@sourav
can O(n) solution possible like this
n=1245120124587412112544735685458545512522524554745669698784552
k=362541245
1. pivot = n[i] where n[i++]=k[j++] for the first time.
2. while(j<= length of k[] and i<= length of n[])
{ if (n[i]==k[j])
{ j++;i++;
}
i++;
}
if( j > length of k[] and i<= length of n[])
return the window size (pivot to i)
On Mon, Oct 25, 2010 at 12:22 PM, ankur aggarwal
<ankur.mast....@gmail.com>wrote:
> If the order is same then the nishaanth is right..
>
> let have similar problem with no constrain on order
>
> eg main string 12561465
>
> and query string is 1 2 6 5
> then output should be "1 2 5 6"
>
>
> On Sat, Oct 23, 2010 at 11:23 PM, nishaanth <nishaant...@gmail.com> wrote:
>
>> It is nothing but a common subsequence problem...isnt it ?
>>
>>
>> On Wed, Oct 13, 2010 at 3:42 PM, Mridul Malpani
>> <malpanimri...@gmail.com>wrote:
>>
>>> @ ankit agarwal, you are right. thanx man.
>>>
>>> On Oct 13, 11:37 am, prodigy <1abhishekshu...@gmail.com> wrote:
>>> > Let I,Q be input array,query array respectively.
>>> >
>>> > 1. Sort query array. O(klogk)
>>> > 2. Allocate an array A of size N.
>>> > 3. Fill A such that A[i]= position of Q[i] in I, -1 if not present in
>>> > I. O(nlogk)
>>> > 4. Allocate an array B of size k with all elements initiated to -1.
>>> > 5. for(counter=0,i=0,counter<n,i++)
>>> > {
>>> > if(B[i]==-1)
>>> > counter++;
>>> > if(A[i]!=-1)
>>> > B[A[i]] = i
>>> > }
>>> > 6. Build min-heap of B.(use an auxiliary array C to keep track of
>>> > position of last occurence of an element of Q in min-heap B.)
>>> > 7. for(diff=i-B[1] ; i<n; i++)
>>> > if(A[i]!=-1)
>>> > B[C[A[i]] = i
>>> > //percolate up or down if needed
>>> > diff=max(diff,i-B[1]);
>>> >
>>> > 8. print diff
>>> >
>>> > On Oct 7, 1:20 pm, RAHUL KUJUR <kujurismonu2...@gmail.com> wrote:
>>> >
>>> >
>>> >
>>> > > @prodigy: how is it coming O(nlogk) can u explain???
>>>
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>>
>>
>> --
>> S.Nishaanth,
>> Computer Science and engineering,
>> IIT Madras.
>>
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>
>
>
> --
>
> With Regards
> Ankur Aggarwal
> +91-7838289304
>
> Software Engineer
> Slideshare
> Delhi
> INDIA.
>
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