Please note:
- Your OO design counts more marks, so, try to make use of well-known
design principles
- Your code should all easy modification. Eg: Dont use IF-ELSE.
- Change management and feature extension in your code should be
flexible.
In, Coding round you will be paired with
@aditya and dipankar: the ans given ws 500ns that is why i posted dis
ques...i ws getting the same ans...
On Tue, Aug 9, 2011 at 9:45 AM, ankit sambyal ankitsamb...@gmail.comwrote:
@aditya : can u explain how u got c part as the answer for question 2
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Please send a link where i can find hard permutations and combinations and
probability questions. Please help. If you guys have good e books or
materials for apti questions, please mail me. Thanks in advance.
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Algorithm
#includestdio.h
#includeconio.h
int main()
{
struct value
{
int bit1:1;
int bit3:4;
int bit4:4;
}bit={1,2,2};
printf(%d %d %d\n,bit.bit1,bit.bit3,bit.bit4);
getche();
return 0;
}
the above code gives output : -1 2 2
any idea why???
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I think it is because bit1 is only 1 bit fwide and whn u initialize it to
1,since MSB is 1 it treats it as a negative integer...Plz correct me if i am
wrong...
On Tue, Aug 9, 2011 at 12:24 PM, Rohit Srivastava access2ro...@gmail.comwrote:
#includestdio.h
#includeconio.h
int main()
{
please explain
On Tue, Aug 9, 2011 at 9:07 AM, Dipankar Patro dip10c...@gmail.com wrote:
ans : n + 2nk - k^2 - k
On 9 August 2011 08:43, Raman raman.u...@gmail.com wrote:
Yes, the answer is n + 2nk - k^2 -k
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Int bit3:4 will be read as lower order 4 bits of bit3 and this will be
treated as int (signed). Thus lower order bit of bit3 which is 2, are
0010 which is 2
try with
1) int bit3:2, output will be -2
2) unsigned int bit3:2, output will be 2.
I hope it is cleared now
On 8/9/11, Rohit Srivastava
hey guys thanks got it!!
On Tue, Aug 9, 2011 at 12:49 PM, sanjay ahuja sanjayahuja.i...@gmail.comwrote:
Int bit3:4 will be read as lower order 4 bits of bit3 and this will be
treated as int (signed). Thus lower order bit of bit3 which is 2, are
0010 which is 2
try with
1) int bit3:2,
In an RDBMS with m attributes, how many superkeys are possible?
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On a little-endian machine, bit structure will be represented as:
0x00 00 00 45
which is bit.bit4 = 0010 (2 in decimal) bit.bit3 = 0010 (2 in decimal)
bit.bit1 = 1 (1 in decimal)
since bit.bit1 exists at the rightmost position, while displaying data as
integer, compiler just repeats the same
@dave: yes it seems so that 17/18 is correct...I deduced it from the cond
prob formula..
I have a minor doubt in general why prob( 2nd toss is a head given that
a head occurred in the first toss ) doesnt seem same as p( head in first
toss and head in second toss with fair coin) +p(head in
2^m-1...??
Im not sure tho
On Tue, Aug 9, 2011 at 1:09 PM, krishna meena krishna.meena...@gmail.comwrote:
In an RDBMS with m attributes, how many superkeys are possible?
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@saurabh: are u referring to bit map sort?
On Mon, Aug 8, 2011 at 5:51 PM, saurabh singh saurab...@gmail.com wrote:
If you limit the magnitude of numbers,it is
On Mon, Aug 8, 2011 at 9:02 PM, siva viknesh sivavikne...@gmail.comwrote:
is there any possible solution for O(1) space and
@aditi did they mention any thing about miss penalty also in the question ??
On Tue, Aug 9, 2011 at 11:48 AM, aditi garg aditi.garg.6...@gmail.comwrote:
@aditya and dipankar: the ans given ws 500ns that is why i posted dis
ques...i ws getting the same ans...
On Tue, Aug 9, 2011 at 9:45 AM,
@Rohit:No...i directly copy pasted the ques...nothing else mentioned...and
what if thr ws sm miss penalty?what wud be the change in soution?
On Tue, Aug 9, 2011 at 1:37 PM, Rohit Srivastava access2ro...@gmail.comwrote:
@aditi did they mention any thing about miss penalty also in the question
yup !!! it will increase the avg access time
On Tue, Aug 9, 2011 at 1:39 PM, aditi garg aditi.garg.6...@gmail.comwrote:
@Rohit:No...i directly copy pasted the ques...nothing else mentioned...and
what if thr ws sm miss penalty?what wud be the change in soution?
