SNAKEEAT - Editorial

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 N-prev.

EXPLANATION -

Some key observations before we jump to the solution -

1. The most optimal choice would be to feed the largest snakes shorter than K[i] on the smallest snakes and to not feed the snakes greater than K[i] in length in i^{th} query.

2. If we get some X smallest snakes killed for some query K[i], then, for all queries where K[j] > K[i] for some j, we get Y smallest snakes killed where the condition (Y \geq X) will be satisfied.

There are many solutions to this problem some of which are discussed below -

ONLINE SOLUTION -

Let us sort the lengths array as the order doesnâ€™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 (K-l[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[x-1] + (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] - (cur-j)*(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 doesnâ€™t matter. Also, let us sort the queries array along with their indices in increasing order of K values.
Hence, a 2-pointer 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
[(cur -
prev) * (K[j+1] - K[j])]
. The need for the prefix is that it is the number of snakes required to be killed to make the length of snakes in [prev+1, cur] to be exactly equal to K[j] and which is the demand of the question to maximize the number of snakes of length K[j].

Remember, that we sorted the query and length arrays both in increasing order.

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

2 Likes

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?

What do you mean by online and offline solutions?

Online solution is a solution in which you are answering a query simultaneously as you are accepting them as input OR you are answering each query before accepting next query as input.

Offline solution is a solution in which you are first accepting all the queries as input, then do something with that and then answer queries in one go.

1 Like

So that means I have always written online solutions. But itâ€™s specifically said â€śOnlineâ€ť or â€śOfflineâ€ť sometimes like in GPD. Does it mean anything? Does it add additional constraints? I canâ€™t possibly see any difference in run time myself.

Offline and Online solutions will usually involve completely different strategies. The GPD question mandated an online solution to make sure the only possible strategy followed is that of persistent data structures. And yes, if your code got the value of K and immediately printed its answer, before getting the next K value , it was an online solution!

Common examples of offline processing are questions involving moâ€™s algorithm.
On a side note GPD can be solved online without persistent data structures.

can you provide some examples of offline solution ?

why TLE in my code,please somebody explain
https://www.codechef.com/viewsolution/13773134

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

@jeetu86044 As mentioned in my comment Moâ€™s algorithm almost always involve offline solutions.

Question â€“ DQUERY

Explanation for moâ€™s algorithm and a bit of offline queries is given in this blog.

2 Likes

@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))

well i guess u missed to mention that if there are repeatative elements , for eg consider a series of 11 13 13 13 14 and query being 14 then binary search will retutn the mid 13 instaed of the 13 b4 14, so u need to take care of that according to your solnâ€¦

hen 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]

can someone explain me more clearly this point

You are sorting the array repetitively unnecessarily every time after each query, thus TLE.

â€śThe 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.â€ť

â€śFor every query having value K, we first find the value of cur by binary searching on length array as it is in increasing order.â€ť

According to the definition of cur mentioned in the editorial, for array
[11,13,13,13,14], cur would be 4^{th} element.

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.

@vivek1_007

In most naive solution is the condition i>prev correct?
It wonâ€™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 N-i.

@vivek1_007

What exactly we are trying to compute in presum array please explain in more detail. why 109- l[i]?

Whatâ€™s wrong with my python codeâ€¦Any kind of a help is most welcomed
https://www.codechef.com/viewsolution/13774246