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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