Given N points(in 2D) with x and y coordinates. You have to find a
point P (in N given points) such that the sum of distances from
other(N-1) points to P is minimum.
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determine whether the given string is of form pZq.
p and q can contain only X and Y.
like if p=XXYX then z will be XYXX
for ex XXYXXXYXX is valid
the limitation is that "you can read only the next character at each point" .
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that input. You are missing
> numbers which include the digit zero, such as 14610, 30278, and 52056.
>
> Don
>
> On Dec 13, 11:37 am, tech coder wrote:
> > I tried the problem and written the code for it . it is in java. it is
> > printing all the possible numbers
>
I tried the problem and written the code for it . it is in java. it is
printing all the possible numbers
I am treating the differences ans an array of integers.
here is the code
public class Main {
public static void main(String[] args)
{
int digit[]={3,2,5,1};// array of absolu
which other group u people are talking about, i would like to join that
group.
On Tue, Dec 13, 2011 at 9:21 PM, sunny agrawal wrote:
> @aayush
> Lots of member are here but that doesn't mean that you should start
> posting the things which are strictly banned on this group. I hope you will
> take
we have to traverse in zig zaz or spirally. so we need two stack or two
queus.
On Mon, Dec 12, 2011 at 12:23 AM, atul anand wrote:
> @Gene : if i am not wrong , level order traversal can be done using only 1
> queuewhy 2 queue???
>
>
> On Sun, Dec 11, 2011 at 9:53 PM, AMAN AGARWAL wrote:
>
>>
use of log.Moreover log (or any
> floating point operations) take lot of hardware time as they are emulated
> on the floating point environment on most machines.Thirdly precision
> problem for higher values of n may cause your solution to give wron g
> answers,,,
> >
> > &g
about drishtisoft
test pattern
1: technical + aptitude 2 hours abt 60 questions c 20 apti 30 and
others 10
2round coding 4 to 5 questions
3 round 1 technical interview , they wiil ask u write working code , list
and tree problems.
On Fri, Dec 9, 2011 at 12:44 PM, aayush jain wrote:
>
there was small mistake. correct code is
int count=0;
void fun(int a[],int index,int n,int i)
{
if(index==n)
{
count++;
cout<<"\n";
for(int k=0;k wrote:
> how to arrive at such a formulation ?
> can any one derive this formula ?
>
>
> On Mon, Sep 26, 2011 at 12:09 AM, asdqwe wrote:
>
>> t
what about this log(base 3 n) is of integral type then n is a power of 3
On Mon, Dec 5, 2011 at 11:36 PM, Dave wrote:
> @Carl: You can generate the constants the first time the procedure is
> called. There is no need to do them every time. So the first call
> would be O(wordsize) but subsequ
loop through the elements of array. keep two vales firstlargest and
2ndlargest and update accordingly.
loop i=1 to n
{
if(a[i]>1stlargest)
{
2ndlargest=1stlargest;
1stlargest=a[i];
continue;
}
if(A[i]>2ndlargest)
2ndlargest=a[i];
}// loop c
:
> @vikas : to do BFS ..first you have to create tree . so what basis will
> you create a tree ? a dictionary can contains thousandss of word , just by
> taking arbitrary word from the dictionary and creating tree i guess
> will take lot of time.
>
> @tech coder : target will be
i think edit distance algorithm can not be used here because in edit
distance problem we have a target string and a source string. Here we dont
have any target word.
I think trie can be used with some preprocessing.
On Thu, Nov 24, 2011 at 11:59 PM, atul anand wrote:
> http://blog.notdot.net/2007
@ Chunyuan Ge
*have u checked ur solution . ur solution is to find the submatrix all
filled with 1 , but the question say that 1 can be at boundaries.
*
On Wed, Nov 23, 2011 at 3:00 PM, kumar raja wrote:
> @Dark prince : what is meant by Allones(i,0.k) what subsquare he is
> considering here??
>
please upload the books "cracking the coding interview 5th edition" and "
data structures and algorithms made easy", if somebody have.
thanks.
