@sourabh,
In addition to your solution, If there is any cycle(loop) exist in the link
list your algo will fail.
To solve this problem first detect this cycle if there is any and count the
element in the cycle, and then you can do the mathematics.
On Thu, Jan 6, 2011 at 6:51 PM, sourabh jakhar
a Y-shaped structure
On Thu, Jan 6, 2011 at 7:29 PM, vishal raja vishal.ge...@gmail.comwrote:
@sourabh,
In addition to your solution, If there is any cycle(loop) exist in the
link list your algo will fail.
To solve this problem first detect this cycle if there is any and count
@juver, ofcourse , and that's not a big deal.
On Wed, Dec 29, 2010 at 5:11 PM, juver++ avpostni...@gmail.com wrote:
Yes, it's true. But we should process DP in the following order not to take
one element more than once.
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] = count[i-1][j] if P(i-1,j) ==1
count[i][j] = count[i-1][j-a[i]] if P(i-1,j-a[i]) ==1
else count[i][j] = 0
On Thu, Dec 30, 2010 at 11:42 AM, vishal raja vishal.ge...@gmail.comwrote:
yeah, My bad.
Missed that.
On Wed, Dec 29, 2010 at 10:52 PM, Wladimir
There were 20 green eyed people and they will commit suicide on 20th day
altogether.
See, There is atleast one green-eyed man.
So let's take the case if there was only one gree-eyed man.
than that man could've seen the colour of all other blued-eyed people and
commit suicide on the very first day.
add up all the elements in array A say sumA and array B say sumB ,substract
the sumA from sumB... You'll get the element.
On Tue, Sep 21, 2010 at 5:36 AM, Anand anandut2...@gmail.com wrote:
Two unsorted arrays are given A[n] and B[n+1]. Array A contains n integers
and B contains n+1 integers
take out that one A pill that's there in the jar. take the half of all the
four pills, that's how u'll make sure that u've had 1 of 'A' and 1 of 'B'
pill.
On Tue, Sep 14, 2010 at 2:52 PM, bittu shashank7andr...@gmail.com wrote:
You are on a strict medical regimen that requires you to take two