On Tue, Aug 9, 2011 at 1:37
*In a crew of 12 members rowing team, 6 members are placed on either side
of the boat. 3 men row on the right and 2 men row on the left. Find the
number of ways of arranging the crew members on each side.*
*
*
*Give the answer in terms of P and C. Example, 12C5 * 12P5+2P1*2C2
*
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so you should use absolute values in second part
On Tue, Aug 9, 2011 at 2:20 AM, Dave dave_and_da...@juno.com wrote:
@Prakash: Yeah, but doesn't the max take care of that?
Dave
On Aug 8, 3:07 pm, Prakash D cegprak...@gmail.com wrote:
yeah.. but the smaller numbers must be negative in
yeah, yours is enough :) tried with various combinations.. thanks
On Tue, Aug 9, 2011 at 2:20 AM, Dave dave_and_da...@juno.com wrote:
@Prakash: Yeah, but doesn't the max take care of that?
Dave
On Aug 8, 3:07 pm, Prakash D cegprak...@gmail.com wrote:
yeah.. but the smaller numbers must be
12C6* 6c3 * 6C2
On Tue, Aug 9, 2011 at 2:10 PM, programming love
love.for.programm...@gmail.com wrote:
members
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can any dce student give an idea whats the procedure and what are the
questions asked by directi recruitment group
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@prakash
it will be 2*12C6*2*6C3*2*6C2
2 each for either side, left and right
On Tue, Aug 9, 2011 at 3:00 PM, Prakash D cegprak...@gmail.com wrote:
12C6* 6c3 * 6C2
On Tue, Aug 9, 2011 at 2:10 PM, programming love
love.for.programm...@gmail.com wrote:
members
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I think it shud be 2*12C6*6C3*6C2
initial 2 is for once we have selected 6 members out of 12, we can either
place them on left side or right side.
@sagar: Could u please explain why have you used 2 two times more?
On Tue, Aug 9, 2011 at 3:51 PM, sagar pareek sagarpar...@gmail.com wrote:
@prakash
The procedure was like this:-
First they had a written test of 45 questions of 90 mins. Depending on the
performance in the test they divided selected students into two buckets. one
for software dev and other for sys ops.
Then there was coding round.
For sys ops one has to extract some information
left- 12C6*6C2
right- 12C6*6C3
Why is 6C2*6C3 combined in a single expression??
On Tue, Aug 9, 2011 at 4:27 PM, programming love
love.for.programm...@gmail.com wrote:
Here you are only selecting the crew members. How will you arrange them?
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got it!
@sagar there shudn be 2's anywhere in the expression.
The different combinations formed on the left hand side by choosing 6 out of
12 will ensure different combinations of other 6 people on the right. So
2*12C6 is not required.
Example:
P1,P3,P5,P7,P9, P11
on left will leave
P2, P4, P6,
thnx
can you please tell some of the written questions so that i will have an
idea what kind of questions are being asked by them
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given a large string of characters lets call it Sbig. Then there is a small
set of characters lets call it Ssmall. You have to find smallest substring
of Sbig which contains all characters in Ssmall.
For example you are given Sbig = hello what are you doing
Ssmall= eo
answer is ello
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why j and k point different location?
#includestdio.h
main()
{
int a=10,*j;
void *k;
j=k=a;
k=(int *)k;
k++;
j++;
printf(%u %u\n,j,k);
}
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Rajesh Kumar
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yeah
sorry one 2 will be used for either side then auto left and right will be
fixed.
thanks for correction
On Tue, Aug 9, 2011 at 4:39 PM, programming love
love.for.programm...@gmail.com wrote:
got it!
@sagar there shudn be 2's anywhere in the expression.
The different combinations
its because void pointer is incremented by 1, when we do k++
whereas integer pointer is incremented by 4, when we do j++
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thanx Ankit
On Tue, Aug 9, 2011 at 5:27 PM, ankit sambyal ankitsamb...@gmail.comwrote:
its because void pointer is incremented by 1, when we do k++
whereas integer pointer is incremented by 4, when we do j++
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@aditi
For the first question.
The answer will be ((0.97 *2) + (0.03*2) +(0.03*100)) i.e 5 ns.
The reason is that cache is accessed in the case of miss as well, so
we need to include that time as well.
