@Wladimir, yeah I have heard about that. Another way of populating primitive
pythagoreans is, for any natural number m > 1 (m^2 - 1, 2m, m^2 + 1) forms
a pythagorean triplet. This is useful in populating pythagorean tiplets but
here the problem is to search such triplets from a given int array.
@
if integers are positive,then go on a cycle... like a[2]goes to its final
position, the element in a[2]'s final position goes to its final position,
and so on... each time on visiting an element, put some marker on it...
like make it negative... finally after an element comes to position of a[2]
Convert an array "a1 a2 a3...an b1 b2 b3...bn c1 c2 c3...cn" to "a1b1c1
a2b2c2...anbncn", inplace
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Note that the problem says that the pile has AT LEAST i stones not exactly i
stones. So it can for sure have more than i.
On Thu, Oct 13, 2011 at 4:50 PM, Gaurav Kumar wrote:
> I don't see this code considers the case when after throwing i stones, the
> pile is still left with (Si-i) stones. For
I don't see this code considers the case when after throwing i stones, the
pile is still left with (Si-i) stones. For example, let say pile 10 had 25
stones, now even after throwing 10 stones, pile 25 would be left with 15
stones, which could again be thrown by the next person. Am I missing
somethi
This solution doesn't work.
It prints a blank string.
I think because sometimes the condition becomes true only when a space is
encountered. After which when we shift the value what happens to the value
at the position where we are shifting from. It will still contain the char.
This would be wron
@rahul...How do u choose z and x for computing z^2 -x^2 ?
On Thu, Oct 13, 2011 at 11:34 PM, rahul wrote:
> You can create a hash with sqrt(z2-x2). This will make it o(n). The
> interviewer just made it lil tricky. That's all
>
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Does anyone have an idea about how to extract content from a file
irrespective of file format???
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You can create a hash with sqrt(z2-x2). This will make it o(n). The interviewer
just made it lil tricky. That's all
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BTW can we solve this by hashing..That is the only feasible solution which
comes to my mind to reduce the time complexity ?
On Thu, Oct 13, 2011 at 11:20 PM, Ankur Garg wrote:
> Dude this is nothing but 3 sum problem
>
> http://en.wikipedia.org/wiki/3SUM
>
> Ask interviewer to check this link an
@Sunny, good catch, k X k approach wont work
lets try to use matrix itself as heap
heapify(int cr, int cc, int Nr, int Nc){
mr = cr; mc = cc;
if(cr+1 < Nr)
mr = cr+1 mc = cc;
if(cc + 1 < Nc && min > A[cc+1][cr])
mr = cr; mc = cc+1;
if(A[cr][cc] > A[mr][mc]){
swap(&A[cr][cc] ,
N2 would me minimum
On 13-Oct-2011 11:08 PM, "ravindra patel" wrote:
> Hi,
> Another question I faced in Amazon F2F.
>
> Given an unsorted array of integers, find all triplets that satisfy x^2 +
> y^2 = z^2.
>
> For example if given array is - 1, 3, 7, 5, 4, 12, 13
> The answer should be -
>
Dude this is nothing but 3 sum problem
http://en.wikipedia.org/wiki/3SUM
Ask interviewer to check this link and say he has gone mad!! :P
Regards
Ankur
On Thu, Oct 13, 2011 at 10:29 PM, ravindra patel
wrote:
> Hi,
> Another question I faced in Amazon F2F.
>
> Given an unsorted array of inte
@Sunny, good catch, k X k approach wont work
lets try to use matrix itself as heap
heapify(int cr, int cc, int Nr, int Nc){
mr = cr; mc = cc;
if(cr+1 < Nr)
mr = cr+1 mc = cc;
if(cc + 1 < Nc && min > A[cc+1][cr])
mr = cr; mc = cc+1;
if(A[cr][cc] > A[mr][mc]){
swap(&A[cr][cc] ,
I thinking in this property but i dont know how to use :(
Euclid, in his book Elements, demonstrated that there is a infinnity of
suits early Pythagoreans. Moreover, he found a formula that generates all
primitive Pythagorean suits. Given two natural numbers m> n, the suit (a, b,
c), where:
Hi,
Another question I faced in Amazon F2F.
Given an unsorted array of integers, find all triplets that satisfy x^2 +
y^2 = z^2.
For example if given array is - 1, 3, 7, 5, 4, 12, 13
The answer should be -
5, 12, 13 and
3, 4, 5
I suggested below algo with complexity O(n^2) -
- Sort the
@pradad
1. Never put strlen in the condition of a loop. That instantly makes
an O(n) loop into O(n^2).
2. Be sure to put the null terminator at the end of the compacted
string. Your current code stops one short of doing that.
3. Put {} around the body of the for loop if the body contains more
than
I should have noted that this can handle inputs up to about 2^32 / (10
* x). Run time is proportional to the number of 1's. You can also add
a bit of code to discover the digits of the multiplicand.
I was able to verify with lisp bignums that: 25,514 1's is equal to
76543 * ( a 25,509 digit numb
Nopes
Still it does not ensure that duplicates will not be there in the priority
queue
what i mean is you cannot write directly
do k times{
data = pop()
// let i,j are row and col of data
push(i+1,j);
push(i,j+1);
}
you need to add the following checks
if((i+1,j) is not in the heap) push(i+1,j)
struct data
{
int row, col,
int val;
};
priority_queue heap;
now fine?
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