the code is simply utilizing the xor property as u know xor sets on odd one
so, if any array would have numbers odd times repeated
their bit will only stay in xor operation else will get nulify.
so in first loop bit of 1 and 3 are set, to seperate them we need to
divide them using any of their se
an useful link
http://stackoverflow.com/questions/7338070/finding-an-element-in-an-array-where-every-element-is-repeated-odd-number-of-tim
On Fri, Apr 19, 2013 at 11:30 PM, kartik n wrote:
> Hi Nishant i did not understand the code can u please describe a bit
>
> Thanks
>
>
> On Fri, Apr 19, 2
Hi Nishant i did not understand the code can u please describe a bit
Thanks
On Fri, Apr 19, 2013 at 10:48 AM, rahul sharma wrote:
> search the previous posts before posting
>
> search for
> [algogeeks] Amazon Interview Question
> you will get this
>
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> You received this message because you
search the previous posts before posting
search for
[algogeeks] Amazon Interview Question
you will get this
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int main() {
int a[] = {1,2,2,3,3,3,4,4};
int size = sizeof(a)/sizeof(a[0]);
int xorr=0;
for(int i=0;iwrote:
> In an array, some numbers occur only once, some numbers occur twice, only
> one number occur thrice. Find the number occuring thrice ? Space complexity
> O(1) Time Complexity O(n). We s
you are right.
k is the edit distance we are searching for and a critical parameter. In
short you can say- k represents how much error(in terms of edit-distance)
you want to tolerate for between document word 'w' and your suggestion.
since our data structure can answer queries for e.g. "Find all wo
the question mentioned is as it isi just copy pasted it here.
@saurabh thanx for the explainaton of the cube problem i guess that is an
appropriate soln for the question.
and for the other question on detection of typos and suggestion i would
like to know to know what 'k' in your explaination
Firstly, that question is missing a lot of details.
In absence of those details I'm going to make soem assumptions:
1. cube is odd lengthed, so that we can define a unique center of cube.
2. While traversing from a cell(x, y, z) we can only move into any of the 6
adjacent cells[x(+-)1, y(+-)1, z(+-
could you please share the link? coz at first glance a Trie looks like a
bad choice for this task.
I'd go with the Levenshtein distance and a kd-tree.
First implement the Levenshtein distance algorithm to calculate the edit
distance of two strings.
Second, since Levenshtein distance qualifies as a
Sorry ,posted the wrong question initially actually i needed the algo for
this question. Thanx.
On Saturday, October 27, 2012 7:04:10 AM UTC+5:30, raghavan wrote:
>
> By any chance did you read the new blog post by Gayle Laakmaan..
>
> I guess to detect typos we can use some sort of Trie imple
@payal why u need this..??...:P:P
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By any chance did you read the new blog post by Gayle Laakmaan..
I guess to detect typos we can use some sort of Trie implementation..
On Fri, Oct 26, 2012 at 7:50 PM, payal gupta wrote:
>
>Given a cube with sides length n, write code to print all possible
> paths from the center to the su
//A <-area of the cell of histogram
// volarr[] <-holds the vol of water already present at particular index
void fn(int index,int volpoured)
{
int vcapacity=A*heightarr[index];
if(volpoured+volarr[index]vcapacity)
{
volpoured-=vcapacity-volarr[index];
volarr[index]=vcapacity;
// if the height of t
hope this works..
http://ideone.com/XSJRJ
On Fri, May 18, 2012 at 8:20 AM, Bhaskar Kushwaha <
bhaskar.kushwaha2...@gmail.com> wrote:
> It depends on which column you are pouring the water.
> For example
> If you choose the shortest column to pour the water then only that column
> will be filled w
It depends on which column you are pouring the water.
For example
If you choose the shortest column to pour the water then only that column
will be filled with water.
Please correct me if I am wrong.
On Thu, May 17, 2012 at 11:27 AM, Nikhil Agarwal
wrote:
> Imagine that you have an histogram sto
@anup an example what the question meant
if(x==1||x==0)
{
/*stuff*/
}
On Wed, Nov 30, 2011 at 11:07 AM, Anup Ghatage wrote:
> If I have understood the question correctly, Infinite :P
>
> You can have infinite ways to express 0 or 1 given that the ways are
> differentiable amongst themselves.
>
If I have understood the question correctly, Infinite :P
You can have infinite ways to express 0 or 1 given that the ways are
differentiable amongst themselves.
An even number can be considered as a 0 and an odd number as a 1... etc
On Wed, Nov 30, 2011 at 8:37 AM, Nitin Garg wrote:
> *What a
from where does the index starts, 0 or 1 ? in this, array to be moved is
{7, 5, 8} ?
and source array the destination
| |
{9, 7, 5, 8, 1, 5, 4, 8, 10, 1}
please explain move_set o
needs explicit function specialisation. be careful with constant strings.