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The typecasting tells the compiler that the void pointer is now pointing to
an integer and when we use this pointer to access the integer it takes value
from 4 bytes. But when we try to increment that pointer, it will point to
the next byte. Try taking k as pointer to double instead of void, u
typecast only temporarily changes the pointer type of LHS but cannot change
that of RHS or even LHS permanently
On Tue, Aug 9, 2011 at 5:36 PM, ankit sambyal ankitsamb...@gmail.comwrote:
The typecasting tells the compiler that the void pointer is now pointing to
an integer and when we use this
In memory heirachy, whenever some data is to be read, first data is to
be searched in M1, if the data is found in M1 then system returns the
value from there and counted as a hit. If the value is not found in
M1 i.e. Miss for M1 then it accesses M2 and returns the value from
there. Since nothing
@aditi since in the hierarchy m1 comes first so if it is a hit then we dont
need to check m2 but in case of a miss 0.03*2 seconds are wasted while
searching in m1 then that value is looked for in m2
so i suppose the best answer is given by rohit jain. unless there is some
missing data
On Tue, Aug
@Rohit and Rohit : Thnk u so much:)
On Tue, Aug 9, 2011 at 5:50 PM, Rohit Srivastava access2ro...@gmail.comwrote:
@aditi since in the hierarchy m1 comes first so if it is a hit then we dont
need to check m2 but in case of a miss 0.03*2 seconds are wasted while
searching in m1 then that value
binary values are stored in reverse order of byte by byte.
in your case, bytes will be stored as,
10101100 0100
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pls help me..its very urgent
need a program to divide a file into equal parts(segments)
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@Kunal: in the written test wer the ques technical or aptitude??
On Tue, Aug 9, 2011 at 4:42 PM, NIKHIL JAIN
nikhil.jain.shali...@gmail.comwrote:
thnx
can you please tell some of the written questions so that i will have an
idea what kind of questions are being asked by them
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Can any one explain question number 8?
On Tue, Aug 9, 2011 at 12:19 PM, Akash Mukherjee akash...@gmail.com wrote:
@prashant
devcpp gives an error cannot convert `char (*)[5]' to `char*' in
initialization
ne ideas??
On Mon, Aug 8, 2011 at 12:58 PM, Prashant Kulkarni
@wastrel sorry for delay as i was busy some other stuffs!! , i think i have
handled the case for -ive number as well , just try out for different test
cases :) do notify me via mail if anything wrong ?
Thanks
Shashank Mani
Computer Science
Birla Institute of Technology,Mesra
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I read somewhere that arithmetic operations can't be performed to void
pointers?? How does v++ does'nt giv error??
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It's simple arithmetic..
n + 2(n-1) + 2(n-2) +...2(n-k)
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@srinivas if any of the characters (i,n,d,a) appear den it is no longer
readso gujr
On Tue, Aug 9, 2011 at 6:04 PM, Srinivas Varanasi srinivas4...@gmail.comwrote:
Can any one explain question number 8?
On Tue, Aug 9, 2011 at 12:19 PM, Akash Mukherjee akash...@gmail.comwrote:
@prashant
In written test
all the basics
like apti(major), OS, networking(major), shell programming(major) were there
i m not sure about C programming in written
and in coding i too faced that same log ACCESS program...i was unable to
fetch system's current time stamp
but still my logic was correct so i
it is (1/5)/( (4/5 *(1/2)^6) + (1/5 * 1)) = 80/85 = 16/17
On Aug 7, 10:54 pm, Nitish Garg nitishgarg1...@gmail.com wrote:
Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80
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Given an array of characters, change the array to something as shown in the
following example.
Given : ddbbccae
O/P : 2d4a2b2c1a1e
Do it in the most efficient manner both in terms of time and space ...
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go through the posts before posting anything :)
On Tue, Aug 9, 2011 at 6:29 PM, arpit.gupta arpitg1...@gmail.com wrote:
it is (1/5)/( (4/5 *(1/2)^6) + (1/5 * 1)) = 80/85 = 16/17
On Aug 7, 10:54 pm, Nitish Garg nitishgarg1...@gmail.com wrote:
Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80
--
ans is 16/17 + 1/2*1/17 = 33/34
On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote:
A bag contains 5 coins. Four of them are fair and one has heads on
both sides. You randomly pulled one coin from the bag and tossed it 5
times, heads turned up all five times. What is the probability
time : O(n)
space : O(1)
simple bruteforce
On Tue, Aug 9, 2011 at 6:29 PM, ankit sambyal ankitsamb...@gmail.comwrote:
Given an array of characters, change the array to something as shown in the
following example.
