Does the pattern comes in this way? HT,TH,TT or HT(X)TH(X)TT ??
Let me know.
--
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On Sat, Aug 11, 2012 at 11:24 PM, Piyush pkjee2...@gmail.com wrote:
How can I find the expected number of tosses , required to obtain a
{HT,TH,TT} , by using random variables??
On Friday, December
if you meant to calculate the E[x] for [HT,TH,TT]. It can be solvable but
very lengthy/boring.
I shall give you an example which should help you.
Let E[X] = x be the expected no. of coin flips to get [HT]
1) if first flip is a tail, we have wasted one flip hence. E[X1] = 1/2*(1+x)
2) if first
4 boys , 3 girls ..
7 children
b: no. of boys
g: no of girls
b-1=g (1st condition)
b=2(g-1) (2nd condition) gives the answer
On Thu, Oct 6, 2011 at 3:42 PM, shady sinv...@gmail.com wrote:
7, try thinking by yourself...
if anyone has some different answer only then post
On Oct 6, 3:05
let no of boys be x and no of girls be y.
then,
x=y+1
2(y-1)=x
solving these we get x=4,y=3
so,x+y=7
there are 7 children.
am I right
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GP
2011/9/24 яαωαт Jee anuragrawat1...@gmail.com
^^ ans will be n*30 mins
On Sep 24, 8:40 am, яαωαт Jee anuragrawat1...@gmail.com wrote:
sum of GP..
a=1
common ratio=4
sum is given=5.6 billion.. find n
simple enough
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lol :P
On Wed, Aug 10, 2011 at 11:35 PM, $hr! k@nth srithb...@gmail.com wrote:
Tie the rope at the top of the tower
Climb down with the help of the rope up to 100 mt peg possItion
Tie the rope to that peg, Climb up to the top of the tower with that rope.
Now release the rope at the top and
varun: can u explain it little further..
On Wed, Aug 10, 2011 at 7:49 PM, varun pahwa varunpahwa2...@gmail.comwrote:
make two ropes 50m and 100 meter. make a loop kind of thing with that now
you have two 50 mtr ropes so get down to 100 mtr point and tie loop rope in
downward now cut the loop
Cut the rope in 50mtrs and 100mtrs length.
Make a small loop(of negligible length at one end of the 50 mtrs rope)
Tie the other end of the rope at the top and from the loop end side pass the
100mtrs rope
such that you have both the ends of 100mtrs rope in your end.
now get down at 100mtrs peg
i hope now it clear:
[image: Screenshot.png]
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Check this out:
Tie it at the 200th meter mark. Throw the 150mt rope down.
Climb down to the 100th meter pole. Tie the rope there from the middle, and
not the end.
So what you have is a 150 mt rope that is tied at 200 mt mark, 100 mt mark
and 50 mts of the rope from 100 mt marks is hanging.
only M is married.
On Sat, Aug 20, 2011 at 7:53 PM, Arun Vishwanathan
aaron.nar...@gmail.comwrote:
@DK:if L is married to M according to you finally , then what does the
third if then statement according to you mean when it is given that if L is
not married then M is married?
On Fri, Aug
consider the last two cases
N married L not married
L not married M married
so now tak M and N
compare it with first case
M married N not married
therfore,only m married
On Sun, Aug 21, 2011 at 1:06 PM, Tushar Bindal tushicom...@gmail.comwrote:
@arun
if L is not married, then M must be
@DK:if L is married to M according to you finally , then what does the third
if then statement according to you mean when it is given that if L is not
married then M is married?
On Fri, Aug 19, 2011 at 10:35 PM, Dave dave_and_da...@juno.com wrote:
@DK: What in the statement of the problem led
make two ropes 50m and 100 meter. make a loop kind of thing with that now
you have two 50 mtr ropes so get down to 100 mtr point and tie loop rope in
downward now cut the loop at 100 mtr you have 100 mtr rope then move down
with the help of that. i hope i am clear.
On Mon, Aug 8, 2011 at 1:52 PM,
Tie the rope at the top of the tower
Climb down with the help of the rope up to 100 mt peg possItion
Tie the rope to that peg, Climb up to the top of the tower with that rope.
