@atul.. let me tell you ., i told that we can send
u took two array
arr[]={2,0,0,0}
arr2[]={2,-2,1,0};
take 2 as a base in 1st array , calculate sum (exclduing 2 ), calculate
base^arr[i] that gives sum1= 3
sorting not allowed , just search linearly 2 in 2nd array e.g. search the
same elem
@ WgpShashank :
considering your latest comment abt you algo...
/*
i didn't get how my approach will fail , can u check for the exmple u said
? if u sum the 1st array using 2 as base then sum will be 3 (*exculding 2 ,
although it won't metter* ) , then u search that elemnt in 2nd array , u
won;
@Shashank: can you look at the first post from you?
you are calculating 3^a[i] & adding it to the sum,
can you write a pseudocode so that none of gets confused.
(also, if you are saying this.
for each element, raise element to the power of 2 (2^a[i]), (like you said
in above post), or 3, like
@sravan . u missed man ..read again what i said , u r calculating 3^a[i] ,
where is 3 comes in to picture , no array has has 3 ?? correct , its should
be 2^a[i] , then we search 2 in 2nd array if found 2 the go ahead, read
the algo suggested above , its should work . let me know whats your
con
What about this approach:
First we will scan the first array and find the smallest number.
if it is -ve then we increment all numbers in both the arrays by that
number .
This ensures that every integer in first array is >= 0
some integers in 2nd one maybe -ve if it is,then two arrays are
not
@Shashank:
from your code, i looked at this part.
for j=0 to m
sum2+=3^b[j]
i assumed 3^b[j] as power(3,b[j]);
==> (2,0,0,0) -> 9+1+1+1 && (1,1,1,1) => 3+3+3+3 & both equals 12.
and.. i didn't understand what you meant by base here. did you mean
anything else?
or did I miss anything?
(how can w
@sravan i didn't get how my approach will fail , can u check for the exmple
u said ? if u sum the 1st array using 2 as base then sum will be 3
(exculding 2 , although it won't metter ) , then u search that elemnt in
2nd array , u won;t find & u return -1 , say these array are not similer .
c
create AVL tree using elements of array 1... with each node of AVL tree
maintain a count variable...
if an element occurs more than once,increment the count... (this step is
not compulsory though,we can simply insert the new element in tree)
go through the second array,for each element in array, fi
But the Previous post is not the correct solution as it is similar to
hash table of storing the frequency counts ..
I was thinking a some Statictical Approach here . We are dealing with
some points which scattered in a space and we need to find that the
points are at same points or not . That mea
@All
Sry for late reply .. I was offline for sometime .
Just wanted to brief why I had come up of having a cummulative sum of
each elements of the array . As I mentioned the Frequency distribution
of the numbers ..
I meant that to the frequency the elements of the array starting from
1 to elemen
@shashank: your approach fails for (2,0,0,0) & (1,1,1,1)
but.. from any of the above approaches seen, we couldn't be 100% sure of
the solution,
but, from shashank's approach, the probability of finding correct soultion
can be improved by using some random prime numbers.
(running tests for more
@atul..no need to sorting , complexity will be O(nlogn) only , sorting
makes searching easy so said to sort , u can linearly search that elemnt in
2nd array it will take O(m) in above case . finally complexity will be
O(nlogn) only
Let me know if anything wrong or missed something ?
Thanks
Some very nice approaches have been presented but I still feels for
practical situations its better to sort and compare..All other
algorithms presented above restricts the max value for an element in the
array.In case the maximum value is small,we can simply count sort,and the
algorithm will st
@Shashank : as i have mentioned in the question , no sorting allowed.
if question would have allowed sorting then why not sort both array and
compare it would be much simpler and no need of doing costlier operation
like finding power.
