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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regards,
Bhaskar Kushwaha
Student
Final year
CSE
M.N.N.I.T. Allahabad
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