Now release the rope at the top and hold it. It ll take you down.:P
On Wed, Aug 10, 2011 at 7:49 PM, varun pahwa
@Dave oh i thought some logical concept willl be applied in that
case...it is ok!!!
thanks:)
On Fri, Aug 5, 2011 at 1:47 AM, Dave dave_and_da...@juno.com wrote:
@Himanshu: That is easy for any boy scout. :-) Tie the rope at the top
of the tower. Then tie a sheepshank knot of a comfortable
tie the rope to the peg and hold the rope at a little less than 100m point.
Then jump.
On Mon, Aug 8, 2011 at 1:19 PM, Himanshu Srivastava
himanshusri...@gmail.com wrote:
@Dave oh i thought some logical concept willl be applied in that
case...it is ok!!!
thanks:)
On Fri, Aug 5, 2011
I guess anubhav soln seems ok
On Thu, Aug 4, 2011 at 8:50 PM, ankit sambyal ankitsamb...@gmail.comwrote:
@aditi:Thats a non uniform rope. The 1st half may burn faster than 2nd
half.
btw Priyanka's solution is correct.
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of this double, half is kept inside the well, and the other half is used as
input to the 2nd well
half mean 1/2 or 50%
how can we assume it to be 100???
if we take it to be 1/2, the question goes wrong, so ur concept is valid but
then question should have been framed correctly
On Fri, Aug 5,
@dave...im not burning half of the rope or anything...my idea is jst to
increase the rate of burning..by folding it in the middle and then lighting
it from both ends...im burning the entire rope wid 4 times the rate of
burning...shud take 15 mins
On Thu, Aug 4, 2011 at 9:52 PM, Dave
@aditi:Thats a non uniform rope. The 1st half may burn faster than 2nd half.
btw Priyanka's solution is correct.
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Please check this : http://www.techinterview.org/post/526313890/bad-king
On Tue, Jul 19, 2011 at 8:43 PM, sagar pareek sagarpar...@gmail.com wrote:
hey guys pls tell any other better solution ...
On Tue, Jul 19, 2011 at 6:41 PM, sagar pareek sagarpar...@gmail.comwrote:
Question :-
Once
thanks
its almost same :)
i was hoping for a diff answer (if exists)
On Fri, Jul 22, 2011 at 4:25 PM, Rajeev Kumar rajeevprasa...@gmail.comwrote:
Please check this : http://www.techinterview.org/post/526313890/bad-king
On Tue, Jul 19, 2011 at 8:43 PM, sagar pareek sagarpar...@gmail.comwrote:
thanks sagar for this wonderful shortcut
but can you please explain it better. in what cases can we use this
approach?
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Well you can find it in WILLIAM STALLINGS's book of cryptography.
or foundation of cryptography by wenbo mao :) :)
On Sun, Jul 17, 2011 at 9:02 PM, Tushar Bindal tushicom...@gmail.comwrote:
thanks sagar for this wonderful shortcut
but can you please explain it better. in what cases can
thankyou :)
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@Tushar Bindal
No need of long calculations :)
here is a shortcut, actually in O(1) time :)
for calculating chances of any two entities to collide in given different
species is just take underoot of it.
here underoot of 365 is approx 19.he he enjoy the solution.
For more details just go
This question was asked by ST micro for hiring intern in my college . Here's
the solution :
Let the bottles of alcohol named 0 to 5 then -
No.Binary Value
00 0 0
10 0 1
20 1 0
30 1 1
41 0 0
51 0 1
Mice - a b c
Now make the mice drink alcohol
Got it...Thanks..
