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4 -11 3 4 8}
>
> You have to print {from index, to last index} so for above example {0,
> 3}; {0,8}; {7,8} is the answer
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least one of them to move forward).
>>
>> eg :-
>>
>> 10 , 20 , 30, -10 , -50 , 40 , -50, -1, -3
>>
>> Output : 10+20+30-10+40-1 = 89
>>
>>
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, it did not work. But i think the brute force solution is possible
in O(n^3) solution. We have O(n^2) combination of end points. we can check
for the maximum possible even length palin string in O(n). So that will
give O(n^3). Anyone has solution about O(n^2)?
On 5 June 2014 22:25, Saurabh Paliwal
Ok! So I guess now we are talkng my solution. What i do is maintain two
pointers i and j, i is the end of the first string and j is the beginning
of the second. If both the character match, I calculate the answer for
pointers i-1,j+1 and add 1 to the answer for i,j. If they don't match, I
simply
by memoization.
On Thu, Jun 5, 2014 at 11:37 PM, Saurabh Paliwal
saurabh.paliwa...@gmail.com wrote:
Ok! So I guess now we are talkng my solution. What i do is maintain two
pointers i and j, i is the end of the first string and j is the beginning
of the second. If both the character match, I
is the answer
So time complexity = space complexity = O(n^2).
Correct me if i am wrong
On 5 June 2014 23:44, Saurabh Paliwal saurabh.paliwa...@gmail.com wrote:
And now I get what you meant when you said palindrome. You should have
explained that if that was not exact palindrome
So yes
Ws
On 27 Jan 2014 17:02, saurabh singh saurab...@gmail.com wrote:
^ No its not invalid. It just represents an equation with infinitely many
correct solutions depending on the domain of x.
Saurabh Singh
B.Tech (Computer Science)
MNNIT
blog:geekinessthecoolway.blogspot.com
On Mon, Jan 27,
care that 3 adjacent are
colored
different.
could you please clarify my doubt.
On Thu, Dec 26, 2013 at 5:25 PM, Saurabh Paliwal
saurabh.paliwa...@gmail.com wrote:
@atul anand :- no, we need not use all the colors.
@kumar raja :- sorry dude for replying this late. Continuing
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, kumar raja rajkumar.cs...@gmail.com wrote:
Saurabh your solutions seems right , but can u explain me how did u
arrive at the time and space complexity with some proof /pseudocode/
explanation?
On 15 December 2013 09:47, Saurabh Paliwal
saurabh.paliwa...@gmail.comwrote:
what is the issue
is minimized.
Regards,
Kumar Raja.
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:
@pankaj : how can you maintain increasing sub-sequence using
heapyour soln is for finding finding K largest element in the
array...so it wont work.
On 10/24/13, Saurabh Paliwal saurabh.paliwa...@gmail.com wrote:
check for {5,2,3} and K = 2.
On Thu, Oct 24, 2013
.87fri...@gmail.comwrote:
@Saurabh Paliwal : yes
On 10/24/13, Saurabh Paliwal saurabh.paliwa...@gmail.com wrote:
Do you mean
*of all the increasing subsequences of length k in this array, find the
one
with maximum sum ?*
On Wed, Oct 23, 2013 at 10:52 PM, atul anand
atul.87fri
nothing
insert 55 (before inserting 50 delete the root of heap as count of heap ==
3), deleted root was - 5
so the output will be
{10,30,55}
and if k is 4
than
{5, 10, 30 , 55)
On Thu, Oct 24, 2013 at 6:20 PM, Saurabh Paliwal
saurabh.paliwa...@gmail.com wrote:
You must be mistaken
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I submitted the solution on hackerrank and got Accepted. There is a change
though, there may be more terms like Term (r+1)
but this can also be done using O(1).
Term (r+2) : (2*n+1 - (r+2)*k)/2 and so forth until numerator becomes zero.
You can use the same technique to get this sum as well.
,
Pankaj Kumar Joshi
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there is a minor correction in definition of r, actually r is the maximum
of all the numbers i such that *i*k-1=n.*
On Fri, Oct 18, 2013 at 2:38 PM, Saurabh Paliwal
saurabh.paliwa...@gmail.com wrote:
I think I have an O(1) solution to this problem.
I think we can use the idea of summing
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the elements should be in sorted order.
For example:
for the above 2-D array, the output should be:
A [ ] = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 14, 15, 16, 21, 22, 35,
40, 58 }
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