Aarthi, i think you are assuming the K keys are 0 to K-1 so that you can use
the array index as its value to count. I doubt of a solution of O(K) space
and O(N) time!
./Channa
On Wed, May 19, 2010 at 4:39 PM, Aarthi Thangamani <
aarthi.thangam...@gmail.com> wrote:
> If you are using an extra spa
Project Euler (http://projecteuler.net) has an exciting set of questions.
Please don't spoil the fun of it by discussing them here. Once you solve a
problem in Project Euler, you get access to the discussion thread of the
problem.
./Channa
On Sun, Oct 11, 2009 at 10:59 AM, Anil C R wrote:
>
> P
@eSKay, @Ankur, et al.,
Please be aware that there are non-Indians too in the group.
Hi All,
Let me try and define the problem precisely (as far as I can).
The ATM machines, which we generally see in India have the following
behavior when an user tries to withdraw money from their Bank account.
Elegant.. I think it can't be better than this. Identifying that each of
them are on different sides of S/2 was the key!
On Tue, Aug 4, 2009 at 10:05 AM, Prunthaban Kanthakumar <
pruntha...@gmail.com> wrote:
> Here is the right answer:
>
> Find the sum of missing numbers. Call it S (this is a eas
There is something wrong with it. I tried your code like this.
int i, n = 8, a[9] = {-1,3,5,1,2,9,10,8,6}; //a[0] is never touched in the
logic
for(i=1; i<=n ;i++ ) //for ease of understanding starting the
array with 1.
{
if(a[ i ] > n )
continue;
else
a[ a [ i ] ]*=n;
}
int x[2], m = 0;
Good catch, Vikram. I had a wrong assumption that the pair (i, sumofXY-i)
would uniquely match with xorofXY. Here is a counter example to disprove my
solution. Suppose X is 5 (binary 101) and Y is 6 (b 110), whose XOR would be
3 (b 011). But, moving unit bit from 5 to 6, we have a pair 4 (100) and
Here is a pseudocode for one of the solutions
sumN = sum of all the elements of the arraysumN2 = sum of 1 to (n+2)
sumofXY = sumN2 - sumN /*sum of the interested integers, say x, y*/
xorN = XOR of all the elements of the array
xorN2 = XOR of 1 to (n+2)
xorofXY = xorN XOR xorN2 /*XOR of x, y*/
for
Please don't post these mails in such groups. How on earth this mail is
relevant for algogeeks? Moreover, this mail is years old, and most of these
facts are no more true.
On Sun, Jan 25, 2009 at 12:56 PM, deepanshu shukla <
deepanshus.it...@gmail.com> wrote:
> FACTS TO MAKE EVERY Indian PROUD
>
; Pratyush Tewari
>
>
>
> On Mon, Jan 5, 2009 at 12:30 AM, Vijay Venkat Raghavan N
> wrote:
> > Guys this is wrong. Lucky number is basically a prime number if you
> observe
> > the definition carefully, and this approach of determining primality
> > obviously won&#
Here is the C-version of the same. It tells you whether given n is lucky or
not.
void luckyOrNot(long int n){
long int n2 = n;
int i= 2;
while(n2>=i){
if(n2%i == 0){
printf("%ld got unlucky when i was %d\n", n, i);
break;
}
n2 = n2 - (n2/i);
i++;
}
if(n2
wrote:
> first n will be the nth number
It's necessary to handle the scenario of i=0. So, if the node has to
be inserted in the beginning of the list, the pointer p to the list
has to change. With your suggested solution, it wouldn't be possible
to see the changed value of p in the calling function. If you assume
you would never insert
Hi MD,
In your last approach, watch out for arbitrarily large values. What
would 'N' be? By the way, 'for' loop there should be for 'n', which is the
size of array 'a'. A simple array like {20, 5, 123456789} would kill your
memory.
Hashing approach would be great if we've some idea on nature of th
0, -20
Now apply the Improvement1 algo and you'll get a sum of 65. Talking about
complexity of this method little complex, so i would leave it till it
becomes necessary to talk about.
Regards,
Channa
On 10/17/07, Muntasir Azam Khan <[EMAIL PROTECTED]> wrote:
>
>
>
>
&
Kannan must have meant for "subsequence". Without looking into Kadane's
algo, I was trying to put constraints away from brute-force.
1. Terminal numbers of the subsequence has to be non-negative. Proof is - if
a terminal number is negative, just exclude that and you'll get a higher
sum.
Regards,
C
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-- Thanks & Regards,Channa Bankapur[CHANNABASAVANNA B.S.]Bengalooru, INDIA.My country, right or wrong; if wrong, to be set right; if right, to be
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