complexity = O(nogn) + O(mlogm) + O(n)
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Must be both have same length also test some other test cases, :)
Let me share things striking in mind , if you calculate the sum of power of
1st array by taking any number as base , remaining all the numbers as
exponent so if take 3 as base & calculate
@Ankur : i guess this would work but to your test add one more sizeof(arr);
On Wed, Jan 4, 2012 at 10:00 PM, Ankur Garg wrote:
> sorry it shud be
> sum of squares and xor and sumof elements
>
> I think this shud work
>
> Regards
> Ankur
>
>
>
>
> On Wed, Jan 4, 2012 at 9:52 PM, atul anand wrote:
sorry it shud be
sum of squares and xor and sumof elements
I think this shud work
Regards
Ankur
On Wed, Jan 4, 2012 at 9:52 PM, atul anand wrote:
> @ Karthikeyan :
>
> sum of cubes fails:-
>
> arr1={2,3,0,-3} = 4
> arr2={1,1,1,1} = 4
>
> On Wed, Jan 4, 2012 at 6:53 PM, Karthikeyan V.B wro
@ Karthikeyan :
sum of cubes fails:-
arr1={2,3,0,-3} = 4
arr2={1,1,1,1} = 4
On Wed, Jan 4, 2012 at 6:53 PM, Karthikeyan V.B wrote:
> Hi,
>
> Consider, arr1={1,2,3} and arr2={-1,-2,-3}
>
> using sum of squares method sum(arr1) = 14 ans sum(arr2) = 14 (since
> square of 1 and -1 is 1)
>
> so
@ankur :
arr1={0,2,0,2);
arr2={1,1,1,1};
xor ans sum condition satisfy.
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What if we check these 2 conditions
XOR and sum of elements and sizeof array same
I cudnt find any counter example
Regards
Ankur
On Wed, Jan 4, 2012 at 6:53 PM, Karthikeyan V.B wrote:
> Hi,
>
> Consider, arr1={1,2,3} and arr2={-1,-2,-3}
>
> using sum of squares method sum(arr1) = 14 ans sum
Hi,
Consider, arr1={1,2,3} and arr2={-1,-2,-3}
using sum of squares method sum(arr1) = 14 ans sum(arr2) = 14 (since square
of 1 and -1 is 1)
so it won work with this case
1.better take the square and negate it before adding
or
2.take sum of cubes
pls correct me if i'm wrong
Regards,
Karthike
can't be use squares of no's as said by rahul patil as said in
previous comment??
On Jan 4, 10:57 am, atul anand wrote:
> @sharad : after checking the link provided by u...it seem like complexity
> will be O(n^2) { not sure } + saurabh point is also valid.
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@sharad : after checking the link provided by u...it seem like complexity
will be O(n^2) { not sure } + saurabh point is also valid.
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Btw an array can have both positive and negative number.
On 4 Jan 2012 11:00, "rahul patil" wrote:
>
> @samm: Rather than adding numbers could we add squares(or cube) of numbers
> which could also be done in linear time?
>
> On Wed, Jan 4, 2012 at 10:56 AM, rahul patil <
> rahul.deshmukhpa...@gma
@sharad Your approach limits the size of array to be very small(as well as
the elements of the array to be small).Else the product will become too big
to be held in an array.Same applies with samm's solution too though in his
case we can be more liberal with element's value..
Saurabh Singh
B.
Can we use concept of prime number as fundamental theorem of arithmetic i.e
every number has a unique factorization into primes (
http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic) and then
multiply them together.
e.g
A1[3] = { 2,3,4} => secondprime*thirdprime*forthprime = 2 * 3 *
@samm: Rather than adding numbers could we add squares(or cube) of numbers
which could also be done in linear time?
On Wed, Jan 4, 2012 at 10:56 AM, rahul patil
wrote:
> @samm: Ur solution is great. It could be used to tell that arrays are not
> similar, in linear time. But cant tell that they ar
@samm: Ur solution is great. It could be used to tell that arrays are not
similar, in linear time. But cant tell that they are 100% similar
ur solution fails for the simple case.
arr1: 3,4
arr2: 5,1
On Wed, Jan 4, 2012 at 10:49 AM, SAMMM wrote:
> No it's not if u use the AP series mathematical f
No it's not if u use the AP series mathematical formula n(n+1)/2..
Then it will be of O(n).
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