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Adobe Interview Question for Software Engineer / Developers






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0
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Let we have given fun: int fun6040()
if i will execute this function 5 times.... 3 times it will give o/p = '0' and 2 times o/p='1'.

take an Array[6]

Array[0]=fun6040();
Array[1]=fun6040();
Array[2]=fun6040();
Array[3]=fun6040();
Array[4]=fun6040();
Array[5]=1;

now apply randomize on Array[6] you will get any random no. which will have probability 50% for '0' as well as '1'.

- PKT February 28, 2011 | Flag Reply
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0
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Well PKT ... you got it wrong.. We always consider probability for limit -> infinity... now here you taking the limited case of 5 function calls.. ur probability just tells about the chance ...it never talks about the outcome..hence here it can give 0 on all these five calls.. but it will maintain the probability for infinity events..

For eg :
lets following be the return of the some first 15 fun6040() of infinite calls..

0 0 0 0 0 1 1 1 1 0 0 1 1 0 0 ........
Here if we take first five calls then for 0 its 100 percent probability but on an avg i.e..15 or infinity it is 60 % i.e.. So your answer is wrong... Saurabh got it right I guess...

- thear February 28, 2011 | Flag
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0
of 0 votes

I think u have some misconception about Probability...

I have a coin
i flip it
wt is the probability of getting "Head" ?????

I am saying 50%

and u r saying we can't sure abt it until infinite flip.... which is wrong.

for a dice {1,2,3,4,5,6}

wt is the probability of getting 1 ??????
1/6 = 16.66% isn1t it??

ok wt is the probability of getting {1,2,3}
3/6 = 50% ????

still doubt?

- PKT March 01, 2011 | Flag
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0
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I dont have any doubt.. You said everythin right.. But u got it wrong on the number of events..probability of head is 1/2 defintly..but if i toss a coin for first two times..I cant say that i wud get one head and one coin..coz probabilty covers infinite events...and there may be case i wud get two heads on first two tosses becoz its just a probabilty ..we are just guessin the measure of truth in the statemnt..thats it.. I hope you understand the probability..

- thear March 01, 2011 | Flag
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0
of 0 votes

Anyways If u still not clear..lemme explain u in a more broader way..

Do the followin example practically
Take a coin.. and toss it six times..
Can u be sure that it wud give 3 heads and 3 tails.. I dont think it will give me this combination everytime(inspite both head and tail having 1/2 probability)..it may be 4 heads 2tails...and multiple cases...

So buddy we can not just take probability on these 6 tosses..rather practically we take it for infinite cases..

I hope now u get it..

- thear March 01, 2011 | Flag
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0
of 0 vote

Maybe we can store an boolean array containing 6 elements, where there are 5 ones and 1 zeros. Every time we run the given function, if it gives zero, then output zero; if it is one, one element from the stored array is output. In this way, the chance of outputting one is 60%*(5/6)=50%.

bool UNIFORM() {
static int count = 0;
bool temp[6] = {1,1,1,1,1,0};
bool input = NONUNIFORM();
if ( !input ) {
return 0;
}
else {
return temp[(count++)%6];
}
}

- Daru February 28, 2011 | Flag Reply
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0
of 0 votes

Oh, sorry... I did the case for 40% for 0 and 60% for 1. My fault

- Daru February 28, 2011 | Flag
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0
of 0 votes

Nice solution...easy to follow...good job

- Anshul April 13, 2011 | Flag
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0
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f1 = 0 with 60% probability.
= 1 with 40% probability.
If we call f1() two times and depending on the return values say that :
0,1 = 0 (60% * 40%)
1,0 = 1 (40% * 60%)
(0,0) and (1,1) = continue.
we will have equal probability for 0 and 1 for the return value of the composite function.


int function f2()
{
int ret = -1;
while(ret == -1)
{
int res1 = f1();
int res2 = f1();
if(res1 == 0 && res2 == 1) ret = 0;
else if(res1 == 1 && res2 == 0) ret = 1;
}
return ret
}

- Administrator February 28, 2011 | Flag Reply
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0
of 0 votes

Nice one saurabh..!!

- thear March 01, 2011 | Flag
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0
of 0 votes

while loop will never be entered and it will always return -1 lol

- Anonymous March 02, 2011 | Flag
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0
of 0 votes

Good logic saurabh.

- Tapesh March 02, 2011 | Flag
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0
of 0 votes

great job saurabh.What if we need 30% (1) and 70% (0) ?

- Lio March 02, 2011 | Flag
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0
of 0 votes

didn't get the logic

- ajeet March 07, 2011 | Flag
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0
of 0 votes

Can you please elaborate the logic a bit more. Thanks!

- abc March 19, 2011 | Flag
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0
of 0 votes

it gives 24% probability to return 1 24% to return 0 and 52% to recurse. ..so there is no zero probability of your function to enter an infinite loop.

- anand May 15, 2011 | Flag
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0
of 0 vote

I think that the 2 pairs (0,0) and (0,1) will appear more often then the other two pairs, because 0 has 60% probability. If that's true, (0,1) will appear more than (1,0) and we don't have 50/50. Please correct me if I'm wrong. Thanks.

- Anh March 01, 2011 | Flag Reply
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0
of 0 vote

Why not just call f1() two times and invert the output on the second call. Then randomly choose between the two outputs.

- mplode March 02, 2011 | Flag Reply
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0
of 0 vote

It is related to "Rejection sampling", a simple way to implement it in C is like this:

int even_prng() {
while (1) { if ((a = f()) ^ f()) return a;}
}

See wiki: en.wikipedia.org/wiki/Rejection_sampling

- Eric Xu March 03, 2011 | Flag Reply
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0
of 0 vote

Just flip the result half the time.

int my_rand(void)
{
	static int flip=0;
	return f()^(flip^=1);
}

- Anonymous March 22, 2011 | Flag Reply
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0
of 0 vote

1.Flip the coin 5 times
2. Count number of 0s and 1s obtained. Decrement 1 from the number of 0s.
3. Now compare number of 0s and number of 1s. In number of 0s > number of 1s return 0,
If number of 1s > number of 0s return 1. If both are equal repeat from 1

- schintan April 23, 2011 | Flag Reply
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0
of 0 vote

static int ret=0;
if(ret==f())//this will execute once in 10 times.
{
ret=!f();
}
else//Will execute most of the time
{
ret=f();
}
return ret;

- S May 04, 2011 | Flag Reply
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0
of 0 vote

while (1)
{x=givenfun();
y=givenfun();
if (x==y)
return x;}

- asd May 10, 2011 | Flag Reply


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