Re: [algogeeks] BT
if the root of T2 is duplicated in T1,this code may give a wrong
answer(as there is no provision of backtracking in case there is a mis
match after some matchings).given above KMP is the best possible answer
i think..
On Wed, Jan 12, 2011 at 11:28 AM, pacific pacific wrote:
> Here is my pseudo code :
> check( node t1 , node t2 )
> {
> if ( t1->data == t2->data)
> {
> return check( t1->left , t2->left ) && check(t1->right, t2->right) ;
> }
> else
> {
> return check(t1->left , t2) || check(t1->right , t2);
> }
> }
>
> time complexity : o(n) because each node in t1 needs to be visited once.let
> me know if this works.
>
>
> On Sun, Jan 9, 2011 at 1:30 PM, Harshal wrote:
>
>> @Nishaanth
>> T1 has millions of nodes. Suppose all the nodes of T1 are equal to root of
>> T2. Then u will have to check every where in T1. Putting height as
>> constraint, u will have to check only those nodes whose height is equal to
>> T2. It will reduce time complexity.
>>
>> Well m not able to think of better time complexity, another way would be:
>> Find Height of T2 ... O(k) //k is no. of nodes in T2
>> Find Height of each node in T1... O(N) //N is no. of nodes in T1
>>
>> now if p nodes in T1 have height same as T2, then we can find if a subtree
>> rooted at any of those p nodes are identical to T2 in O(pk) time.
>>
>> Thus total time complexity: O(N) + O(k) + O(pk).
>> correct me if I am wrong..
>>
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Re: [algogeeks] Re: ThreeListSum
Thanx Dave...same logic as finding a sum in a sorted array..fool to miss it On Tue, Jan 11, 2011 at 8:40 AM, Dave wrote: > @Anoop: It can be done in O(n^2): > > Sort arrays B and C > For i = 1 to n Do >j = 1 >k = n >While( j <= n and k >= 1) >If( A(i) + B(j) + C(k) < 0 ) Then j = j + 1 >Else if( A(i) + B(j) + C(k) > 0 ) Then k = k - 1 >Else Return TRUE >End While > Return FALSE > > The While loop is iterated at most 2N times for each i. > > Dave > > On Jan 11, 4:02 am, anoop chaurasiya > wrote: > > does any better solution exists than O(N^2logN)?? > > > > On Tue, Jan 11, 2011 at 2:56 AM, juver++ wrote: > > > There is no need to sort first two arrays. > > > > > -- > > > You received this message because you are subscribed to the Google > Groups > > > "Algorithm Geeks" group. > > > To post to this group, send email to algoge...@googlegroups.com. > > > To unsubscribe from this group, send email to > > > algogeeks+unsubscr...@googlegroups.com > > > > . > > > For more options, visit this group at > > >http://groups.google.com/group/algogeeks?hl=en. > > > > -- > > Anoop Chaurasiya > > CSE (2008-2012) > > NIT DGP > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- Anoop Chaurasiya CSE (2008-2012) NIT DGP -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] ThreeListSum
can't u presort one of them in O(nlogn)...??? On Tue, Jan 11, 2011 at 5:05 AM, dinesh bansal wrote: > I think for unsorted lists, it will be O(n^3). > > > > On Tue, Jan 11, 2011 at 1:26 PM, juver++ wrote: > >> There is no need to sort first two arrays. >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algoge...@googlegroups.com. >> To unsubscribe from this group, send email to >> algogeeks+unsubscr...@googlegroups.com >> . >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> > > > > -- > Dinesh Bansal > The Law of Win says, "Let's not do it your way or my way; let's do it the > best way." > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Anoop Chaurasiya CSE (2008-2012) NIT DGP -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] ThreeListSum
does any better solution exists than O(N^2logN)?? On Tue, Jan 11, 2011 at 2:56 AM, juver++ wrote: > There is no need to sort first two arrays. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Anoop Chaurasiya CSE (2008-2012) NIT DGP -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] find triplets in an integer array A[] which satisfy condition: a[i]^2 + a[j]^2 = a[k]^2
@fenghuang try this array:
a[]={3,3,3,3,3,3,4,4,4,4,4,4,5}
so asq[]={9,9,9,9,9,9,16,16,16,16,16,16,25}
here as u can see the total number of requisite triple pairs are 6*6=36,
in general for above array total number of pairs is (n/2)*(n/2) i.e. n^2/4
where n is the size of the array.
by using O(n) algo and since you are choosing them one by one,u can't
include all of them as they are of order O(n^2).
so removing repetitions is the only option i think..
