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
1.scan the array and find the maximum digits of integer. lets say m..
2.Again scan the array and pad the intergers whose digits are less that m (
m) with zero's .. copy the new integers in a new array
3.sort the new array in desc order and carry the same even for original
array.
4. Now the
@$ did not understand how the original array is sorted to give the number!
On Sat, Aug 13, 2011 at 12:50 PM, *$* gopi.komand...@gmail.com wrote:
1.scan the array and find the maximum digits of integer. lets say m..
2.Again scan the array and pad the intergers whose digits are less that m (
m)
@$ the ans should be 9823191 and not 98231910
On Sat, Aug 13, 2011 at 12:59 PM, Nikhil Veliath nve...@gmail.com wrote:
@$ did not understand how the original array is sorted to give the number!
On Sat, Aug 13, 2011 at 12:50 PM, *$* gopi.komand...@gmail.com wrote:
1.scan the array and find the
hi nikhil..
correct ... just type error..
but as per my algo .. as we need to consider the elements of original
array... the last element is 1 , but not 10.
On Sat, Aug 13, 2011 at 1:01 PM, Nikhil Veliath nve...@gmail.com wrote:
@$ the ans should be 9823191 and not 98231910
On Sat, Aug 13,
import java.util.Arrays;
import java.util.Comparator;
public class NewSort implements ComparatorInteger {
@Override
public int compare(Integer o1, Integer o2) {
String s1 = o1.toString();
String s2 = o2.toString();
String com1 = s1 + s2;
String com2 = s2 + s1;
if (com1.compareTo(com2) 0 ) {
we will sort the new array in desc order .. and we will carry the same order
even to original array..
so after sorting the new arraay will be 90,80,23,19,10 .. so corresponding
original array is 9823191
On Sat, Aug 13, 2011 at 12:50 PM, *$* gopi.komand...@gmail.com wrote:
1.scan the array and
1 Sort the Array
2 Put it to the String Accordingly..
Prem
On Fri, Aug 12, 2011 at 6:04 PM, Yasir Imteyaz yasir@gmail.com wrote:
An array of integers is given and you have to find the largest possible
integer by concatenating all elements:
example:
array: 87 36 52
answer: 875236
Kindly check it with both the examples. It won't work.
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radix sort the digits wrong way (left most digit first), and then
concatenate
On Fri, Aug 12, 2011 at 6:12 PM, Yasir yasir@gmail.com wrote:
Kindly check it with both the examples. It won't work.
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Algorithm
Amazing ,
Kindly Follow me up :-
1 Given Array of Random Value ... say int a[]= {20,365,299,50,67,21};
Apply Any Sorting Algo on it... which Results into somewhat like
a={365,299,67,50,21,20,.. blah..}...
2 Well No U hv to put those values to a string variable.
see the second example...if a single digit is there...tht will be at the
very end of the array...and if it is 9 it must come at first place...there
it goes wrong...
On Fri, Aug 12, 2011 at 6:23 PM, Prem Krishna Chettri hprem...@gmail.comwrote:
Amazing ,
Kindly Follow me up :-
1 Given
use something like radix sort, but from numbers left to right.
in given example: 87 36 52
87
36
52
place 87 then 52 than 36
if input is something like
9
87
36
52
place 9875236
this you can do with minor modification of radix sort.
thanks
Rajesh
On Fri, Aug 12, 2011 at 6:31 PM,
I have use bactracking
http://codepad.org/rF4Sr3zk
it works for only 1 and 2digit number
On Fri, Aug 12, 2011 at 6:04 PM, Yasir Imteyaz yasir@gmail.com wrote:
An array of integers is given and you have to find the largest possible
integer by concatenating all elements:
example:
I think tis works
On Fri, Aug 12, 2011 at 6:16 PM, Nitin Nizhawan nitin.nizha...@gmail.comwrote:
radix sort the digits wrong way (left most digit first), and then
concatenate
On Fri, Aug 12, 2011 at 6:12 PM, Yasir yasir@gmail.com wrote:
Kindly check it with both the examples. It
int compare(int a, int b) {
string s = Integer.tostring(a);
string y = Integer.tostring(b);
if (s +y y+s)
return -1;
else
return 1;
}
use this to write quick sort , you can get the answer.
On Fri, Aug 12, 2011 at 8:34 PM, Yasir Imteyaz yasir@gmail.com wrote:
An array of integers
@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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