ohh ok i got it.. thanx :)

On Sat, Jul 2, 2011 at 12:48 AM, sunny agrawal <sunny816.i...@gmail.com>wrote:

> it is (2,4] not [2,4].
> open interval, close interval ......... consider (i,j] as [i+1,j]
>
> because a[i] = sum of values from 0 to i
> and a[j] = sum of values from 0 to j
> if a[i] = a[j] then it means sum does not changes between a[i] to a[j]
>
> so from i+1 to j there are equal no of 1's and -1's
> or in other words equal no of 0's and 1's
>
>
> On Sat, Jul 2, 2011 at 12:42 AM, Anika Jain <anika.jai...@gmail.com>wrote:
>
>> @sunny: in a[2,4] has 2 1s and one 0 then how is it a solution? i mean i
>> didnt get wen a[i]==a[j] then a[i,j] is a solution case..
>>
>> On Fri, Jul 1, 2011 at 4:13 PM, sunny agrawal <sunny816.i...@gmail.com>wrote:
>>
>>> Take an array of size of the length of the string.
>>> fill the array positions with one where string contains 1, and -1 where
>>> it is 0
>>>
>>> Now for each i (1,n-1) perform the following operation
>>> a[i] = a[i-1] + a[i]
>>> now a[i] will contains sum of values from a[0] to a[i] in original array.
>>>
>>> Now only thing remains to find is two indexes in this array such that
>>> a[i] = a[j]
>>> where ever this condition is met ( i,j ] is a solution
>>>
>>> or if for some i a[i] = 0 in this case [0,i] will be a solution
>>>
>>> this can be done using hashing
>>> hash the values in the array of size 2*n+1. as range of values is -n to n
>>> and keep track of max interval.
>>>
>>>
>>>
>>>
>>> On Fri, Jul 1, 2011 at 3:22 PM, Anantha Krishnan <
>>> ananthakrishnan....@gmail.com> wrote:
>>>
>>>> Given a string containing 0's and 1's. One would need to find the
>>>> longest sub-string such that the count of 0's is equal to the count of 1's
>>>> in the largest sub string.
>>>>
>>>> Regards
>>>> Anantha Krishnan
>>>>
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>>>
>>>
>>>
>>> --
>>> Sunny Aggrawal
>>> B-Tech IV year,CSI
>>> Indian Institute Of Technology,Roorkee
>>>
>>>
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>
>
>
> --
> Sunny Aggrawal
> B-Tech IV year,CSI
> Indian Institute Of Technology,Roorkee
>
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