Given : ddbbccae
O/P : 2d4a2b2c1a1e
Do it in the most efficient manner
@shady : I think there is a catch in that approach. Plz post ur code
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Mapchar,int charCountMap = new HashMap();
for(String char : characters.length){
if(charCountMap.containsKey(char){
charCountMap.get(char)++;
}else{
charCountMap.put(char,1);
}
}
later read the map and print the count and key it might suffice.
On Tue, Aug 9, 2011 at 6:37 PM, ankit sambyal
@raghavan: ur approach uses O(n) space
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For
@Arpit: No. The probability of getting 6 consecutive heads is
1/5 * 1 + 4/5 * (1/2)^6 ) = 17/80,
while the probability of getting 5 consecutive heads is
1/5 * 1 + 4/5 * (1/2)^6 ) = 9/40.
Thus, the probability of getting a head on the sixth roll given that
you have gotten heads on all five previous
google on RLE (run length encoding) its almost similar!!
On Tue, Aug 9, 2011 at 6:46 PM, ankit sambyal ankitsamb...@gmail.comwrote:
@raghavan: ur approach uses O(n) space
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@Arun: The probability of getting a head on the first toss is
1/5 * 1 + 4/5 * (1/2) ) = 3/5,
while the probability of getting 2 consecutive heads is
1/5 * 1 + 4/5 * (1/2)^2 ) = 2/5.
Thus, the probability of getting a head on the second roll given that
you have gotten a head on the first roll is
@dave- calculation mistake on my part - method is right.
getting 17/18 only thanks anyways.
On Aug 9, 6:26 pm, Dave dave_and_da...@juno.com wrote:
@Arun: The probability of getting a head on the first toss is
1/5 * 1 + 4/5 * (1/2) ) = 3/5,
while the probability of getting 2 consecutive heads
The probability of getting n consecutive heads is
P(n heads) = 1/5 * 1 + 4/5 * (1/2)^n,
Thus, the probability of getting a head on the n+1st roll given that
you have gotten heads on all n previous rolls is
P(n+1 heads | n heads) = P(n+1) / P(n)
= ( 1/5 * 1 + 4/5 * (1/2)^(n+1) ) / ( 1/5 * 1 + 4/5 *
#includestdio.h
#includestring.h
#includestdlib.h
char * squeez(char *str)
{
char temp1[100];
char * temp;
temp=(char *) malloc(strlen(str)+1);
int start=0,cursor=0;
char mychar;
while(str[start]!='\0')
{
mychar=str[start];
while(str[start]==str[start+cursor]) cursor ++;
if(cursor==1)
@dave - method is right, calculation mistake on my part, getting 17/18
only. thanks anyways.
On Aug 9, 6:26 pm, Dave dave_and_da...@juno.com wrote:
@Arun: The probability of getting a head on the first toss is
1/5 * 1 + 4/5 * (1/2) ) = 3/5,
while the probability of getting 2 consecutive heads
++ on a void* will increase the address by 1 byte. ++ on a type* will
increase the address by sizeof(type) bytes. As if the initial pointer
were an array of type
On Aug 9, 2:49 pm, Rajesh Kumar testalgori...@gmail.com wrote:
why j and k point different location?
#includestdio.h
main()
{
Simple ankit
#includestdio.h
#includestring.h
main()
{
int i,j=0,l,count;
char str[100],tmp;
printf(Please enter the string\n);
fgets(str,100,stdin);
l=strlen(str);
tmp=str[0]; count=1;
for(i=1;il;i++)
{
if(tmp==str[i])
count++;
else
{ if(count==1) str[j++]=tmp; else{
I applied to Texas Instruments last year for software profile and these are
some of the questions that were asked in interview round :
1) How to find least common ancestor of two nodes in a tree ?
2) Some questions on virtual table and vptr ? Is Vtable created per class or
per object ?
3) Some
my solution is in O(n) time and in O(1) space:D :D
On Tue, Aug 9, 2011 at 7:21 PM, sagar pareek sagarpar...@gmail.com wrote:
Simple ankit
#includestdio.h
#includestring.h
main()
{
int i,j=0,l,count;
char str[100],tmp;
printf(Please enter the string\n);
@sagar: for abcd, ur pgm gives abcd. I was trying a pgm which gives
1a1b1c1d.