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*"Life is a Game. The more u play, the more u win, the more u win , the
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use a max heap of size k,
On Tue, Nov 22, 2011 at 11:38 PM, Aamir Khan wrote:
>
> On Tue, Nov 22, 2011 at 8:43 PM, Dave wrote:
>
>> @Ganesha: You could use a max-heap of size k in time O(n log k), which
>> is less than O(n log n) if k < O(n).
>
>
> We can always ensure that k <= n/2.
>
> If k >
here is an O(n) approach using a stack.
problem can be stated as " find the 1st smaller element on the right.
put the first element in stack.
take next element suppose "num" if this number is less than elements
stored in stack, pop those elements , for these pooped elements num will
be the r
@ sravanreddy001
complexity is O(N^2) whether tree is balanced or not doesn't matter
For each level it's visiting elements. all elements upto n-1 level .
i dont know from where u got the concept of logn , the code is not
making any decision to go in left or right , it is going in left and right
the complexity is N^2
for each level you are traversing all the nodes below that level. for nth
level , you have to traverse all n-1 nodes.
so O(N^2)
better to use queue , we can traverse level order in O(n).
On Sat, Nov 19, 2011 at 4:58 PM, shady wrote:
> this doesn't seem like level order prin
using a max heap and min heap we can find the median
On Tue, Nov 8, 2011 at 10:59 AM, Arun Vishwanathan
wrote:
> Hi to find running median from a stream of random generated numbers I have
> heard of the 2 heap ( min and max heap ) solution but I fail to understand
> it...could someone please e
this is the solution . it also prints all the possible as well as the total
count
int count=0;
void fun(int a[],int index,int n,int i)
{
if(index==n)
{
count++;
cout<<"\n";
for(int k=0;k wrote:
> how ?
> elaborate on the solution part ? how to arrive at that
> formulation ?
>
> On Se
people just dont read the question properly and post the answer
On 9/17/11, prasanth n wrote:
> @sinjanspecial:
>
> i think your code is for an 1 D matrix..but i have to find for a 2D matrix..
>
> On Sat, Sep 17, 2011 at 7:33 PM, sinjanspecial
> wrote:
>
>> hii I think this code will work
>>
>> #
+1 to don
On Thu, Sep 15, 2011 at 9:40 PM, Don wrote:
> Not really. Usually you would need a second parameter indicating the
> size of the input. In theory it might be possible to put a marker
> value at the end of the array.
>
> Most implementations of malloc store the size of the memory block
1: use bit array(hashing) then only it is possible in O(n)
On Fri, Sep 16, 2011 at 12:04 PM, tech coder wrote:
> @ akshat read the question properly before posting such solution
>
>
> On Wed, Aug 31, 2011 at 12:27 PM, Akshat Sapra wrote:
>
>> Solution:
>>
>
@ akshat read the question properly before posting such solution
On Wed, Aug 31, 2011 at 12:27 PM, Akshat Sapra wrote:
> Solution:
>
> arr[n],sum = 0;
>
>
> for ( int i = 0 ; i < n; i++ ) {
>sum ^= arr[i];
> }
>
> print sum; // required number
>
> --
>
>
> Akshat Sapra
> Under Graduat
@ prateek absolutely wrong , dear u need to brushup ur basics d is a pointer
,
it's can be 2 or 4 depebding on compiler
On Wed, Sep 14, 2011 at 6:26 AM, PRATEEK VERMA wrote:
> 9 and 3
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
>
an interseting problem
for a fibonacci series
the recurrence relation is
t(n)=t(n-1)+t(n-2)+O(1)
on solving it gives upper bound of O(n^2)
but when draw tree for the recurcsion we see that it is growing
exponentially giving a complexity of O(2^n).
so what is the complexity for fibonaacci series
we can also use dfs and find if there exist path between given two
nodes(profiles here).
if yea , there is a connection b/w two profiles.
On Wed, Sep 14, 2011 at 1:58 PM, Azhar Hussain wrote:
> Union Find Algorithm would do
>
> -
> Azhar.
