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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