eg : original : 0 1 2 3 4
corrupted : 3 4 0 1 2
this can be done as below..
compare the middle element with first and the last element..
if middle < first
chk for first half
else
chk for second half
above eg.
middle = 0
first = 3
last = 2
as 0<3
consider 3 4 0
middle = 4
first
for interview ques 1 ..
fink ak 's position using normal binary search in original list ..log
n
postion of a1 in corrupted list is n-k+1 .. O(1)
eg : original : 0 1 2 3 4
corrupted : 3 4 0 1 2
n=5, k=3 , hence the ans 2...correct me if m wrong
@ anup .. can u explain solution through
Informatica sucks.I have topped in its offcampus written test still they din
call me.
On Wed, Sep 14, 2011 at 7:05 PM, ravi maggon wrote:
> In Thapar Criteria was
> BE CSE and I think MCA was also eligible but for interviews only BE CSE
> were shortlisted.
> Package: 7.75 lpa
>
> On Wed, Sep
many i know what's the criteria and package for the company...
regards
sushaanth
BE-computer science
Madras Institute of technology
On Sep 14, 3:02 pm, Ankit Agarwal wrote:
> 2 question
>
> numbers are (a1+1)*a2*a3... an = a1*a2*a3...an + a2*a3...an
> the first term is same...
> for second term
In Thapar Criteria was
BE CSE and I think MCA was also eligible but for interviews only BE CSE were
shortlisted.
Package: 7.75 lpa
On Wed, Sep 14, 2011 at 6:56 PM, sush57 wrote:
> many i know what's the criteria and package for the company...
>
> regards
> sushaanth
> BE-computer science
> Madra
2 question
numbers are (a1+1)*a2*a3... an = a1*a2*a3...an + a2*a3...an
the first term is same...
for second term is (a1*a2...an)/(a1)
now we have to find max of ( ((a1*a2..an)/a1), (a1*a2...an)/a2)
so the question of max becomes min of( a1, a2, a3... an)
--
Ankit Agarwal
Computer Science
Yeah..agree with this... Just find minimum no and increment that..! Any
counter-example??
On Saturday, 10 September 2011 01:19:41 UTC+5:30, hashd wrote:
>
> For question 2 I guess finding the minimum element's index should suffice
> (considering all elements are positive integer). No need to ev
interviewer also made addition to this ques of increment by 1 that you can't
do multiplication.
On Sat, Sep 10, 2011 at 10:08 AM, Neha Singh wrote:
> i agree with hashd.
>
> --
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> "Algorithm Geeks" group.
> To post to this
i agree with hashd.
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For question 2 I guess finding the minimum element's index should suffice
(considering all elements are positive integer). No need to even calculate
n! as it might cause overflow in case the arrary is big.
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