Re: [algogeeks] Re: candies - interviewstreet -- how to go about solving this problem

2012-07-10 Thread Sumit Agarwal
@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 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.comwrote:

 does ur sol seems lyk incerasing 1 if next number is greater that prev n
 decreasing 1 if less..???

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Re: [algogeeks] Re: candies - interviewstreet -- how to go about solving this problem

2012-07-10 Thread bala bharath
can u explain ur algorithm for the sequence
*
  5 4 3 2 1*

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Re: [algogeeks] Re: candies - interviewstreet -- how to go about solving this problem

2012-07-10 Thread Sravan Kumar Reddy Ganta
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 algorithm for the sequence

 *
   5 4 3 2 1*

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Re: [algogeeks] Re: candies - interviewstreet -- how to go about solving this problem

2012-07-09 Thread sanjay pandey
does ur sol seems lyk incerasing 1 if next number is greater that prev n
decreasing 1 if less..???

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Re: [algogeeks] Re: candies - interviewstreet -- how to go about solving this problem

2012-07-09 Thread Anshu Mishra
@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.comwrote:

 does ur sol seems lyk incerasing 1 if next number is greater that prev n
 decreasing 1 if less..???

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Re: [algogeeks] Re: candies - interviewstreet -- how to go about solving this problem

2012-07-09 Thread Bhaskar Kushwaha
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 wrong !
I mean that the subarrays solution are not independent!


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 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.comwrote:

 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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Re: [algogeeks] Re: candies - interviewstreet -- how to go about solving this problem

2012-07-09 Thread Mr.B
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 44 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.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 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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