any one given Amzon online test recently?
Plz share quetions and interview experience.
Thanks in advance.
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@shiva: Nice thinking, man... :)
@Yq Zhang: this similar to base 26 apporch, i have tested below code for
boundary cases ,
0, 26(z), 27(aa), 26*26(yz), 26*27(zz)
public String getColName(int id) {
char ch[] = { 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k',
'l',
@vivek
ur code is give true for 5
The first few numbers of series are
3, 7, 13, 31, 37, 43, 67, 73, 79, 127, 151, 163, 193.
On Mon, Aug 13, 2012 at 10:15 AM, vivek rungta wrote:
> bool is_lucky(int x){
> int pos=x;
> inr count=2;
> while(count if(pos%count)
> return 0;
> pos=pos-pos/cou
@All *Here is the working code: test on input {-1,5,3,-8,4,-6,9} and
{1,-1,2}*
Algo:
increment current till first +ve number(p) and decerement end till last
-ve number(n)
now consider only array between [p..n]
If current is negetive, increment current
If current is positive, swap it with end and
congrats..:)
and do share you experience..
On Sat, Aug 20, 2011 at 10:16 AM, ankur pratik wrote:
> thanx to all members of algo geeks... i got a job... this goup is
> awesome for any discussion
>
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> "Algorithm Geeks"
@all
can some plz suggest the source for where i can fine the dates and news
regarding amazon off-campus Placements.
On Fri, Aug 12, 2011 at 6:36 AM, Akash Mukherjee wrote:
> Hi,
>
> Can anyone suggest from where I should practice for the online coding round
> for amazon off-campus placements.
@Decipher
Microsoft is coming in over college.. can you Give some more question which
are recently asked ...in MS test...
@everone.. do share question of MS if some body has given the recently this
Test..
Algorithm:
1. maintain min_diff so far
2. maintain max_element so far
complexity is O(n) , a
@anand
http://anandtechblog.blogspot.com/2010/07/longest-increasing-subsequence.html
there is problem with your blog... plz check it . its very nice.. and
very informative... but it no opening check it plz...
On Fri, Dec 31, 2010 at 10:18 PM, Anand wrote:
> @Nikhil,
>
> I searched through th
yes O(n) solution posible..
1. traverse the first array..
2. for(i=1;i<=n;i++)
{ X[a[i]] = i ;
}
3. print array X[];
On Sat, Jan 1, 2011 at 10:42 PM, Soumya Prasad Ukil
wrote:
> Is the second array really required, assuming intersection of both A & B is
> equal to either A or B?
>
> How
can some body provide Sql interview Question, for bigger companies like
Microsoft , SAP..etc..
Like here on Algogeeks we find all range of Algorithms question...
Pls suggest some links for tricky question in SQL.. and Unix(basics)..
thankyou in advance...
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*Divesh*
(¨`·.·´¨) Always
`·.¸(¨`·.·
@alexsolo
what is Kd-tree .. plz explain a little bit you algorithm...
On Sun, Dec 5, 2010 at 7:28 PM, alexsolo wrote:
> use kd-tree
>
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@jai gupta
how can you find solution in O(n)
in O(n) you can just find the distance of all the points form origin..
then find the k smallest numbers(distance) in O(k log n + n ) using heap
short..
On Sun, Dec 5, 2010 at 7:28 PM, alexsolo wrote:
> use kd-tree
>
> --
> You received this message
@Nikhil
run you algo ..
on test case
index -> 1 2 3 4 5
value -> 1 2 3 4 5
ouput is mid+1= 3+1=4
but it should be 5...
correct me if i am wrong... and u have assumed all are positive, hence base
index should be 1
On Sun, Dec 5, 2010 at 4:41 PM, Nikhil Agarwal wrote:
> If All the elements are un
@shreyas VA
you are using O(n) extra space...
On Sat, Dec 4, 2010 at 8:46 PM, Shreyas VA wrote:
> Given the size of the input array,
> construct array1 = {0, 1, 0, 1} till n elements
>
> traverse through input array checking sum of 1's n 0's.
>
> at the end if both sums are equal return a
here base index is 1 and total no. of elements are n.
k=1;
i=1;
while(i wrote:
> I hope the following approach will work.
> Let 'a' be the input array with size n.
>
> int i = 0, j = 1;
>
> while( true)
> {
> while( i < n && a[i] == 0) i += 2;
> while( j < n && a[j] ==
@dave
plz run gene's code for input 0xAD...or send me some link of bitwise
programing which involve simultaneous many opearation... like above...
i am always confused with bitwise programing...
On Sun, Nov 21, 2010 at 11:11 PM, Dave wrote:
> @Coolfrog$: Don't forget the bi
@gene
plz explain .. what is going on... by taking example. i am unable to run
a test case
1. x=0xAD (1010 1101)
2. x<<1 ===>01011010
|
x>>1 >01010110
x =0100
how we will get
answer as ( 0101 1011).??
