By vamaddur, history, 11 months ago, ,

We are given an array of size N and Q queries (1 <= N, Q <= 100000). Each query has 4 parameters: index i, index j, length L, and differences D (0 <= D <= 1). We must answer "YES" if a permutation of the subarray from arr[i] to arr[i+L-1] differs from a permutation of the subarray from arr[j] to arr[j+L-1] in at most D places, and "NO" otherwise.

This sounds like an offline segment tree problem, but I am not sure how to start implementing it.

Can someone give me some hints to get me started and moving in the right direction?

UPD: Just for the sake of it, I tried submitting a naive N^2 solution, which didn't make it in the time limit of 2 seconds, as expected.

UPD2: Original problem statement is here.

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 » 11 months ago, # |   0 Auto comment: topic has been updated by vamaddur (previous revision, new revision, compare).
 » 11 months ago, # |   +8 Can you give me link to problem statement? This statement is unclear
 » 11 months ago, # |   0 I didn't understand the permutation part clearly, but you can do the queries for any array (No need to be permutation) in O(D * lg(n)) by hash.
•  » » 11 months ago, # ^ |   0 Can you further describe this approach?
•  » » » 11 months ago, # ^ | ← Rev. 2 →   0 First you will hash the whole array : hash[i] = hash[i - 1] * BASE + a[i] .Now you can calculate the hash for each subarray in O(1) and two subarrays are different if and only if their hashes are different numbers.Also you can find LCP (Longest Common Prefix) of two subarrays in O(lg(n)) by using binary search.Now you can solve the problem by finding LCP, D + 1 times : cin>>l1>>r1>>l2>>r2; while(d+1) x=LCP(l1,l2); l1+=x+1; l2+=x+1; d--; return (l1>=r1+2 && l2>=r2+2); Tell if I made a mistake or completely misunderstood the problem :)
•  » » » » 11 months ago, # ^ |   0 @Sa1378 That solves the problem for D = 0, but I am kind of unclear on how to compute the hash in constant time.
•  » » » » » 11 months ago, # ^ |   0 That code I sent works for D>0 too.The hash for subarray [l,r] would be this : hash[r] - hash[l - 1] * BASE(r - l + 1) .
 » 11 months ago, # |   -35 I give you a code very different from your approach. The problem with your code is that you are using the original string in order to add periods and remove Do My Essay vowels and so produce the output string. What you have to do is produce an output string having no vowels defined according to your assignment and having a period prior to each consonant. In case you find a consonant you have to append + consonant to your output string else omit output.
 » 11 months ago, # |   0 Auto comment: topic has been updated by vamaddur (previous revision, new revision, compare).
 » 11 months ago, # |   +28 How to compare two multisets A = {a1, ..., ak} and B = {b1, ..., bk}? Take a random function f. If , most probably A and B are the same multisets. Otherwise they are different.How to check if A and B differs only by one element? Suppose that and . We can get y - x by comparing sums. We can also get y2 - x2 by comparing square sums. Now we know x and y, and we can compare and .
•  » » 11 months ago, # ^ |   +3 What kind of function are we talking about here? I assume a polynomial with at least degree 2 and prime coefficients?
•  » » » 11 months ago, # ^ | ← Rev. 3 →   0 In the first part, he is talking about functions in a general way. A constant function or any other that would not allow us to distinguish two numbers would not work, so we are looking for a bijection or a "almost" bijection.On the second part of the answer he specifies two functions f(x) = x and f(x) = x*x. These functions let you find x and y. Of course you can use other functions if they allow you to solve the equations for x and y.
•  » » » » 9 months ago, # ^ |   0 How can we prove what rng_58 has proposed ?
•  » » » » » 9 months ago, # ^ |   +4 In general, you can't. He is using hash functions to compare sets. So there's always a probability that his method fails.
•  » » » 11 months ago, # ^ | ← Rev. 2 →   0 One simple approach that would hopefully work is Zobrist hashing. Assign each element x a random (64-bit) number zob[x]. The hash of a range a[i..j] is zob[a[i]]^zob[a[i+1]]..^zob[a[j]]. Use a segment tree over the xorsum of the zobrist values, to calculate the hash of an arbitrary range in logarithmic time. Edit: Actually, you don't even need a segment tree, a prefix array should be enough (thanks to the properties of the xor operator, i.e. hash(a[i..j]) = hash(a[0..j]) ^ hash(a[0..i-1])).
 » 11 months ago, # | ← Rev. 2 →   0 quite similar to this problem:https://www.codechef.com/JUNE17/problems/CLONEMEYou can take a look of the AC codes. Hope it helps.
 » 11 months ago, # | ← Rev. 2 →   0 A O(n sqrt(n)) solution.(No hash)Step 1:divide the elements into sqrt(n) intervals.Step 2:divide the queries into O(n) parts according to the intervals the leftmost&rightmost elements are inStep 3:deal with the queries like this: vector > query[SQRT_N][SQRT_N]; //...... int l=0,r=0; //[l,r) for(int i=0;icl)l--,ADD_ELEMENT(l); while(r>cr)r--,DEL_ELEMENT(r); //...... Now,we just need to ADD_ELEMENT/DEL_ELEMENT in O(1) timeDeclare an array diff[MAX_ARR],which shows the difference of the times each numbers appears between arr[l,l+L) and arr[r,r+L) (possible negetive),and a variable DIFF which equals to the sum of all of the ABS(arr[k])),and it can also mean the minimum elements that 'a permutation of the subarray from arr[i] to arr[i+L-1] differs from a permutation of the subarray from arr[j] to arr[j+L-1]'When adding/removing elements,we just change two of them,so the change of arr[] and DIFF can all be O(1) (I'm sorry I don't have a complete code)Sorry for my poor English :)UPD: I've found the English name of this algorithm: Mo's algorithm
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•  » » 10 months ago, # ^ |   -15 This problem is really complex. I will need hours to solve this one by myself. If it takes more time, then I will surely have to get the help form other people too. help with dissertation proposal
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 » 8 months ago, # |   0 Take the primary detail. You've were given your first subarray. Do you've got a second detail? Add that one too, there you have got a 2d one. If there may be a third upload that one, and so on. We do that till we've got the very last subarray it is technically the whole array of length n. But now we go returned two positions, and skip the n-1 detail and you do add the nth detail... You get a new subarray. Write My Essay For Me
 » 8 months ago, # |   0 Thank you for uploading this because I had been doing research on comparing permutations and I was unable to find a good information regarding this. Academic Writing