@atul : suppose if all element of input array is zero then i think hashing
will also produce n^2 output. so worst case time complexity would be O(n^2).

On Sun, Sep 2, 2012 at 9:18 PM, Carl Barton <[email protected]>wrote:

> No you're correct. I was doubting it would work :P
>
> For hashing you would need perfect hashing to ensure O(n) wouldn't you?
>
>
> On 2 September 2012 23:21, atul anand <[email protected]> wrote:
>
>> @carl : if it would work only for entirely positive integers , then i
>> aux[] array created will never contain 0 or repeated elements...
>> correct me if i am wrong...
>>
>> On 9/2/12, atul anand <[email protected]> wrote:
>> > @navin : hashMap can be used to do it in O(n) time.
>> >
>> > On 9/2/12, Carl Barton <[email protected]> wrote:
>> >> Your approach sounds like the optimal (and linear) algorithm for the
>> >> submass finding problem. But if I recall correctly that only works in
>> >> linear time if the input is entirely positive integers?
>> >> Maybe I'm being stupid but wont checking that array be quadratic?
>> >>
>> >> On 2 September 2012 20:02, Navin Kumar <[email protected]>
>> wrote:
>> >>
>> >>> @pradip: Finding same element in temp array is little bit tricky. For
>> >>> displaying item from i+1 to j , you have to make sure that there is
>> >>> equal
>> >>> element at i and j index in temp array. How will you ensure it in O(n)
>> >>> time?
>> >>>
>> >>>
>> >>> On Sun, Sep 2, 2012 at 3:16 PM, Pradeep Mishra
>> >>> <[email protected]>wrote:
>> >>>
>> >>>>  This algorithm is *O(n)*.
>> >>>>
>> >>>> Given an int[] input array, you can create an int[] tmp array where
>> >>>> tmp[i]
>> >>>> = tmp[i - 1] + input[i]; Each element of tmp will store the sum of
>> the
>> >>>> input up to that element.
>> >>>>
>> >>>> Now if you check tmp, you'll notice that there might be values that
>> are
>> >>>> equal to each other. Let's say that this values are at indexes j an k
>> >>>> with j < k, then the sum of the input till j is equal to the sum
>> tillk
>> >>>> and
>> >>>> this means that the sum of the portion of the array between j and k
>> is
>> >>>> 0! Specifically the 0 sum subarray will be from index j + 1 to k.
>> >>>>
>> >>>>    - NOTE: if j + 1 == k, then k is 0 and that's it! ;)
>> >>>>    - NOTE: The algorithm should consider a virtual tmp[-1] = 0;
>> >>>>
>> >>>> Correct me if i am wrong Thanx
>> >>>>
>> >>>> --
>> >>>> Pradeep Kumar Mishra
>> >>>> IIIT Allahabad
>> >>>> B. Tech 3rd Year ( IT )
>> >>>> Another Mail - [email protected]
>> >>>>
>> >>>>
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