Amazon Interview Question: Question regarding how to ret... | Glassdoor

## Interview Question

Software Development Engineer Intern Interview

# Question regarding how to retrieve the second highest

repeated number in array the second question was how to print the nodes of binary tree level wise
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technical

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I answered verbally the first one well. But i could not code(I panicked). The first interviewer was chilled out. He asked me regarding space complexities and stuff.

The second one I was able to write the solution on time but I could explain him the complexity issues. And I spoke some rubbish. I freaked out. I think that was the only reason i got rejection.

Interview Candidate on Feb 8, 2014
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Retrieve the second highest repeated number in array
* Assuming no output if there aren't at least two pairs of repeated numbers
Create a vector of pairs of boolean and integer values
Using two variables, highest and second highest repeated values:
- Initialize to some unreachable value (-1 or so)
For each number in array:
- Check if the number is in the vector:
- If not, add it to the vector and set the boolean to false
- If it is and the boolean is false, set the boolean to true and:
- Compare the value to the highest repeated value:
- If it is higher, set the second highest to the old highest, and
set the new highest to the number we've found
- If it is lower, compare it to the second highest repeated value:
- If it is higher, set the second highest to the number we've found
Upon completion of array iteration, second highest repeated value.
* This answer may not be perfect or optimal, but always give a fast solution
* and then improve upon it if it's found to be necessary.

Print the nodes of binary tree level wise
* Assuming pointers are accessible:
Have a helper function that gets called for all recursion that takes in depth
For all children (right or left) call the helper function with an incremented depth
- If the depth matches, call a print function (or if POD, use printf or std::cout, etc)
- If the depth exceeds, return

Anonymous on Jan 8, 2015