On Wed, Dec 1, 2010 at 11:37 PM, fenghuang wrote:
> @anoop
> in fact, it always work even if there are repeated elements, because they
> don't change the decision.
> in detail, assume ii, ,jj, kk is one of the answers, then
> a[ii]+a[jj]=a[kk]. since the array is sorted, so a[ii-1]+a[jj] <= a[kk] and
> a[ii] + a[jj+1] >= a[kk].
> so when you try the pair of 'ii-1' and 'jj', the next step must be
> calculate a[ii] + a[jj] as long as a[ii-1]+a[jj] is not equal to a[kk]. the
> same to the pair 'ii' and 'jj+1'.
> the algorithm is correct and in the whole procedure, repeated elements
> don't affect the decision.
> I'm sorry for my poor English.
> Thank You!
>
> On Thu, Dec 2, 2010 at 12:14 PM, anoop chaurasiya <
> anoopchaurasi...@gmail.com> wrote:
>
>> sorry for the interruption,we can make it work even if the elements are
>> repeated, by removing the duplicacy in linear time(as the array is already
>> sorted) and taking a count of no. of duplicates in the seperate array.
>>
>> On Wed, Dec 1, 2010 at 9:37 PM, Senthilnathan Maadasamy <
>> senthilnathan.maadas...@gmail.com> wrote:
>>
>>> A small correction to the algorithm above. In Step 3, instead of finding
>>> *any* pair (a,b) such that a+b = x we need to find *all* such pairs.
>>> However the complexity remains the same.
>>>
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>>
>>
>>
>> --
>> Anoop Chaurasiya
>> CSE (2008-2012)
>> NIT DGP
>>
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Re: [algogeeks] find triplets in an integer array A[] which satisfy condition: a[i]^2 + a[j]^2 = a[k]^2
sorry for the interruption,we can make it work even if the elements are repeated, by removing the duplicacy in linear time(as the array is already sorted) and taking a count of no. of duplicates in the seperate array. On Wed, Dec 1, 2010 at 9:37 PM, Senthilnathan Maadasamy < senthilnathan.maadas...@gmail.com> wrote: > A small correction to the algorithm above. In Step 3, instead of finding > *any* pair (a,b) such that a+b = x we need to find *all* such pairs. > However the complexity remains the same. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Anoop Chaurasiya CSE (2008-2012) NIT DGP -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: find triplets in an integer array A[] which satisfy condition: a[i]^2 + a[j]^2 = a[k]^2
@DEV,your idea is nice.but i think it wont work in case the array is having repeated elementsam i right On Wed, Dec 1, 2010 at 1:51 PM, nishaanth wrote: > I have a O(n^2) solution.. > > Hash all the squares.0(n) > > Now for every pair find the sum, see if its there in the hash table.O(n^2) > > Total complexity : O(n^2) > > > On Wed, Dec 1, 2010 at 11:00 PM, fenghuang wrote: > >> yeah, you're right. thank you and is it the optimal solution about this >> question? >> >> >> On Thu, Dec 2, 2010 at 1:08 AM, Dave wrote: >> >>> @Fenghuang: No. You don't have to search for b for every a, or more >>> precisely, you don't have to search for a j for every i. Something >>> like this should work for step 3: >>> >>> i = 0 >>> j = k-1 >>> repeat while i <= j >>>if asq[i] + asq[j] < asq[k] then i = i+1 >>>else if asq[i] + asq[j] > asq[k] then j = j-1 >>>else break >>> end repeat >>> if i <= j then you have found an i, j, and k satisfying the desired >>> relationship; >>> otherwise, no such i and j exist for this k. >>> >>> This loop is O(n). Surround this with a loop over k and precede that >>> loop with a sort to get Senthilnathan's algorithm. So, as he says, the >>> whole task is O(n log n + n^2) = O(n^2). >>> >>> Dave >>> >>> On Dec 1, 10:11 am, fenghuang wrote: >>> > should it be O(n^2*lgn)? for each x in n, it's O(n), and for each a, >>> > it'sO(n), and searching b, it's O(lgn), so it's O(n*n*lgn),Thank You! >>> > >>> > On Wed, Dec 1, 2010 at 11:02 PM, Senthilnathan Maadasamy < >>> > >>> > >>> > >>> > senthilnathan.maadas...@gmail.com> wrote: >>> > > Hi, >>> > > How about the following approach? >>> > >>> > > Step 1: Replace each array element with its square - O(n) >>> > >>> > > Step 2: Sort the array - O(n*log(n)) >>> > >>> > > Step 3: For each element x in the array >>> > > Find two elements a, b in the array (if any) such >>> that >>> > > a+b = x. >>> > > Since finding two such elements can be done in O(n) time >>> in a >>> > > sorted array, the complexity of this step 3 is O(n^2). >>> > >>> > > While reporting the output you can take the square root and print it >>> out. >>> > >>> > > Total complexity is O(n^2). >>> > >>> > > -- >>> > > You received this message because you are subscribed to the Google >>> Groups >>> > > "Algorithm Geeks" group. >>> > > To post to this group, send email to algoge...