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Given a list of numbers ( fixed list) Now given any other
list, how can you efficiently find out if there is any
element in the second list that is an element of the
first list (fixed list)
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Answer Posted / karan verma
Above method will be most efficient in terms of time
complexity that is O(n).
If we desire space complexity O(1)
--> sort the two lists O(nlogn)
--> find the missing no. O(n)
O(n+nlogn)=O(nlogn)
space complexity=O(1)
| Is This Answer Correct ? | 11 Yes | 7 No |
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