@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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