Amazon Interview Question Software Engineer / Developers


Country: United States
Interview Type: Phone Interview


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2
of 2 vote

Walk second list and insert all elements in a hash.
Walk first list and if an element is already in hash it is the common element.

- Anonymous on October 29, 2011 | Flag Reply
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0
of 0 votes

This is if the lists are unsorted, as I think they are. You could do it without extra space in O(n) if the lists were sorted.

- eugene.yarovoi on October 31, 2011 | Flag
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1
of 1 vote

1) Find lengths of two linked list O(n1) + O(n2)
2) Find the long list and skip the (long list length - short list length) nodes
3) Now walk the two linked lists together (in parallel one step/move-next at a time), and compare the values of two nodes (just compare node address if the lists are already joined). Return the node value/address where the euality check succeeds. O(min(n1, n2))

- Anonymous on November 02, 2011 | Flag Reply
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0
of 0 votes

???

Why would you blindly skip nodes in the longer list? The question asks you to look for the intersection between the two lists. What guarantee do you have that the first (LongerList.Count - ShorterList.Count) Nodes in the LongerList are going to be non-intersecting elements?

- Anonymous on November 02, 2011 | Flag
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0
of 0 votes

Because when intersection point is somewhere after these nodes only. like 1-2-3-4-5 and 3-4-5. first 2 nodes can be skipped.in a Y shaped intersection.

- Amol on November 22, 2011 | Flag
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0
of 0 votes

cant it be like 1-2-3-4-5 and 1-2-3. it's still a Y shaped intersection

- Anonymous on November 23, 2011 | Flag
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0
of 0 vote

List1.Intersect(List2).First();

Of course that's not an acceptable answer.
So sort the lists
Insert elements of one list into a HashTable.
Walk the second list and check each element to see if it's in the HashTable.
Return the first or null.

- Anonymous on November 02, 2011 | Flag Reply


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