>
>
> On Wed, Sep 14, 2011 at 1:42 PM, JITESH KUMAR wr
for 16 bit compiler ans is 2 3
for 32 bit compiler ans is 4 3
pls correct me if i m wrong
On Wed, Sep 14, 2011 at 11:36 AM, tech coder wrote:
> +1
>
> On Wed, Aug 31, 2011 at 9:27 PM, SANDEEP CHUGH wrote:
>
>> its 4 3 i think
>>
>>
>> On Wed, Aug 3
+1
On Wed, Aug 31, 2011 at 9:27 PM, SANDEEP CHUGH wrote:
> its 4 3 i think
>
>
> On Wed, Aug 31, 2011 at 9:25 PM, aditi garg wrote:
>
>> 8 3
>>
>>
>> On Wed, Aug 31, 2011 at 9:22 PM, rohit wrote:
>>
>>> output??
>>>
>>> int main()
>>> {
>>> char *d = "abc\0def\0";
>>> printf("%d %d",sizeof
Which one is better to work for ThoughtWorks or Samsung.
ThoughtWorks package 5.86
Samsung 6.25.
Those who r working in samsung or thoughtworks please reply
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if somebody has the preparation material and videos provided by
Careercup.com. Then please send me.
thanks.
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this is possible in o(n^2) with n^2 addidtional space .
On Sun, Sep 11, 2011 at 11:33 AM, Neha Singh wrote:
> @victor: Can u provide the algo ??
>
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please send the 5th edition if someone have thanks
On Sun, Sep 4, 2011 at 11:18 PM, Sanjay Rajpal wrote:
> Send it to me also.
>
> Thanx in advance .. :)
> Sanju
> :)
>
>
>
> On Sat, Sep 3, 2011 at 9:58 AM, UMESH KUMAR wrote:
>
>> hi
>> anybody do you have CrackCoding ebook pls send .
>>
>> than
can u elaborate
On Tue, Aug 30, 2011 at 11:13 AM, sukran dhawan wrote:
> kadane 's algo ?
>
> On Tue, Aug 30, 2011 at 1:19 AM, tech coder wrote:
>
>> given a matrix with +ve and -ve numbers find the sub matrix with max sum
>>
>> --
>> You received t
oh yes d will give error.
On Tue, Aug 30, 2011 at 9:36 AM, tech coder wrote:
> c and d will be affected.
>
>
> On Tue, Aug 30, 2011 at 8:51 AM, nidhi jain wrote:
>
>>
>>
>> yes, D vl give an error ,as increment operator cannot be used as lvalue
>> for assi
c and d will be affected.
On Tue, Aug 30, 2011 at 8:51 AM, nidhi jain wrote:
>
>
> yes, D vl give an error ,as increment operator cannot be used as lvalue for
> assignment.
>
> NIDHI JAIN
> INFORMATION TECHNOLOGY(FINAL YEAR)
> NIT,DURGAPUR
>
> --
> You received this message because you are subsc
++rahul
see a constructor cant recieve an srgument of the same class type(by value),
although it can recieve a reference of same class type, because the process
become recursiv eand go infinite till stack overflow
On Tue, Aug 30, 2011 at 9:28 AM, rahul sharma wrote:
> a. default constructor
> c
the sqrt of 32 bit number can't be more than 16 bits.
have an array of 2^16 elemnts wtih elemts 1 2 3 4 5 32768 .
now apply binary search
i=a[mid]where mid=(lower+upper)/2
if(i*i==num)
i is the sqrt
increment lower and upper accordingly as we do in binary search
so order is Olognw
convert a string into another in minimum number of steps. insertion of a
new character , deletion will be considerd as a step.
for ex.
to convert "map" into man requires 1 step.
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an example
array 35,36,37,36,37,38,39,40,39,40,41,42,43,42 etc
array need not to be sorted.
the complexity of Dave algo is O(N) in worst case , when element is not
present in the array.m i right Dave
On Tue, Aug 30, 2011 at 11:02 AM, Anup Ghatage wrote:
> @tech
> Could you give an example
There is an array and the distance between any two consequent elements is
one(+1 or -1) and given a number. You have to check whether the number is in
array or not with minimum complexity.