But now i think this problem is wrong, because in this case it exceeds the
size of the array if we try to o/p as 1a1b1c1d. Hence we require a new array
for it.
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*typedef struct {**
*
*char * a;**
*
*nodeptr next;**
*
*}*nodeptr;**
*
*
*
*what does nodeptr stand for??*
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it will give error in line 3 because nodeptr is undefined till that point..
On Aug 9, 2011 8:03 PM, programming love love.for.programm...@gmail.com
wrote:
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@ankit for abcd the output shud be abcd onlythe whole idea is
compressing the string...
On Tue, Aug 9, 2011 at 7:52 PM, ankit sambyal ankitsamb...@gmail.comwrote:
@sagar: for abcd, ur pgm gives abcd. I was trying a pgm which gives
1a1b1c1d.
But now i think this problem is wrong, because
We have an array of integers, we need to find the element a[i],a[j] and a[k]
values where.. a[i]^2 + a[k]^2 = a[k] ^2
what would be the fast algorithm to find ?
- Samba
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ya got it now. I misunderstood the question
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yep it will !!
any ways nodeptr is another name for pointer to the same structure so you
dont need to write structname *nodeptr
you can simply write nodeptr *p(say) to declare pointer to the structure
On Tue, Aug 9, 2011 at 8:09 PM, rohit jangid rohit.nsi...@gmail.com wrote:
it will give error
1. Square each element of the array and then sort it---O(nlogn)
2. for(i=0;i(size-3);i++)
{
j=i+1; k=size-1;
while(jk)
{
if(a[[i] + a[j] == a[k])
printf(\n%d %d %d,sqrt(a[i]),sqrt(a[j]),sqrt(a[k]));
else if(a[i] + a[j] a[k])
j++;
hello
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saurabh
test my program
please tell me if any bug is there
Thank you,
Siddharam
On Tue, Aug 9, 2011 at 8:18 PM, saurabh singh saurab...@gmail.com wrote:
The code failed for all test cases I tried.
On Tue, Aug 9, 2011 at 8:15 PM, ankit sambyal ankitsamb...@gmail.comwrote:
ya got it
Can anyone give me the recursive algo to find closest ancestor of two nodes
in a tree(not a BST).
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#includestdio.htypedef struct {char * a;
}*nodeptr;
main(){nodeptr *h;h-a=programming;printf(hi %s\n, h-a);}
this gives an error. Please explain the concept behind this.
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we can do it in logn by using binary search approach found
n is the number whose square root has to be
if(n==1)
return 1;
if(n==0)
return 0;
int low=0,high=n/2,mid,temp;
while(1)
{
mid = (low+high)/2;
is it giving an error at *nodeptr declaration?
On Tue, Aug 9, 2011 at 8:26 PM, programming love
love.for.programm...@gmail.com wrote:
#includestdio.htypedef struct {char * a;
}*nodeptr;
main(){nodeptr *h;h-a=programming;printf(hi %s\n, h-a);}
this gives an error. Please explain the
Hi
On 9 August 2011 20:26, programming love love.for.programm...@gmail.comwrote:
#includestdio.htypedef struct {char * a;
}*nodeptr;
main(){nodeptr *h;h-a=programming;printf(hi %s\n, h-a);}
this gives an error. Please explain the concept behind this.
You have used typedef incorrectly
Go through the concept of Information Overload and how to deal with it ?
Please make appropriate posts.
People posting same answer again and again. Saying hello to others.
http://en.wikipedia.org/wiki/Information_overload
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dont use *h just use h
On Tue, Aug 9, 2011 at 8:31 PM, siddharth srivastava akssps...@gmail.comwrote:
Hi
On 9 August 2011 20:26, programming love
love.for.programm...@gmail.comwrote:
#includestdio.htypedef struct {char * a;
}*nodeptr;
main(){nodeptr *h;h-a=programming;printf(hi %s\n,
this gives an error. Please explain the concept behind this.
try using something like this:
#includestdio.h
#includemalloc.h
typedef struct {
char * a;
}* nodeptr;
main(){
nodeptr h = (nodeptr*)malloc(sizeof(nodeptr*));
h-a=programming;
printf(hi %s\n, h-a);
}
though
typedef struct{
I don't think that this function is doing what you want it to do. If
you ask for a^b, it returns a^1 in most cases.
