a very good counter example. for the approach. even thought you didn't 
solve as per my solution.

(1 2 3 4 5 6 3 2) (6 9 10 12 6 5 4 3 2 1)
 
A small change to the original algorithm. The candies to max. element in 
each trivial array is
max(elements_before_it + 1 ,elements_after_it) + 1

And, start with 2 in each subarray except the first one, where we start 
with 1.
and.. keep increasing until max is reached. for the decreasing sequence. 
start with (number of elements in decreasing seq. and reach until 1.

(1 2 3 4 5 6 3 2) (6 9 10 12   6 5 4 3 2 1) 
 1 2 3 4 5 6 2 1   2 3  4 (5/7) 6 5 4 3 2 1

so, candies for 12 will be *max (3+1, 6) + 1*

 1 2 3 4 5 6 *2 1*   2 3  4 (5/7) 6 5 4 3 2 1 
you gave a wrong assignment as (5,4) instead of (2,1) underlined above. 
this was the point where your did a mistake with my solution.


On Monday, 9 July 2012 11:04:40 UTC-4, Bhaskar wrote:
>
> take a test case:
> 1 2 3 4 5 6 3 2 6 9 10 12 6 5 4 3 2 1
>
> the subarrays then are:
> (1 2 3 4 5 6 3 2 ) (6 9 10 12 6 5 4 3 2 1)
>  1 2 3 4 5 6 5 4    4 5  6   7  6 5 4 3 2 1  -->candies allotment on 
> solving subarrays..
>                     --------
>                 here both are given same candies which is wrong !
> I mean that the subarrays solution are not independent!
>    
>
> On Mon, Jul 9, 2012 at 3:58 PM, Anshu Mishra <anshumishra6...@gmail.com>wrote:
>
>> @sanjay it's not like that
>>
>> e.g : (3 5 6 7 8 4) 7
>>         1 2 3 4 5 1  2 
>> Yes we have to increase just by one, but while decreasing choose the 
>> lowest possible such that each trivial component, if it is in decreasing 
>> phase, should end with 1.
>>   
>> On Mon, Jul 9, 2012 at 12:53 PM, sanjay pandey <sanjaypandey...@gmail.com
>> > wrote:
>>
>>> does ur sol seems lyk incerasing 1 if next number is greater that prev n 
>>> decreasing 1 if less..???
>>>
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>> Anshuman Mishra | Software Development Engineer | Amazon
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>
>
> -- 
> regards,
> Bhaskar Kushwaha
> Student
> Final year
> CSE
> M.N.N.I.T.  Allahabad
>
>

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