Design verification engineer Interview Questions

Trail 1: At random weigh two stones vs. two stones (3 sitting on the side) A: Of the 4 on the scales if one side weighs more then the other then weigh one on each side (since one of them must be heavier) B. If the 2 vs 2 are equal then at random weigh 2 (one on each side) of the three left on the side. If they are the same then the 3rd one that never got weighed is the heaviest. Simple case of process of elimination by grouping (Divide and Test)

I would first weigh in one stone, say stone 1, and assume the weight is say 2 lbs(try 1). Then separate the 6 remaining stones into 2 piles, 2,3,4 and 5,6,7. Weigh in either 2,3,4 or 5,6,7, it doesn't matter. Say 2,3,4, if the sum of these 3 is 6 then the lighter stone has to be in the 5,6,7(try 2). Weigh in 5,6, if the sum of the two is 4 then the lighter is stone 7(try3). If sum is less than 4 then weigh in either 5 or 6 to find out. So, the maximum number of tries is 4 and least is 3.

Needs two weighing at most: 1. Put {1, 2, 3} on LHS and {4, 5, 6} on RHS. 2. If LHS and RHS are equal 7 is the lighter one. else discard heavier of previously weighed group. Now we have a group 3 stones left. Lets call them A, B, C. 3. Put A on LHS and B on RHS. 4. If LHS and RHS are equal C is the lighter one. else lighter or LHS or RHS is the lighter one. Voila!

0-100 have constant addition. add the numbers in the array and subtract it from the constant sum.

Assuming that a number only appears once, find the sum and subtract from the total possible value (1+2+3+....+100) and subtract to find the missing number. But it would be smart to ask if a number can appear multiple times or not since that may increase or decrease the complexity of the problem, but shows that you're really thinking about all the problem parameters.

This should be done in 2 iterations...keep 3 in each hand...higher weight indicates the heavier marble is in that set...now with 3 marbles select 2...if weights are equal then the marble kept aside is the heavier one..

2 iterations. Take any 4 out of 6 and weigh 2 vs 2. If it is equal, weigh the other 2 as 1vs 1 and find the heaviest. If the 2 v 2 is uneven, take the uneven marbles and weigh them as 1 vs 1 and find the heaviest one.

What exactly does this mean? Did they ask you to write all the phases along with their syntax or what exactly is the question here?

Can you please give more details on the questions asked during your different rounds if you can recollect them?

Did you get any reply from qualcomm?

were you fresher or experienced?

2. You can implement LIFO using a combination of queue.push_back +queue.pop_back OR queue.push_front+queue.pop_front

2. You can implement LIFO using a combination of queue.push_back +queue.pop_back OR queue.push_front+queue.pop_front

In parallel computing, a barrier is a type of synchronization method. A barrier for a group of threads or processes in the source code means any thread/process must stop at this point and cannot proceed until all other threads/processes reach this barrier. Place a barrier between the instruction that works on them memory and the instruction that set the flag indicating that work is completed on that memory location.