maintain 2 sums : incl and xclud.
incl is the sum of the elements including the previous element.
xclud is the sum of elements excluding the previous element.

now .. for evry element 1 to n-1.
set xclud = max(incl, xclud)
set incl = xclud (value from previous iteration) + a[i].

finally, print whicheva is the max.

On Tue, Jul 19, 2011 at 5:01 PM, Nitish Garg <[email protected]>wrote:

> Typo in my above post.
> s[i] = max(s[i-2], s[i-2]+a[i], s[i-1]) work?
>
>
> On Tue, Jul 19, 2011 at 4:59 PM, Nitish Garg <[email protected]>wrote:
>
>> Won't this recurrence work:
>> s[i] = max(s[i-2], s[i-2]+a[i], a[i-1]) work?
>> I think it works.
>>
>>
>> On Tue, Jul 19, 2011 at 4:54 PM, oppilas . <[email protected]>wrote:
>>
>>> New constraint:-
>>> What if the array also contains positive and negative numbers?
>>>
>>> On Tue, Jul 19, 2011 at 4:36 PM, sagar pareek <[email protected]>wrote:
>>>
>>>> ok ok ok thank you all
>>>>
>>>>
>>>> On Tue, Jul 19, 2011 at 4:35 PM, ankit sambyal 
>>>> <[email protected]>wrote:
>>>>
>>>>> @Sagar: 13 is the correct answer. (6+4+3)
>>>>>
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>>>> SAGAR PAREEK
>>>> COMPUTER SCIENCE AND ENGINEERING
>>>> NIT ALLAHABAD
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