@googlegroups.com. >>> > > To unsubscribe from this group, send email to >>> > > algogeeks+unsubscr...@googlegroups.com >>> >>> > > . >>> > > For more options, visit this group at >>> > >http://groups.google.com/group/algogeeks?hl=en.- Hide quoted text - >>> > >>> > - Show quoted text - >>> >>> -- >>> You received this message because you are subscribed to the Google Groups >>> "Algorithm Geeks" group. >>> To post to this group, send email to algoge...@googlegroups.com. >>> To unsubscribe from this group, send email to >>> algogeeks+unsubscr...@googlegroups.com >>> . >>> For more options, visit this group at >>> http://groups.google.com/group/algogeeks?hl=en. >>> >>> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algoge...@googlegroups.com. >> To unsubscribe from this group, send email to >> algogeeks+unsubscr...@googlegroups.com >> . >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> > > > > -- > S.Nishaanth, > Computer Science and engineering, > IIT Madras. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Anoop Chaurasiya CSE (2008-2012) NIT DGP -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] palindrome problem
the idea for O(|S|^2) is as follows
Let string be s[1..n]
p[i][j]=1 if s[i...j] is a palindrome
p[i][j]=0 otherwise.
we can precalculate table p[i][j] as follows..in O(|s|^2)
p[i][i]= 1
p[i][i+1]=1 if s[i]==s[i+1], 0 otherwise
for all others..
p[i][j]= (s[i]==s[j] && p[i+1][j-1]==1).
now let m[i] represents the minimum number of palindromes needed to
represent s[1...i],
so m[n] is our required answer..
now...let us define m[i] recursively as follows..
m[0]=0, m[1]=1
m[i]=min { m[j-1]+1 where 1<=j<=i && p[j][i]==1}
what it actually says is that...we can go through all possible palindromes
which can be at the end of s[1...i],and check choosing which one is the best
option for us...
i guess its correct...if there is any problem with it then do suggest
On Tue, Nov 16, 2010 at 7:56 PM, cheng lee wrote:
> how to do that in O(|s|^2)?
>
> 2010/11/16 anoop chaurasiya
>
>> will O(|s|^2) dp workor u need something more efficient than that
>>
>>
>> On Tue, Nov 16, 2010 at 1:51 PM, lichenga2404 wrote:
>>
>>> A palindrome is a string which is identical to itself in reverse
>>> order. For example
>>> \ABAAABA" is a palindrome. Given a string, we'd like to break it up
>>> into the small-
>>> est possible number of palindromes. Of course, any one-character
>>> string is a palindrome
>>> so we can always break a length n string into n palindromes. Design an
>>> e cient al-
>>> gorithm to determine the minimum number of palindromes in a
>>> decomposition for a
>>> given string, and analyze the running time. You may assume you are
>>> given a proce-
>>> dure Palindrome(S) which runs in time O(jSj) and returns true if and
>>> only if S is a
>>> palindrome.
>>>
>>> How to solve this problem with the dynamic programming method?
>>>
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>>>
>>
>>
>> --
>> Anoop Chaurasiya
>> CSE (2008-2012)
>> NIT DGP
>>
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>
>
>
> --
> licheng
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Re: [algogeeks] palindrome problem
will O(|s|^2) dp workor u need something more efficient than that On Tue, Nov 16, 2010 at 1:51 PM, lichenga2404 wrote: > A palindrome is a string which is identical to itself in reverse > order. For example > \ABAAABA" is a palindrome. Given a string, we'd like to break it up > into the small- > est possible number of palindromes. Of course, any one-character > string is a palindrome > so we can always break a length n string into n palindromes. Design an > e cient al- > gorithm to determine the minimum number of palindromes in a > decomposition for a > given string, and analyze the running time. You may assume you are > given a proce- > dure Palindrome(S) which runs in time O(jSj) and returns true if and > only if S is a > palindrome. > > How to solve this problem with the dynamic programming method? > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- Anoop Chaurasiya CSE (2008-2012) NIT DGP -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
