## MM2181

BAN USER- 1of 1 vote

AnswersGiven a list of Contacts, where each contact consists of a contact ID and a list of email IDs. Output a unique list of contacts by removing duplicates. Two contacts are considered to be the same, if they share at least one email ID.

- MM2181 in United States| Report Duplicate | Flag | PURGE

Facebook Web Developer - 1of 1 vote

AnswersYou are given n points (x1, y1), (x2, y2), ..... (xm, ym) of a two dimensional graph. Find 'n' closest points to (0,0) [ n <= m ]. Euclidean distance can be used to find the distance between 2 points.

- MM2181 in United States| Report Duplicate | Flag | PURGE

Facebook Web Developer

Rep**lisafergusona**, Consultant at MyntraI am Lisa from Chicago,I am working as a Show host in the New World. I also work Performs ...

Rep**sushiplarson**, Animator at Achieve InternetHi, I am a creative Assistant Video Editor with experience in all aspects of video production. Working at a post-production ...

Rep**williamland1990**, Associate at AdventGifted in donating shaving cream worldwide. Crossed the country getting my feet wet with the elderly in Salisbury, MD. Have ...

Rep**PatriciaNRowe**, Consultant at ADPHi i am a Freelance Writer and Social Media Manager who helps finance professionals and Fin-tech startups build an audience ...

**CareerCup**is the world's biggest and best source for software engineering interview preparation. See all our resources.

Open Chat in New Window

No. of steps = 3 .

- MM2181 June 20, 2010For 9 balls,

Divide them into 3 groups A, B and C of 3 balls each

1) Compare A and B

2) Compare B and C

if A != B and B = C , A contains the ball

if A != B and B!= C , B contains the ball

if A = B and B!= C , C contains the ball

Now we know the set containing the odd ball and are down to just 3 balls

Let the set contain three balls x,y,z .

[ When we used the balance for detecting the odd set, we came to know whether the odd ball is lighter or heavier ]

3) Compare x and y

if x = y , z is the odd ball

if x!= y , depending upon our observation [ lighter or heavier , we find out the odd ball between x and y ]

So , this variation does not have a best or worst case as we have to carry out steps 1 and 2 to determine whether the odd one is lighter or heavier. So in any case , we will require 3 steps

Also, the 8 ball problem requires 2 steps if the odd one is known to be lighter/heavier . Suppose it is heavier

Divide it into a set of 3 balls A = 1,2,3 B = 4,5,6 and C = 7,8

1) Compare A and B . If A = B then goto step 2 . Else goto step 3

2) Compare 7 and 8 and the heavier is the odd one

3) From the heavier set between A and B , compare any 2 balls. If they are equal , third one is odd . Else the one which is heavier is odd.

Thus, 2 steps