On Wed, Jul 6, 2011 at 11:31 PM, shiv narayan narayan.shiv...@gmail.comwrote:
speed of river=(distance traveled by object in it) / total time it
took to travel
here hat has traveled a distance of 1 KM
and it has taken =5mn+5 min=10 min=10min/60=1/6 hrs;
so speed =
probability that i win standing at second position: 1/365
third position : 364/365*2/365 = 1/365)*(628/365)
fourth position : 364/365*363/365*3/365
4th : 364/365*363/365*362/365*4/365
nth position:
(365-1)*(365-2)*(365-3)*(365-4)*(365-5).*(365-(n-2))*(365-(n-1))*(n)*(1/365)^n
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Sorry for the previous post
the last line was incorrect
it should have been (n+1)th position
I was just writing roughly and pressed send instead of save.
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probability that i win standing at second position: 1/365
probability that i win standing at third position : 364/365*2/365 =
1/365)*(628/365)
probability that i win standing at fourth position : 364/365*363/365*3/365
probability that i win standing at 4th position :
364/365*363/365*362/365*4/365
Sory once again for that incomplete answer.
The complete one is here.
probability that i win standing at second position: 1/365
probability that i win standing at third position : 364/365*2/365 =
1/365)*(628/365)
probability that i win standing at fourth position : 364/365*363/365*3/365
Ans :- 3
let bottles be1,2,3,4,5,6
and mice be a,b,c.
separate bottle 6
make pairs P(1,2,3) ; Q(2,4) ; R(3,4,5) and given to mice a,b,c resp.
if poison is inbottle mice who dies
1 a
2 a,b
3
the solution is given
herehttp://www.thecareerplus.com/?page=resourcescat=150subCat=10qNo=2
but can anyone lease explain it better
please give a original solution
and stop making rude comments about answers posted genuinely.
If you have an original solution, please post it.
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And what about binary search?
On Wed, Jul 6, 2011 at 12:26 PM, 991 guruprakash...@gmail.com wrote:
Sorry abt the previous post ( and this one ) if it ended up as a spam.
I just saw the question and left the place. When I finished posting,
ppl hav already given replies...
On Jul 7, 12:12 am,
We have two eggs,so have only two chances of missing.SO its about a
combination of binary and linear search.
On Thu, Jul 7, 2011 at 9:09 AM, Aakash Johari aakashj@gmail.com wrote:
And what about binary search?
On Wed, Jul 6, 2011 at 12:26 PM, 991 guruprakash...@gmail.com wrote:
Sorry
How AP(ans=14) solution is satisfying the constraints?
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only ONE mouse ...consume each sample of bottles of bear with a difference
of one hour
and calculate time..
sry if is thr any stupidity in this answer..but i think it may be right
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@Bhavesh
NO there is No stupity
just a mistake in reading the question
mice die within 14 hrs.Not exactly 14 hours :)
3 is correct answer.
On Mon, Jun 27, 2011 at 10:51 PM, Bhavesh agrawal agr.bhav...@gmail.comwrote:
only ONE mouse ...consume each sample of bottles of bear with a
ok , yeah 3 is the correct answer ..
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For
4
@amit what's the answer ?
On Mon, Jun 27, 2011 at 12:40 AM, shiv narayan narayan.shiv...@gmail.comwrote:
can u please explain how is it 3?
On Jun 26, 11:18 pm, D.N.Vishwakarma@IITR deok...@gmail.com
wrote:
3 mice .
On Sun, Jun 26, 2011 at 6:13 PM, ArPiT BhAtNaGaR
3
think in binary.. :)
On Mon, Jun 27, 2011 at 12:56 AM, Arpit Sood soodfi...@gmail.com wrote:
4
@amit what's the answer ?
On Mon, Jun 27, 2011 at 12:40 AM, shiv narayan
narayan.shiv...@gmail.comwrote:
can u please explain how is it 3?
On Jun 26, 11:18 pm, D.N.Vishwakarma@IITR
first make two group of 3 bottle each
one mice for each group
make mixture of 3 bottle and put for mice .
do same for other group
only one mice will die
. then select group of dead mice .
beak it into three group
one bottle each
now we can use old mice which is not dead and one more for two bottle
you cant use the old mouse again because time he has mentioned is 14
hours... so you will have to wait for another 14 hours which exceeds the
given time limit of 24 hours... so it is 4.
