PREREQUISITES  Sorting, Binary Searching, Sliding Window QUICK EXPLANATION  Sort the lengths array. For each query K, binary search for the largest element smaller than K and name it cur. Then binary search again on the prefix difference sum array (which will give the number of snakes to be fed to snakes in [prev+1, cur] to make their length at least K) in the region [1, cur] for finding the least index of snake to be killed and let it be prev. The answer is Nprev. EXPLANATION  Some key observations before we jump to the solution 
There are many solutions to this problem some of which are discussed below  ONLINE SOLUTION  Let us sort the lengths array as the order doesn't matter for the answer. Let us solve the problem for a query having its value as K. The most naive solution would be to keep iterating forward from 1 till we reach an element just smaller than K and name this index as cur. Let us keep one more pointer/variable prev which denotes the number of smallest snakes eaten, initially at 0. We will start feeding from $cur^{th}$ snake. Now, to make it of length K, we need to feed it on (Kl[i]) smallest snakes available, where i is initially cur. We increase the prev by (K  l[i]) and decrease i by 1 to reflect this change. We do this process of increasing prev and decreasing i while the condition (i > prev) remains satisfied. Then we print the answer to be simply N  i  1. Complexity  $O(Q*N)$ This solution is inefficient but can be passed in TL if we optimise the solution by binary searching the cur and prev as both functions are monotonic in nature. Now, let us also make a prefix sum array defined as follows  $$presum[x] = presum[x1] + (10^9  l[x]),\hspace{4mm} \forall \hspace{4mm}1 \leq x \leq N$$ For every query having value K, we first find the value of cur by binary searching on length array as it is in increasing order. Then, we also find the largest index prev which is going to get killed. In the naive solution also, we were taking the summation of (K  l[i]) from cur by iterating backwards and set the prev equal to that summation and it can be observed that the function (i > prev) is monotonic as i is decreasing and prev increasing. prev can be found simply by binary searching method on sample space [1, cur]. Binary searching method works because we are finding the smallest j for which the condition $$presum[cur]  presum[j]  (curj)*(10^9  K) \hspace{4mm} \leq \hspace{4mm} j$$ is satisfied and this function is monotonic with increasing j and then we set prev = j. We want the shortest j as this is the demand of the question to minimize the snakes being eaten OR maximize the snakes of length $ \geq K$ which is (N  j). The need for the presum array is that it is the number of snakes required to be killed to make the length of snakes in [prev+1, cur] to be more than or equal to K and which is the demand of the question to maximize the number of snakes of length K. After finding prev, answer is (N  prev) for the query having value K. Complexity  $O((N+Q)*log(N))$ OFFLINE SOLUTION  Let us again sort the length array in increasing order as the order doesn't matter. Also, let us sort the queries array along with their indices in increasing order of K values. Hence, a 2pointer offline solution will be to maintain 3 variables  cur, prev, prefix  where cur means that $cur^{th}$ element is the largest element smaller than K value of current query which we are processing. prev refers to the largest index smaller than cur which is killed and can also be said that prev smallest snakes will be killed. Another thing to observe is that cur and prev can only increase if the queries are sorted according to K value of queries according to observation 2. Let us assume that currently we are processing answer for $j^{th}$ query in sorted order. prefix refers to $$ \sum \hspace{2mm} (K[j]  l[h]), \hspace{4mm} \forall \hspace{4mm} cur \geq h > prev $$ To calculate prefix, we keep on increasing cur and update prefix by adding (K[j]  l[cur]) to prefix till we get l[cur+1] to be greater than K[j]. Then, We keep on increasing prev and keep updating prefix by subtracting (K[j]  l[prev]) while (prefix > prev). Then we get the answer for that query to be simply (N  prev) as prev refers to the largest index of snake which will be killed. Before proceeding to the next query, we also update the prefix by adding Complexity  $O(N*log(N) + Q*log(Q))$ There also exists another solution which is to do the naive solution as explained above, but to not do computation again for some K previously encountered. This can be implemented easily by mapping K value of query to its answer and checking everytime before computing answer for a query, that if the answer for same K value has been computed before using a BST (map or set in C++). It can be proved easily that it is also efficient. Offline solution implementation  here
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asked 24 May, 14:35
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@jeetu86044 As mentioned in my comment Mo's algorithm almost always involve offline solutions. Question  DQUERY Explanation for mo's algorithm and a bit of offline queries is given in this blog. answered 24 May, 20:06

getting TLE using binary search and sorting,but i havent store the query result in array and check again for previous encountered query in array thats why i am getting tle? answered 24 May, 17:33

why TLE in my code,please somebody explain https://www.codechef.com/viewsolution/13773134 answered 24 May, 19:57
You are sorting the array repetitively unnecessarily every time after each query, thus TLE.
(25 May, 04:21)

Hi, I tried similar approach. Tried some test cases all passed. Can you give me some test case for which my solution is failing? https://www.codechef.com/viewsolution/13767521 answered 24 May, 19:59

@grajesh you sorting the array repetively for each test in loop for(p=0;p<q;p++).Instead sort the array outside the loop only one time .Hence your solution has time complexity O(q*(nlogn+n)) answered 24 May, 20:32