On Sun, Nov 21, 2010 at 9:59 AM, Gene
Divesh Dixt
sorry last output should be 0xEC...
On Sun, Nov 21, 2010 at 9:11 AM, Divesh Dixit <
dixit.coolfrog.div...@gmail.com> wrote:
> assuming all are 8bit no.
> input = 0x46 (0100 0110)
> output = 0x26 ( 0010 0110 )
> input = 0x75 (0111 0101)
> output = 0xFC (1110 1010 )
>
> Algorithm
@Mukesh
indiabix.com is really a nice on for place ment...
thankyou..
keep posting such linking
On Wed, Nov 17, 2010 at 9:53 AM, Mukesh Kumar thakur <
mukeshraj8...@gmail.com> wrote:
> try website:-.www.placementpapers.com
> indiabix.com
>
>
> On 11/15/10, Bittu B wr
we can reduce space complexity further more by just maintaining one
array...
a[256]={0,0,0,0,0,0,0,0,0};
As character can be non numeric... all possibility in 256 ..
s[] = input string.
for( i=0;i 1)
return s[i] ;
}
On Thu, Oct 14, 2010 at 12:42 A
@sourav
can O(n) solution possible like this
n=1245120124587412112544735685458545512522524554745669698784552
k=362541245
1. pivot = n[i] where n[i++]=k[j++] for the first time.
2.while(j<= length of k[] and i<= length of n[])
{if (n[i]==k[j])
{ j++;i++;
@Dave
yes im also trying to say the same point as Gene
these problem is similar to
"*Least fare for a return trip Algorithm*" where u have suggest
the exactly same algorithm..
@Gene do see these similar problem
"[algogeeks] Least fare for a return trip Algorithm"
here we have
wrong.. don't be annoyed
plz
thankyou in advance... take care..
On Sat, Oct 16, 2010 at 6:21 PM, Dave wrote:
> @Coolfrog$:
>
> 1. No. It should be O(n + k log k) because finding the kth smallest
> element of an array of size n is O(n), using the Median of Medians
> a
of step ii in to
consideration along with
a1+b3,a3+b1,a2+b3,a3+b2 ...
On Fri, Oct 15, 2010 at 8:20 PM, Dave wrote:
> @Coolfrog$: You don't have to sort the arrays. You only have to find
> the k smallest elements in each and sort those k entries. This can be
> done
@amod
as given A->B andB->A are in sorted form...if not sort them in
O(n log n).
then
first suggestion A1+B1
second suggestion MIN( A1+B2 , B1+A2) ===> let it be B1+A2
third suggestion MIN( A1+B2 , A1+B3, A3+B1,A2+B3, A3+B2)> let it be
A2+B3
and so on...
@Dave what
@Dave @Mukesh
yaa got it... so simple... i was needlessly making it complex
thankyou guys .
On Tue, Oct 5, 2010 at 2:09 AM, Dave wrote:
> @Coolfrog$: It sounds like you think there are n items in each list,
> but the problem statement says that the total number of items in the
&
ement in the input sequences" but there are k sequence...
so the loop must run for *k*(n-1) and each iteration of for loop running
cost to O(log k).
overall complexity can be
O(k(n-1)log k)..
plz do correct me Dave
thanks..
On Mon, Oct 4, 2010 at 11:20 PM, Dave wrote:
@Dave
1. if three sequence given are 2,3,4,5
17,20,31,50
6,9,10,15
while running we will get 2,3,4,5 as sequence no.1 vanishes form where to
choose next element. for heap . (algo.??)
2.
Loop until
Similar to majority voting problem...
Assume there exits a majority element
As only one such element is possible...
Algo:A[1..n]
int Major(A[1..n]){
Majority=a[1];
count=1;
for(i=2;i<=n;i++)
{ if(majority ==a[i])
count++;
else
{ if(count==0)
{ m
@ sorurav
yes , the basic logic is required so that the code can be understood in
single Run..
i also request the same to all dear friends.
Regards..
On Tue, Sep 28, 2010 at 8:11 PM, sourav wrote:
> Hi Friends
>
> This is an interesting problem and request you all to give a brief
> i
@Dave
very nice one line solution..
we all are revolving around x>>3 concept...
On Sat, Sep 25, 2010 at 10:17 PM, Dave wrote:
> answer = (x || 7) + 1;
>
> Dave
>
> On Sep 25, 6:56 am, Krunal Modi wrote:
> > Q: Write an algorithm to compute the next multiple of 8 for a given
> > positive i
t; char *s;
> s=(char *)malloc(sizeof(s));
> gets(s);
> xstrrev(s,0,strlen(s)-1);
> puts(s);
> return 0;
> }
>
>
> On Fri, Sep 24, 2010 at 12:15 AM, coolfrog$ <
> dixit.coolfrog.div...@gmail.com> wrote:
>
>> @nishant:
>> what i
@nishant:
what is i and j..???
