update operation, instead of traversing whole array, first decrement all previous indices stored in block and then update the block( see code for further clarification.).

Here is my code getting TLE using map data structure.

Here is my code getting accepted after using array.

1 Like

taran give me some advice to solve medium hard problems on codechef

What’s wrong in my code…can anybody help me out, plz…

Here is my



For CHEFHAM, I tried a different approach, I broke the array in subarrays of length two, now there can be two cases:

  1. Both the elements in a subarray are different, then just swap the elements
  2. If both the elements in a subarray are same, swap the position of this subarray with the nearby subarray.

Take care of the last element left for odd n.
Here is the link to my solution :


1 Like

@taran_1407 can u post a link to your segment tree solution for chefexq