On Mon, Jun 27, 2011 at 1:00 AM, D.N.Vishwakarma@IITR deok...@gmail.comwrote:
first make two group of
thanks dave.
On Mon, Jun 27, 2011 at 1:07 AM, Dave dave_and_da...@juno.com wrote:
3 Mice: Call them mouse #1, mouse #2, and mouse #4 (think binary
code).
Give mouse #1 a mixture of bottles 1, 3, and 5.
Give mouse #2 a mixture of bottles 2, 3, and 6.
Give mouse #4 a mixture of bottles 4, 5,
hw u r gettin 3
i m gettin 4
mine is make 4 grups
1,2,6 no 1
2,3,5 no 2
1,3,4 no 3
4,5,6no 4
nw out of 4 2 mice will die,and in their corresponding groups common bottle
will give you the answer.
correct me if i am wrong
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@Harshit: Check dave's solution... U'll get ur ans :)
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i got it :)
nice @dev!!
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5 mice:
result time complete
bottle to mice1: 14 hour
after 2.5 hour to mice2 : 16.5 hour
after 2.5 hour to mice3 : 19 hour
after 2.5 hour to mice4 : 21.5 hour
after 2.5 hour to mice5 : 24 hour
one of
hey harry.what r u upto?
guys have already shown that answer is three
On Mon, Jun 27, 2011 at 4:45 AM, hary rathor harry.rat...@gmail.com wrote:
5 mice:
result time complete
bottle to mice1: 14 hour
after 2.5
@ross: seems logically correct..nice solution..
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No , you are wrong .. problem statement says how many matches should a
teams win to ensure its qualification , their no word like minimum or
maximum ...
8 gets wrong if following situation arises
1 - 9
2 - 9
3 - 9
4 - 9
5 - 8
6 - 6
7 - 4
8 - 2
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@rishabh :
in your solution u have taken scores of last 4 teams as 6 4 2 0. what if i
take 2 2 2 2 then the ans would be 56-(2+2+2+2)/4 = 12...!!!
and i can also take the scores of last 4 teams as 6 4 4 2 then the ans would
be
56-(6+4+4+2)/4 = 10!!!
so how you can say it would be 11?
On Fri,
@rishabh : now i understand it better... thanks :)
On Fri, May 27, 2011 at 7:22 AM, Rishabh Maurya poofiefoo...@gmail.comwrote:
because we want upper 4 teams to win maximum matches altogether so
to satisfy this criteria .. last team should win 0 , and team 7 must have
lost all its
@Vishwakarma
it is now ok that 11 should be the answer, but why any 4 teams cannot win 12
matches in total...
for that they have to score 12*4 = 48 points out of 56. then wats the
problem.
i know how it is coming 11 now, but i am replying back for the doubt i have
in a line u just mentioned in
@vishwakarma
thanks for rectifying me...
its clear... 12 is not posible, i was in another way :)
On Fri, May 27, 2011 at 7:46 AM, vishwakarma vishwakarma.ii...@gmail.comwrote:
correction---it was typo mistake ...
Team C loses to(one to A and one to B)
On May 27, 7:44 pm, vishwakarma
Consider the following scenario:
On the first trip, the elephant carries 1000 bananas. the elephant walk 250
km consuming 250 bananas left in position 250 (500 bananas). After that, he
goes back over 250 Km eating more bananas 250 bananas. On the second trip,
the elephant carries 1000 bananas
Brute-force algorithm with memoization for this problem!
/*
Autor: Wladimir Araújo Tavares
*/
#include stdio.h
#include stdlib.h
#include math.h
#include string.h
#define min(a,b) ab?a:b
#define max(a,b) ab?a:b
int memo[3001][1001];
int banana(int V, int D){
int total;
int j;
int
@ Dave: Disregard what I wrote!!
The algorithm that I developed after works as follows:
We can recursively define the maximum number of bananas brought by the
elephant by D km starting with V bananas:
banana (V, D) = max (V-D,
banana (V - min (V, 1000), D) + min
1) go to 1000/3 with 1000 babanas, reserves 1000/3 at 1000/3 position
2) same as 1)
now there are 2000/3 bananas at 1000/3 position.
