Given the root of a tree and a target node, find the cousins of the target node (those that are on the same level as the target node, but do not share the same parent).
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Some webpage design problems
The technical question was: You're given an array of strings. Sort it so that the result returns an array of an array of strings sorted into its anagrams. (e.g. input: ["aa", "ad", "da"], output: [ ["aa"], ["ad", "da"] ]
find the sum of two integers represented as strings, return the sum as string, i.e "123" and "456" would return "579".
Design an LRU cache. It's a data struct with a capacity. Beyond this capacity the least recently used item is removed. You should be able to insert an element, access an element given its key, and delete an element, in constant time. Note that when you access an element, even if it's just for a read, it becomes the most recently used.
Remove duplicates. I could not solve it without sorting the array first.
Fizzbuzz but with fibonacci numbers!
Given a string that looks like this: Fall2014 BIO110, return 4 pieces of information: Season - Winter, Spring, Summer, Fall Year - 20XX Subject - BIO, ENGR, etc Course number The season and year cluster may be space separated. Season always comes before year. The subject and course number cluster may also be space separated. Subject always comes before course number. The season/year and subject/course clusters are always space separated. You are guaranteed that season/year cluster comes first. Season can also be given like this: W = Winter Sp = Spring Su = Summer F = Fall Years can be given without their prefix of "20". So if you're given Fall14, you should be able to extract a year of 2014. Subject and course number come as they are. You are guaranteed that subjects are purely alphabetical. Examples: Input: F 13 ENGR 110 Season: Fall Year: 2013 Subject: ENGR Course #: 110 Input: Su2015 BIO340 Season: Summer Year: 2015 Subject: BIO Course #: 340
Deep copy of a linked list, with an extra pointer to a random node.
Determine if a string of parentheses is valid
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