Bloomberg LP Interview Question for Financial Application Engineers


Country: United States
Interview Type: Phone Interview




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2
of 2 vote

if the stream if items is in time order. queue could be used to store the item, and only keep last 60 seconds items

- a0 November 24, 2015 | Flag Reply
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0
of 0 vote

array of structs, sort them according to timestamp and add the 'i's of last 60 seconds (calculated using the difference of timestamps) - O(nlogn)

- ruleZ November 24, 2015 | Flag Reply
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0
of 0 vote

Array is not a nice way to store this, what if a record gets deleted in between somewhere, algorithm goes bust. Linked list (preferrably a doubly in case have to handle past as well next set of records). Doesn't need to be a stack or queue, but for this particular question a stack i more preferred, since last 60 sec data is to be referred, as and when an object is added, that becomes the current timestamp (which needs to be popped) and from there 60 sec backward. So as the item is added, pop that and count 60 sec backwards. If the question was sum of next 60 items, then that kinda points towards a queue. But since we would not be able to answer both these questions together by using a stack or queue, simply implementing a doubly linked list would serve the purpose.

- anon November 26, 2015 | Flag Reply
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0
of 0 vote

I think deque STL container, and only keep last 60 seconds items as "a0" said

Double-ended queues are sequence containers with dynamic sizes that can be expanded or contracted on both ends

they allow for the individual elements to be accessed directly through random access iterators

The complexity (efficiency) of common operations on deques is as follows:
Random access - constant O(1)
Insertion or removal of elements at the end or beginning - constant O(1)

so you can insert the new items at the end of the deque and remove the exceeded items from the beginning
and because the Random access constant O(1) you can count the 60 seconds items very fast with complexity O(n)

- engkfke January 12, 2016 | Flag Reply
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0
of 0 vote

Min-heap with insertion time as "time_t", where one periodically peeks heap to find oldest element and remove if its older than 60 secs (current_time - time_t).
Insert new element.
Keep a sum var which adds every new element inserted and subtracts value of removed node.

- prodigy January 16, 2016 | Flag Reply
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0
of 0 vote

Stack would be a solution. LIFO principle. When you pop items compare it to the time. if less than t-60, keep popping items out. If greater than t-60, stop popping items.

- itsmeamit January 18, 2016 | Flag Reply
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0
of 0 vote

Deque works best here. Assumptions: a) timestamp is filled with current time when we store the Item b) we discard old items. If b) is not valid, I'd use a vector and an index/iterator to the tail within 60 sec window.

- mike800 May 16, 2016 | Flag Reply


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