@g4... : Is the sequence in which children are arranged is fixed or the
teacher can change the sequence to minimize the candies ?
On Mon, Jul 9, 2012 at 3:58 PM, Anshu Mishra anshumishra6...@gmail.comwrote:
@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
@sumit the sequence is fixed
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can u explain ur algorithm for the sequence
*
5 4 3 2 1*
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for your example
5 4 3 2 1
5 4 3 2 1 -- candies assignement.
(since the length of the longest decreasing sequence is 4,
and length of increasing seq. before it is 0.
its max(0+1,4)+1 = 5
--Sravan Reddy
On Tue, Jul 10, 2012 at 8:09 AM, bala bharath bagop...@gmail.com wrote:
can u explain ur
does ur sol seems lyk incerasing 1 if next number is greater that prev n
decreasing 1 if less..???
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@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
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 44 5 6 7 6 5 4 3 2 1 --candies allotment on
solving subarrays..
here both are given same candies which is
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,
Requires review and comments:
My solution:
find the continuous increasing sequences from the input followed by
continues decreasing array.
let there are k such array (continuous increase followed by continuous
decrease)
Now we have the trivial components. find sum for each such array.. and sum