3) go to 1000/3 position with 1000 babanas, then there are (2000/3 +
1000 - 1000/3) = 4000/3 babanas
4) go to (1000/3 + (4000/3 -1000)/3) = 4000/9 position with
533
On Sat, May 21, 2011 at 2:13 AM, Dave dave_and_da...@juno.com wrote:
@Bhavesh: 533-1/3.
Dave
On May 20, 10:47 am, Bhavesh agrawal agr.bhav...@gmail.com wrote:
1 elephant can take 1000 banana at a time and eat 1 banana after each 1km
travel.
total bananas are 3000 and distance have
with 534 , the elephant can travel only 534 Km! I am right?
Wladimir Araujo Tavares
*Federal University of Ceará
*
On Fri, May 20, 2011 at 8:36 PM, Dave dave_and_da...@juno.com wrote:
Upon reading the problem more carefully, the answer is 534 bananas,
not 533-1/3.
Dave
On May 20, 3:43
@Dave: The problem statement says, the elephant can take 1000 at a time.
If he take max 1000, and eat 1 banana in each 1 km travel, he will be having
0 after 1000 Km.
Anuj Agarwal
Engineering is the art of making what you want from things you can get.
On Sat, May 21, 2011 at 9:57 AM, Dave
up vote to
9 + 1 + 1/9
On Thu, Jan 27, 2011 at 6:06 PM, sunny agrawal sunny816.i...@gmail.comwrote:
another one
9*(1+ 1/9)
On Thu, Jan 27, 2011 at 5:40 PM, nhkrishna2...@yahoo.com
nhkrishna2...@gmail.com wrote:
9+1+1/9
On Jan 27, 4:43 pm, ankit agarwal ankitgeniu...@gmail.com wrote:
another one
9*(1+ 1/9)
On Thu, Jan 27, 2011 at 5:40 PM, nhkrishna2...@yahoo.com
nhkrishna2...@gmail.com wrote:
9+1+1/9
On Jan 27, 4:43 pm, ankit agarwal ankitgeniu...@gmail.com wrote:
(9*9-1)/(9-1)
On Thu, Jan 27, 2011 at 4:55 PM, nishaanth nishaant...@gmail.com
wrote:
@neha yeah you can use them as per your choice
On Wed, Jan 26, 2011 at 9:31 PM, Dave dave_and_da...@juno.com wrote:
9/.9 + 1 - 1
On Jan 26, 8:12 am, may.I.answer may.i.answ...@gmail.com wrote:
You have four numbers 1 , 1 , 9 ,9 .
Now using these four and operator + , - , * ,/ and
ankur is right
this problem is similar to the problem of converting a matrix to zero matrix
On Tue, Jan 4, 2011 at 8:36 PM, Ankur Khurana ankur.kkhur...@gmail.comwrote:
how are they similar ?
On Tue, Jan 4, 2011 at 8:31 PM, jennmeedo jennme...@gmail.com wrote:
Generalization algorithm for
The ant needs to cover: 9.403 units. It will need to pass the diagonal of
the side (4 by 5) and go up or down the side 3 units.
3+ sqrt(16+25)
On Fri, Dec 31, 2010 at 4:16 PM, bittu shashank7andr...@gmail.com wrote:
2nd puzzle
An ant has to crawl from one corner of a room to the
No need to enumerate all possible states.
In the final state (2,8,5), each jug is neither full nor empty, while
every valid operation has to fill or empty one jug.
So it is not possible to get this state from any other state by one
valid operation. (As others said, the state before the final
@ dheerraj...u cant measure 8 litre...u hve no additional instrument
@mohit...what do u mean by n th stageplzz elaborate
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@Sharad
let's say that it will take n steps to reach from [15,0,0] to [2,8,5] then
after nth state will be 2,8,5
and (n-1)th state will be say [x,y,z] from which one transfer will lead to
o/p [2,8,5]
hope it's clear
Mohit Ranjan
On Tue, Jun 8, 2010 at 6:54 AM, sharad kumar
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