T Add(T a, T b)
{return a+b ;}
template<>
char* Add char* a, char* b)
{return strcat((char*)a,b); }
surender
On Tue, Jul 19, 2011 at 10:17 PM, Anika Jain wrote:
> here T becomes char *.. u r trying to add two addreses
here T becomes char *.. u r trying to add two addreses here...
On Tue, Jul 19, 2011 at 9:46 PM, nidhi jain wrote:
> Consider a c++ template funtion
> template
> T& Add(T a, T b)
> {return a+b ;}
> if this function is called as T c = Add("SAM", "SUNG"); what will
> happen? What is the problem in t
Navneet,
Your answer is correct, it would have been great if you could have explained
it for others.
I myself took good time to understand it...
Here is the answer
http://exploreriddles.blogspot.com/2011/06/interview-questions-puzzle.html
To maximize the chances of retrieving Red Ball, it is m
Put one red ball in one jar and rest 99 balls in other jar.
Probability in that case is 1/2*1 + 1/2*49//99
On Tue, Jun 14, 2011 at 7:45 PM, Vishal Jain wrote:
> Folks,
>
> This question was asked during a screening process of a product based
> company. Please answer.
>
> http://exploreriddles.b
Ankur, you are rite.
He mentioned that we may not know what the data might be. And yes, I agree
that he kept the restriction because we may not know what the data is. In
that case, I guess bit by bit copying of data from the next node to the
current node (it is possible since both the nodes are of
For the 2nd ques, wat i think is the interviewer has kept the restriction of
not moving the data since you may have a linked list about which you donot
have much information about the structure of the node of the list(only know
about the next pointer) and hence you cannot move the data. For that, y
@ your name last name : please read the question carefully.
*
*
*
*
On Fri, Apr 8, 2011 at 12:11 AM, your name last name wrote:
> Moving the data is unnecessary if in case the whole pointer shifting is
> meant for the entire node and not for individual elements of the node.
> temp ->next = ( temp
Moving the data is unnecessary if in case the whole pointer shifting is
meant for the entire node and not for individual elements of the node.
temp ->next = ( temp -> next ) -> next
the above statement is more than enough to remove the midlle node from the
list...
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@All
yeah my solution does move data ...
I am very interested in knowing a solution that moves no data , someone
please come up with a solution .
On Thu, Apr 7, 2011 at 9:23 PM, balaji a wrote:
> lets say there are two pointers - temp1,temp2 and the start of the node is
> head
> now,
> tem
lets say there are two pointers - temp1,temp2 and the start of the node is
head
now,
temp1=head;
temp2=head->next;
temp1->next=temp2->next;
now temp1 points to the first node and the temp2 points to the second
node(which got deleted from the list).
On Thu, Apr 7, 2011 at 5:43 PM, U
1st case:
if ptr points to the first node, then ptr->next=(ptr->next)->next will do.
2nd case:i think tehre has to be movement of data from third node to
the second node.
On 4/4/11, Umer Farooq wrote:
> Hello friends,
>
> The following question has appeared in two top companies of my city. I'd
>
So, is there any other way of doing this? I did it the way anurag did it;
but the interviewer asked me to do it without moving the data.
On Thu, Apr 7, 2011 at 6:38 PM, Abdul Rahman Shariff wrote:
> @akash ur rite u cannot move data there
>
>
> On Thu, Apr 7, 2011 at 2:35 PM, Akash Mukherjee wrot
in the first case
temp ->next = ( temp -> next ) -> next .
The middle node will become orphan and it won't be deleted, I guess.
in second case
I did it the same way. Then he asked me to solve this without moving the
data?
On Wed, Apr 6, 2011 at 8:03 PM, Anurag atri wrote:
> in case you are giv
@anurag:the second case of yours requires movement of data if I am not
wrong...
@umer farooq:pls calrify what does movement of data mean?
On Wed, Apr 6, 2011 at 8:33 PM, Anurag atri wrote:
> in case you are given a pointer to the first node simply do
> temp ->next = ( temp -> next ) -> next .
> i
That can be moved. Basically, he is trying to convey that move the data of
temp->next to temp. That's perfectly fine.
On Thu, Apr 7, 2011 at 2:05 PM, Akash Mukherjee wrote:
> @ anurag
>
> temp -> data = ( temp -> next ) -> data ;
>
> but data cannot be moved, ryt??