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To po
are u carzy replying such spams and putting ur resume on public forum
On Mon, Aug 29, 2011 at 12:52 PM, Ankur Garg wrote:
> Hi
>
> PFA my resume for the below position
>
> Regards
> Ankur
>
> On Tue, Aug 30, 2011 at 1:00 AM, katie williams <
> katiewilliams...@gmail.com> wrote:
>
>> *URGENT- So
can u elaborate ur answer pls
On Mon, Aug 29, 2011 at 12:47 PM, Brijesh wrote:
> Sort the array and use dynamic programming.. it will take at most n^2
> complexity.!
> BY dynamic programming , i mean make a 3D type array.. which will give all
> the combination which sums to all the nos.. try it!
given a matrix with +ve and -ve numbers find the sub matrix with max sum
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algogee
is there exist an approach to find subset with a particular sum in less than
O(2^n)
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yes .964 easy hai yaar
On Mon, Aug 29, 2011 at 5:47 AM, vishwa wrote:
> if first is hit then the probability will be 0.7he missed 1st chance ,,
> probability with which he missed is 0.3 ... same way he will try 2nd
> chance.. hit chance is 0.6.. but he missed first already.. so total
> p
int add(int a,int b)
{
int sum=a^b;
int carry=a&b;
while(carry!=0)
{
carry<<=1;
a=sum;
b=carry;
sum=a^b;
carry=a&b;
}
return sum;
}
On Sat, Aug 27, 2011 at 12:28 PM, Mohit kumar lal
wrote:
> int add(int a, int b)
> {
>
Given an array A[] and a integer num. Find four no.s in the array whose sum
is equal to given num.
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@ Don exactly waht u write i wanted to say
On Fri, Aug 26, 2011 at 11:52 AM, tech coder wrote:
> @Tech: I'm not sure I understand your algorithm. Let's try it on
> {1,1,2,2,3,4,5,5,6,6,7,7}. The two number occurring an odd number of
> times are 3 and 4. We xor the numbers
@Tech: I'm not sure I understand your algorithm. Let's try it on
{1,1,2,2,3,4,5,5,6,6,7,7}. The two number occurring an odd number of
times are 3 and 4. We xor the numbers getting 7 = 111 in binary. Now
how do we divide the numbers into two groups?
see we come to know that both number differ at bi
it can be done in O(N) by using XOR ing the elements
1: Xor all the elemnts since those elemnts that even freq will nullify each
other we get number taht will tell in which the two required number differ.
2: divide the array in two sets on the basis of bit in which numbers
differ
3:1 element will
thanks to all got it
On Fri, Aug 26, 2011 at 4:54 AM, Neha Singh wrote:
> @tech coder:
> When u execute : s+="c c++";
> a new string object is created with the value of s appended with "c c++".
> Now s is made to hold the reference of this new string object. T
in java a tring object is immutable
but in following code
String s="java";
s+="c c++";
System.out.print(s);
the output is " javac c++"
why this is so
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@ shashank and all
can we an approach that take less time than O(N^2). I mean not in O(N ) or
nlogn but wrote:
> @Sharvan ..Yes We Can do That and yes i forgot that intead of
> locals.add(a[i][j]) , we need tow write locals.add(i+j) Hope its fine now
> :)
>
>
>
>
> *Thanks
> Shashank Mani
> Co
check whether a string is present in matrix grid (either horizonatal(left to
right), vertical(top to boottom) ,reverse horizonatal(right to left)
reverse vertical(bottom to top).
i think the question is clear.
first try to give solution for horizontal and vertical only, just give the
logic
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write a code which will receive as input a matrix (int[][] matrix) and which
should find all local maximum from the matrix. A local maximum is such a
number in the matrix that is greater than all its immediate neighbors. The
method should return the List of locations of all local maximum numbers
fo
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