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answered 24 May, 22:08
"The most naive solution would be to keep iterating forward from 1 till we reach an element just smaller than K and name this index as cur." "For every query having value K, we first find the value of cur by binary searching on length array as it is in increasing order." According to the definition of cur mentioned in the editorial, for array [11,13,13,13,14], cur would be $4^{th}$ element.
(25 May, 04:26)

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answered 24 May, 22:12
It means that the sample space of binary search is [1, cur] and the difference sum array will give the number of snakes to be fed to snakes in [prev+1, cur] to make their length at least K where prev refers to the largest index smaller than cur which is killed and can also be said that prev smallest snakes will be killed. For more clarity, please read the explanation part.
(25 May, 04:30)

What's wrong with my python code...Any kind of a help is most welcomed https://www.codechef.com/viewsolution/13774246 answered 25 May, 08:05

@vivek1_007 According to your third solution, which uses some sort of a BST, can you please provide the proof why the solution will fit in TL supposing all the values of $K$ are distinct ? answered 25 May, 11:46

why is the code showing TLE..plz explain..link to code > https://www.codechef.com/viewsolution/13750899 answered 25 May, 13:17

Can anyone tell me any test case for which my code is failing. I have used sorting and binary search and I tried all combinations of test cases I could think of and all are passing. I got a WA for this code  https://www.codechef.com/viewsolution/13767521 . Any help is welcome. Thanks answered 25 May, 14:08

Can anyone please explain why am i getting TLE in my code. https://www.codechef.com/viewsolution/13706936 answered 26 May, 00:41

I have used suffix sum instead of prefix.I have tried running many test cases but have not been able to spot the problem. I am getting a WA. https://www.codechef.com/viewsolution/13765159 Can anyone please point out what's the problem in my solution? answered 26 May, 15:08

Are there any special test cases we need to consider for the online solution? We use the same approach in our submission (for Java); we also test with our own test cases and it's seemed ok. But the submission is still WA. Here is our code: link text. Thank you in advance. answered 26 May, 17:38
@victor_arsenal, I have run your program against my test files and it gives exactly the same output as my own program, which got AC.
(26 May, 20:50)
wow, strange. When I submitedt it in the contest, it returned WA. Anyway, thank you very much.
(26 May, 21:05)

why TLE in my code pls help me https://www.codechef.com/viewsolution/13686479 answered 26 May, 19:06

I think observation 2 is incorrect. Consider the following case: In the case of query one, K=10, 4 snakes are killed. answered 26 May, 19:19

@sharmarahul liscut and lisrem dont get updated in 3rd case since k1>snakelist[i] for every i hope it helps answered 26 May, 21:39

@vivek1_007 In Online Solution, could you please explain why you defined prefix sum as answered 26 May, 23:18

In case of offline solution how to find largest element smaller than K value of current query which we are processing? answered 30 May, 14:19

As requested, here is a video editorial on this: Snackdown: SNAKEEAT  Video Editorial We use the binary search approach here to solve the problem. Cheers! :) answered 31 May, 23:22
Thank you for your videos. I think it will be great if you do some hard problems or problems based on some rare topic :) Keep up the good work!
(01 Jun, 00:14)
1
Hi @ista2000, Thanks for the feedback! There are some hard and rare problems picked up on the channel: Mo's Algorithm with Updates, Fourier Transform, Persistent Segment Trees etc... You could have a look at them.
(01 Jun, 03:33)

I haven taken a different approach to solve the problem, the idea: for each number of eaten snakes compute the length of the smallest remaining snake and store it in an array. All what needs to be done for a query k is to binary search that array for the fist entry >= k and return nindex. When submitting the solution I encountered the following strange behaviour: I get a TLE for the original solution (see my solution). I produced a WA "on purpose" by not considering the last query in the last testcase (see without last query). BUT I get the WA at a time of 0.23 seconds  hence pretty fast. I came to the last solution but optimizing my first one through many steps and the time when I got the WA on purpose constantly decreased (started with 3 seconds and went through 0.99, 0.3 til 0.23). Can anyone explain this behavior to me? I can't imagine a single query causing the time difference from 0.23 to TLE. Am I missing something? Or are the times for WA not representative? But somehow they are, because times decreased as I optimized my solution. Would be glad for any ideas which give some insight on this! answered 02 Jun, 19:19

@vivek1_007
In most naive solution is the condition i>prev correct? It won't work for the case.
N=9
x x x x x 6 6 6 8
k=8
The answer should ideally be 3, but as per the condition (i>prev) the answer would be 5 if we calculate it as Ni.
@vivek1_007
What exactly we are trying to compute in presum array please explain in more detail. why 10^{9} l[i]?
@vivek1_007
please explain the condition (prefix > prev)in offline solution. What is its significance.
@arpit728
I have followed 1based indexing, so you will get 4 as the answer. But, the actual answer is Ni1, because we are doing an extra iteration. So, you get the answer to be 3. Just updated it.
@vivek1_007
Can you please refer this link, I have posted a question.
https://discuss.codechef.com/questions/98924/doubtinsnakeeateditorialfromsnackdown
I have some more doubts.
I used merge sort and binary search in my code, still I was getting Time Limit Exceeded error My code