Code is here http://codebin.org/view/30e9f2c0. Logic is made clear by the
variable names. Idea is similar to the one which is used to build a queue
using 2 stacks.
On Wed, May 22, 2013 at 8:45 AM, MAC macatad...@gmail.com wrote:
I think this is only possible if you make sure that at push you
I think there is no need for such a complex code. use length() method to
get the size of the stack and return the middle element i.e.
m=S.length()
if(m is even)
return arr[m/2]
else
return arr[m+1/2]
it can be done in O(const) time
On Thu, May 23, 2013 at 12:54 PM, Avi Dullu
Using two stacks to simulate the getmiddle behavior.
Amortized analysis can show each geimiddle is O(1) time complexity.
Best,
2013/5/23 Prateek Jain prateek10011...@gmail.com
I think there is no need for such a complex code. use length() method to
get the size of the stack and return the
I have a list of N teams T1, T2, T3 … Tn. Each of these teams has played a
match against every other team. I have a function displayResult(Team T1,
Team T2), it returns the team which won the match between any two given
teams T1 and T2.
I have to write the teams in an order such the (n-1)th team
This is not necessarily possible.
If you have teams A, B, and C, it is possible that
A beat B
B beat C
C beat A.
Based on the first two games the ranking should be A,B,C. But based on
the third game, C should be ranked higher than A.
Don
On May 23, 11:06 am, Nishant Pandey
@DON,
For example if in a particular order, the teams appeared as T1, T2, T3, T4
… then team T1 had lost to T2, T2 had lost to T3, and T3 had lost to T4… It
may be possible that T3 lost to T1 .. but that need not be taken into
consideration while writing the order. Only the neighboring elements
http://www.careercup.com/question?id=14804702
On Thu, May 23, 2013 at 8:53 PM, Nishant Pandey
nishant.bits.me...@gmail.com wrote:
@DON,
For example if in a particular order, the teams appeared as T1, T2, T3, T4
… then team T1 had lost to T2, T2 had lost to T3, and T3 had lost to T4… It
may
If you create a directed graph where each node is a team and an edge
exists from A-B if A lost to B, then find a Hamiltonian Path in the
graph. That path will be the sequence you need.
Don
On May 23, 12:02 pm, bharat b bagana.bharatku...@gmail.com wrote:
@Don : you are correct but hamiltanion path problem is NPC right?
quicksort algorithm is good solution. I already shared algorithm in above
link.
On Thu, May 23, 2013 at 10:25 PM, Don dondod...@gmail.com wrote:
If you create a directed graph where each node is a team and an edge
exists from
On Sat, May 11, 2013 at 10:29 AM, Aman Jain pureama...@gmail.com wrote:
2. from this no*de u*, again apply* dfs* to find longest distant node
from this node.Let us say that node is v.
Small doubt about the solution. Consider this graph a - b - c - d
You randomly choose vertex 'b' and do a
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