I would like to submit the following question I was asked
recently during my technical interview at Google.
I'm rephrasing the question to make it clear for everyone
to understand:
- You are going on a one-way flight trip that includes
billions of layovers.
- You have 1 ticket for each part of your trip (i.e: if
your trip is from city A to city C with a layover in city
B, then you will have 1 flight ticket from city A to city
B, and 1 flight ticket from city B to city C.
- Each layover is unique. You are not stopping twice in the
same city.
- You forgot the original departure city.
- You forgot the final destination city.
- All the tickets you have are randomly sorted.
Question are:
- Design an algorithm to reconstruct your trip with minimum
complexity.
- How would you improve your algorithm.
Example:
- randomly sorted:
New York->London
San Francisco-> Hong Kong
Paris->New York
London->San Francisco
- sorted:
Paris->New York
New York->London
London->San Francisco
San Francisco-> Hong Kong
Answer Posted / raghvendra
1) Assuming the data in 2 dim array. (2 Columns: FROM
destination & TO destination)
2) Sort the array (any standard sorting logic) using column
1 i.e. FROM destination.
3) Take "TO" destination value of row 1 and use binary
search to search its match in "FROM" destination column
from row 2 to n
If match found, adjust that row as row 2. (Swap
pointers in place of physical data)
Take "TO" destination value of row 2 and
use binary search to search its match in "FROM" destination
column from row 3 to n
continue this logic till as long as a match
is found.
if match not found.
Resort the remaining unadjusted rows
using "TO destination" column.
Take "From" destination value of
row 1 and use binary search to search its match in "To"
column in all unadjusted rows.
Match must be found otherwise data
incorrect. adjust that row as row -1. (Swap pointers in
place of physical data)
Continue this till end of the
array.
Now you have the sorted list.
I am sure there can better ways to achieve the same.
Is This Answer Correct ? | 3 Yes | 0 No |
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