why not the answer (0,7)?

On Nov 1, 4:02 pm, sumant hegde <sumant....@gmail.com> wrote:
> Say diff [i] = (no. of 1s in B[ 0...i ]) - (no. of 1s in A[0...i]). In other
> words the subarray B(0,i) contains diff(i) no. of more 1s than the subarray
> A[0,i]. And  diff[i] < 0 indicates the excess of 1s in A's subarray.
> Ex
> A:  0 1 1  0 0 0 1 1 1 1
> B:  1 0 0  0 1 1 0 1 0 0
> diff:1 0 -1 0 0 1 0 0 -1 -2
> If diff[n-1] were 0, the answer would be whole array (interval 0,n-1).
> but it's actually -2, so I may want to chop off a segment (or 2 segments
> from both ends) such that the retained interval should be the longest.
> For that I look for two instances of an element in diff which yield the
> longest interval. In the example it's two -1's (alternatively two zero's at
> diff[1] & diff[7] also work). Ignore the left-side -1. Thus the final
> interval is (3,8).
> This works because, A[0..8] had one extra 1,(which is what diff[8]=-1 tells)
> which is balanced by removing the segment A[0..2] (which too contained an
> extra 1, again by diff[2]=-1).
>
> On Mon, Nov 1, 2010 at 3:49 PM, sunny agrawal <sunny816.i...@gmail.com>wrote:
>
> > @sumanth
> > can you plz post algorithm in short.
>
> > On Mon, Nov 1, 2010 at 3:45 PM, sumant hegde <sumant....@gmail.com> wrote:
>
> >> see attached file
>
> >> On Sun, Oct 31, 2010 at 4:27 PM, snehal jain <learner....@gmail.com>wrote:
>
> >>> Find longest interval:-
> >>> We are given with two arrays A and B..each of size
> >>> N...elements of array contains either 1 or 0...
> >>> we have to find such an interval (p,q)(inclusive) such that the sum of
> >>> all
> >>> the elements of A (between this interval) and sum of all elements of
> >>> B
> >>> (between this interval ) is equal...
> >>> i.e.
> >>> a[p]+a[p+1]....+a[q]= b[p]+b[p+1]....+b[q]
>
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> > --
> > Sunny Aggrawal
> > B-Tech IV year,CSI
> > Indian Institute Of Technology,Roorkee
>
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