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@atul...
As usual an editing mistake :)
int X[2][N] -> int X[2][N+1];
-
Regarding the new fixed code's explanation:
Basically as explained by praveen..
There are 2 things that need to be done to solve the problem:
1) Pick 2 nos and see if their abs diff is > K or n
@atul..
First of all 6 is not in the heap but its index '0' is..
I think before also u had raised this question of heap stability and i
did explain with an example in one of my previous posts that it won't
affect the checks...
I m repeating the same explanation here with the reason why it won't
a
@all
sorry for my prev post.
On Jan 5, 12:42 am, gaurav bansal wrote:
> your algo wont work for a[]={8,11,2} with k=7.
> acc to u,ans should be 3,but it is 2.
> you are not considering the diff between max and min value
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your algo wont work for a[]={8,11,2} with k=7.
acc to u,ans should be 3,but it is 2.
you are not considering the diff between max and min value
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@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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algogeek
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
Sorry there is some correction
p[] : array of petrol
d[]: array of distance
int FirstPetrolPump(int p[],int d[],int n)
{
int c=0,ThisPetrol=0,FirstPump=0,i;
for(i=0;ihttp://groups.google.com/group/algogeeks?hl=en.
*Playing with memory*
* *
Sindhu and Alistair play a memory game involving of a sequence of random
numbers between *1 *and *10*, inclusive, that is called out one at a time.
Each player can remember up to *5 *previous numbers. When the called
number is in a player's memory, that player is aw
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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When u say char p[]="hello";
memory to array is allocated on stack and "hello" is written into that
array. so u can modify "hello" by saying p[0]= 'y';
=
when u say char *x="hello";
memory to x is allocated on stack, but string "hello" is plac
@Dave : in your pseudo code for B(m,n)
you are not using m ... i guess there is a typo error.
this should be num[i]=m + 1 -i instead of this
num[i] = n + 1 - i
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