Try this instead.
int power(int a, int b)
{
int result = 1;
if (b == 1) result = a;
else if (b1)
{
result = power(a,b/2);
result *= result;
if (b%2) result *=
he is questioning the complexity and not the algorithm... btw, you are right
On Tue, Aug 9, 2011 at 8:41 PM, Don dondod...@gmail.com wrote:
I don't think that this function is doing what you want it to do. If
you ask for a^b, it returns a^1 in most cases.
Try this instead.
int power(int
@Ankuj: Yeah, but he asked for it to be recursive. Yours is iterative.
Dave
On Aug 9, 9:56 am, Ankuj Gupta ankuj2...@gmail.com wrote:
we can do it in logn by using binary search approach found
n is the number whose square root has to be
if(n==1)
return 1;
thanks to all
On Tue, Aug 9, 2011 at 8:45 PM, shady sinv...@gmail.com wrote:
he is questioning the complexity and not the algorithm... btw, you are
right
On Tue, Aug 9, 2011 at 8:41 PM, Don dondod...@gmail.com wrote:
I don't think that this function is doing what you want it to do. If
On 9 August 2011 10:32, Mohit Goel mohitgoel291...@gmail.com wrote:
what the reason behind it how do they differ in functonality..?
Here is a good and short expl:
http://ee.hawaii.edu/~tep/EE160/Book/chap14/subsection2.1.1.8.html
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a new pointer type that can store the address of such structure
for example
nodeptr p;
will declare a pointer to that struct
but it is useless for me.
On Aug 9, 2011 8:19 PM, programming love love.for.programm...@gmail.com
wrote:
@rohith, what if that statement is removed. Now, what will
Here is my code:
http://ideone.com/deosU
Same as that of Sagar's :D
time O(n) and space O(1)
On 9 August 2011 20:25, siddharam suresh siddharam@gmail.com wrote:
saurabh
test my program
please tell me if any bug is there
Thank you,
Siddharam
On Tue, Aug 9, 2011 at 8:18 PM, saurabh
hi
On Tue, Aug 9, 2011 at 8:24 PM, prashant gautam
prashantgautam@gmail.com wrote:
hello
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for software the technical ppr(frst round) is mcq or subjective?
On Tue, Aug 9, 2011 at 7:23 PM, Decipher ankurseth...@gmail.com wrote:
I applied to Texas Instruments last year for software profile and these are
some of the questions that were asked in interview round :
1) How to find least
My bad but it can be made recursive :)
On Aug 9, 8:17 pm, Dave dave_and_da...@juno.com wrote:
@Ankuj: Yeah, but he asked for it to be recursive. Yours is iterative.
Dave
On Aug 9, 9:56 am, Ankuj Gupta ankuj2...@gmail.com wrote:
we can do it in logn by using binary search
#include stdio.h
#include stdlib.h
int main()
{
char inFileName[80];
char outFileName[80];
int numSegments;
int bytesPerSegment;
printf(Enter file name:);
fgets(inFileName,80,stdin);
printf(Enter number of segments:);
scanf(%d, numSegments);
FILE *f = fopen(inFileName, rb);
*Executing code with printf's for each iteration for better understanding.*
#includestdio.h
main(){
int n, i, j, k, t1=0, t2=0, t3, a[30];
printf(Enter the number of elements\n);
scanf(%d, n);
for(i=0; in; i++){
scanf(%d, a[i]);
}
for(i=0; in; i++)
a[i]=a[i]*a[i];
k=n-1;
i=0;
j=k;
#includestdio.htypedef struct {char * a;
}*nodeptr;
main(){nodeptr h;h-a=programming;printf(hi %s\n, h-a);}
@sidharth: thanks a lot for correcting me :)
@aditya : no. there was some mistake;
in the code i pasted above it's giving segmentation fault. Is it cause i'm
initializing h without using
it should be 2^(m-1)-1
because in given m elements 1 element should be primary key, which will
combined with other attributes to form super key.
so there will be 2^(m-1) such subset now we subtract 1 for the phi.so this
should be 2^(m-1)-1.
correct me if i m wrong,,,
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the solution will be 12C6* 6c3 * 6C2
because if you choose 6 people for the left side, then there is no option
for the right side(i.e. we can select only the remaining 6 people for right
side)
also this 12C6 will provide all possible combinations for choosing 6 members
for left or right and
tree closestSharedAncestor(tree root, tree node1, tree node2, int
result)
{
tree returnValue = 0;
if (root)
{
if (root == node1) result += 1;
if (root == node2) result += 2;
int sum = 0;
tree returnLeft = closestSharedAncestor(root-left, node1, node2,
sum);
if
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