See when u xor two same numbers, the result is 0.
So as mentioned in the question, all numbers occur twice, so the result will
be 0 for them and the one occuring once will be left(as 0 ^ number gives
number itself).
Hope u got
Sanjay Kumar
B.Tech Final Year
Department of Computer Engineering
Oh sorry, i didnt read the question carefully:)
Sanjay Kumar
B.Tech Final Year
Department of Computer Engineering
National Institute of Technology Kurukshetra
Kurukshetra - 136119
Haryana, India
On Wed, Aug 17, 2011 at 12:34 AM, Sanjay Rajpal sanjay.raj...@live.inwrote:
See when u
See when u xor two same numbers, the result is 0.
So as mentioned in the question, all numbers occur twice, so the result will
be 0 for them and the one occuring once will be left(as 0 ^ number gives
number itself).
Hope u got it :)
Sanjay Kumar
B.Tech Final Year
Department of Computer
Yes, sry abhishek , i didnt see the question carefully.
But this can be done with hash map requiring O(n) space and O(n) time.
Sanjay Kumar
B.Tech Final Year
Department of Computer Engineering
National Institute of Technology Kurukshetra
Kurukshetra - 136119
Haryana, India
On Wed, Aug 17, 2011
pl give the algo
On Wed, Aug 17, 2011 at 2:50 PM, Sanjay Rajpal srn...@gmail.com wrote:
Yes, sry abhishek , i didnt see the question carefully.
But this can be done with hash map requiring O(n) space and O(n) time.
Sanjay Kumar
B.Tech Final Year
Department of Computer Engineering
National
+1 to dave.xor is the way to go.
On Tue, Aug 16, 2011 at 7:06 PM, Dave dave_and_da...@juno.com wrote:
@Raghavan: But aren't maps implemented as binary search trees? That
would make insertion and searching O(log n), and the overall operation
O(n log n).
Dave
On Aug 16, 4:08 am,
i cudnt understand how is it done here by using xor by chen.. aftergetting F
it wud be the xor of of odd occuring elements, fine, then he wrote if(xor)A1
==0 how is this logic used??
On Wed, Aug 17, 2011 at 8:17 AM, saurabh singh saurab...@gmail.com wrote:
+1 to dave.xor is the way to
Dave tu mahan hai . . . .
-- Forwarded message --
From: Dipankar Patro dip10c...@gmail.com
Date: 14 Aug 2011 23:27
Subject: Re: [algogeeks] Re: array question
To: algogeeks@googlegroups.com
@Dave nice algo. Really like it.
So the whole complexity depends on the sorting.
On 14
thanks guys.
On Mon, Aug 15, 2011 at 1:12 PM, Nikhil Veliath nve...@gmail.com wrote:
Dave tu mahan hai . . . .
-- Forwarded message --
From: Dipankar Patro dip10c...@gmail.com
Date: 14 Aug 2011 23:27
Subject: Re: [algogeeks] Re: array question
To: algogeeks
23:27
Subject: Re: [algogeeks] Re: array question
To: algogeeks@googlegroups.com
@Dave nice algo. Really like it.
So the whole complexity depends on the sorting.
On 14 August 2011 22:58, Dave dave_and_da...@juno.com wrote:
@Dipankar: If extra space is not allowed, I think the optimal
@Dave nice algo. Really like it.
So the whole complexity depends on the sorting.
On 14 August 2011 22:58, Dave dave_and_da...@juno.com wrote:
@Dipankar: If extra space is not allowed, I think the optimal solution
is to sort the two arrays, which takes O(max(m log m, n log n)). Then
the
@Piyush, using stack i guess it can be done in O(n)
On Tue, Jul 26, 2011 at 5:42 PM, Shikhar shikharko...@gmail.com wrote:
@ankit: you are right...sorry about the error
On Jul 26, 5:11 pm, ankit sambyal ankitsamb...@gmail.com wrote:
The O/P of ur example should be 2,2,1,1,1,-1,-1
or am I
a crude algo,
for(i=end to start)
{
while(!stk.empty())
{
if(arr[i]arr[stk.top])
pop();
else
break;
}
if(!stk.empty())
l = arr.length-1;
else
l = stk.top;
output[i]=l-i-1;
stk.puch(i);
}
This will be O(n). Correct me I am wrong anywhere..
On Tue, Jul 26, 2011 at 5:49 PM,
@Akshata : Plz explain ur algo... Its not clear.
Like in the first iteration,
else
l = stk.top;
is getting executed. but stack is empty, so how r u assigning value to l
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sorry for the typo ankit, its
if(stk.empty())
l = arr.length-1;
else
l = stk.top;
On Tue, Jul 26, 2011 at 6:19 PM, ankit sambyal ankitsamb...@gmail.comwrote:
@Akshata : Plz explain ur algo... Its not clear.
Like in the first iteration,
else
l = stk.top;
is getting executed. but
@Shikhar
1) Push the first element to stack.
2) for i = 1 to n-1
a) temp =a[i]
b) while(stack not empty)
int x = pop(stack)
if(xtemp) print(temp);
else
push(x,stack)
break;
c) push(temp,stack)
3) After the
@Anand :Your approach will turn out very crude if elements are something
like 1000, 2000
keeping an array i.e count[1000] is not feasible. I think souravsain's
approach is better.
On Mon, Jun 7, 2010 at 3:57 AM, Anand anandut2...@gmail.com wrote:
Here is my approch which runs in O(n).
The link http://geeksforgeeks.org/?p=1488 has many different solutions and
implementation of hashing method.
On Mon, Jun 7, 2010 at 12:59 AM, Raj N rajn...@gmail.com wrote:
@Anand :Your approach will turn out very crude if elements are something
like 1000, 2000
keeping an array i.e
@souravsain :Your approach works really well..
Here is the Implementation:
http://codepad.org/ricAcQtu
On Sun, Jun 6, 2010 at 11:40 AM, souravsain souravs...@gmail.com wrote:
@divya:go through the elements and keep inserting them in a BST. While
inserting if elements already exists in BST,
@Anand
Depending upon the sequence of data in the input, an insertion/search into
the (unbalanced) BST will take O(n) time causing the overall complexity to
shoot up to O(n^2) for each element counted once. Sourav's approach requires
a balanced binary search tree.
@Divya..
If you know something
output willl be 12 12 5 6 6
On 6 June 2010 18:27, souravsain souravs...@gmail.com wrote:
@divya: Does your problem require the output to be sorted also? What
will be the output required if inout is 12,5,6,12,6? Will it be
12,12,6,6,5 or 12,12,5,6,6,?
Sain
On Jun 6, 12:01 am, divya
Here is my approch which runs in O(n).
http://codepad.org/d3pzYQtW
http://codepad.org/d3pzYQtW
On Sun, Jun 6, 2010 at 7:47 AM, divya jain sweetdivya@gmail.com wrote:
output willl be 12 12 5 6 6
On 6 June 2010 18:27, souravsain souravs...@gmail.com wrote:
@divya: Does your problem
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