>
> On 4/6/11, Anurag atri wr
@akash ur rite u cannot move data there
On Thu, Apr 7, 2011 at 2:35 PM, Akash Mukherjee wrote:
> @ anurag
>
> temp -> data = ( temp -> next ) -> data ;
>
> but data cannot be moved, ryt??
>
> On 4/6/11, Anurag atri wrote:
> > in case you are given a pointer to the first node simply do
> > temp
@ anurag
temp -> data = ( temp -> next ) -> data ;
but data cannot be moved, ryt??
On 4/6/11, Anurag atri wrote:
> in case you are given a pointer to the first node simply do
> temp ->next = ( temp -> next ) -> next .
> if you are given a pointer to the second node do ,
> temp -> data = ( temp
in case you are given a pointer to the first node simply do
temp ->next = ( temp -> next ) -> next .
if you are given a pointer to the second node do ,
temp -> data = ( temp -> next ) -> data ;
temp -> next = NULL ;
correct me if I am wrong .
On Mon, Apr 4, 2011 at 9:48 PM, Umer Farooq wrote:
>
int hasPathSum(struct node* node, int sum) {
// return true if we run out of tree and sum==0
if (node == NULL) {
return(sum == 0);
}
else {
// otherwise check both subtrees
int subSum = sum - node->data;
return(hasPathSum(node->left, subSum) ||
hasPathSum(node->righ
This problem has DP solution in O(n^2) I think.
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I agree with the trie solution. But now how do you generalise it for K. I
mean a word can be made from k other words.
On Wed, Jan 12, 2011 at 12:53 AM, juver++ wrote:
> If you represent dictionary as a hash table, lookup costs you O(L) at
> least!
> Cause you need to calculate hash for the strin
If you represent dictionary as a hash table, lookup costs you O(L) at least!
Cause you need to calculate hash for the string. But for the trie it is O(L)
in a worst case.
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words and...so on...
on dictionaries lookup is O(1) now think abt complexity...
pass1
for (1 to N) example here N=strlen(newspaper)
other passes are also same...
On Tue, Jan 11, 2011 at 8:07 PM, juver++ wrote:
> What do you mean by N?
> If N - size of the dictionary. And L- maximum length o
What do you mean by N?
If N - size of the dictionary. And L- maximum length of the words then above
algo is O(N*L)
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how abt this..
Pass1 = look whether words like n,ne,new,news.exist or not...store
information in some array like n is not found so is ne but new is found and
news is found so array will be like
0,0,1,1.
Pass2 = do d same from from the end...
r,er,per,aper,paper
0,0,1,0,1
P
Wouldn't that be O(n2) . what if n, ne, new, news, newsp etc all are valid
words ? Cant it be optimized?
On Tue, Jan 11, 2011 at 6:27 PM, juver++ wrote:
> Use trie (or similar DS).
> For each word, try to find second part of the target in a linear time
> O(length).
>
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please ignore earlier message...it was supposed to be a private message..
On Tue, Jan 11, 2011 at 6:52 PM, Vikas Singh wrote:
> lalla..bruteforce na bako.. :P
>
>
> On Tue, Jan 11, 2011 at 6:07 PM, manoj lalavat wrote:
>
>> you can optimize BF
>>
>> newspaper
>>
>> first word you will search
>>
>
lalla..bruteforce na bako.. :P
On Tue, Jan 11, 2011 at 6:07 PM, manoj lalavat wrote:
> you can optimize BF
>
> newspaper
>
> first word you will search
>
> n
>
> then
>
> ne
>
> then
>
> new
>
> then news..and so onif at any point of time first word exist in
> dictionary then only see whether
Use trie (or similar DS).
For each word, try to find second part of the target in a linear time
O(length).
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you can optimize BF
newspaper
first word you will search
n
then
ne
then
new
then news..and so onif at any point of time first word exist in
dictionary then only see whether the word with the remaining characters
exist in the dictionary or not.
On Tue, Jan 11, 2011 at 5:57 PM, Vikas Si
problem has many properties like its a BST with parent pointer
we shud think of such a solution which uses its both property
If we will think abt the brute force recursive solution
its complexity will be O(no_of_nodes*height)
which can be made more efficient by putting some restrictions
in each r
any hint for below question ?
Mohit
On Sun, Dec 26, 2010 at 10:38 PM, MAC wrote:
> you are given a bst where each node has a int value , parent pointer , and
> left and right pointers , write a function to find a path with a given sum
> value. Path can go from left subtree tree , include roo
I guess this is just a *program* problem, not even including any
algorithm problem.
Even from viewer of c++ programmer, I think you should refer any
classic c++ tutorial for it,
instead of asking it online. It is an interview problem, and it can be
find on any c++ book.
Just use some time to find i
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