@
On Thu, Sep 23, 2010 at 1:20 PM, Nishant Agarwal <
nishant.agarwa...@gmail.com> wrote:
> void xstrrev(char *s , int i , int j)
> {
> char t;
> if(i {
> t=s[i];
> s[i]=s[j];
> s[j]=t;
> xstrrev(s,i+1,j-1);
>
> }
> }
>
|s>temp2)
{return;}
*temp = *s;
*s=* temp2;
*temp2=*temp;
temp2++;
s++;
strrev(*s,*temp2)
}
On Thu, Sep 23, 2010 at 1:15 PM, coolfrog$
wrote:
> char* temp, temp2;
> char* s="Nitin";
> for(temp2=s;*temp2='\0';temp2++ );/*just to calc
char* temp, temp2;
char* s="Nitin";
for(temp2=s;*temp2='\0';temp2++ );/*just to calculate the length of s*/
void strrev(char * s,char* temp2)
{ if (s==temp2 ||s>temp2)
{return;}
*temp = *s;
*s=* temp2;
*temp2=*temp;
temp2++;
s++;
strrev(*s,*temp2)
}
But it
@aswath :
nice solution... very simple too.
it gives the desired solution as required..though it is a case of indirect
recursion... but still a nice aproach...
keep it up
thanks.
Regards
Divesh
On Thu, Sep 23, 2010 at 12:45 PM, aswath G B wrote:
> #include
> void f1(int);
> void f2(
@nishaanth
for n=1 N=0 as k=1 and k wrote:
> @Davefor n=2 ur formulae would give 1 but the result is 3
>
>
> On Thu, Sep 23, 2010 at 6:53 PM, Apoorve Mohan wrote:
>
>> what is the data type of 'j' ?
>>
>>
>> On Thu, Sep 23, 2010 at 6:49 PM, Krunal Modi wrote:
>>
>>> for(k=1 ; k>> j=k;
>>> wh
@Dave
what is the complexity of this algorithm... it given *O*(*n*(log*n*)(loglog
*n*))
but i think it should be O(n^2)
On Thu, Sep 23, 2010 at 10:39 AM, Dave wrote:
> Following up to my own posting... Sorry. I answered the wrong
> question, "What is the sum of the first 10,000 primes?"
>
>
@Greed
in windows.. output is infinite loop and printing...
0
0
0
0
0
... infinite loop..
On Thu, Sep 23, 2010 at 9:57 AM, Greed wrote:
> You can use setjmp ,longjmp or goto to get desired result.
> #include
> #include
> void print(int v)
> {
>int i=0;
>
>jmp_buf env;
@ umesh kewat
suppose A,B,C,D and D is celebrity
A,B C,D
how would u eliminate one form A,B even if u can ... it will be order
of O(n logn) ...
Regards ...
Divesh
On Thu, Sep 23, 2010 at 4:00 AM, umesh kewat wrote:
> Use divide and conquer strategy of algorithm by diving in 2-2 g
@kartheek
i am getting. this prudent approach
but what is "add what remained to the remainder."
suppose u have A,B,C,D and B is celebrity ...
if( A knows B)
{ A is not celb.
if( B knows C){}
else{ C is not celb.
if (C knows B )
{ if(
@ Dave
can u explain ur answer...plz..
On Thu, Sep 23, 2010 at 8:55 AM, Dave wrote:
> @Krunal: N = n * ceiling(log_2(n)) - 2^ceiling(log_2(n)) + 1,
>
> where a^b is a to the power b.
>
> Dave
>
> On Sep 23, 8:19 am, Krunal Modi wrote:
> > for(k=1 ; k > j=k;
> > while(j>0){
> >
How can more than one element exist of 2n/3 times repeated..
> @dave: can u add that for loop and send as i tried but could not succeed
>
> On Wed, Sep 22, 2010 at 6:23 PM, coolfrog$ <
> dixit.coolfrog.div...@gmail.com> wrote:
>
>> solution in o(n log n)
>> can be (
solution in o(n log n)
can be ( as if solution exit only one element cam be a majority element in
the given array)
1. sort the array in O(nlogn)
2. x = a[2n/3]
if(a[0]==x)
{ if(x== a[(2n/3])+1)
return (x)
}
On Wed, Sep 22, 2010 at 5:53 AM, Dave wrote:
;
> > > What's const? The point of this isn't that it's a difficult prob to
> > > solve. Point lies in working with the design to make this close to log
> > > n.
> >
> > > Define what value "const" holds.
> >
> > > On Sep
There is a 2 dimensional array with each cell containing a 0 or 1 ,
Design an algorithm to
find out which row has the maximum number of 1's , Your algorithm should
have O(n2)
time and no space complexity.
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its dictionary means shorted ordered arry.
let low = 1; and high= const.(10^const)
Boolean isWord(String word)
{ while(low <= high)
{ mid = (low+ high)/2;
if(word = getWordAt(mid))
return true;
if( word > getWordAt(mid))
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