Coding Interview Questions

Gowtami K. - "select a.playername as player1, b.playername as player2, abs(a.level-b.level) as level_disparity from players a cross join players b on a.playername < b.playername where abs(a.level-b.level) <=5 order by level_disparity `"See full answer

Flora K. - "USING RECURSION"See full answer

Alireza K. - "import random def coin_flip(): x=4*[0]+[1] res=[] for i in range(20): res.append(random.choice(x)) return res res=[0,0] # [head,tail] for j in range(1000): temp=coin_flip() res[0]+=sum(temp) #head res[1]+=(20-sum(temp)) #tail"See full answer

Ina K. - "HashMap supports insert, search, delete and retrieve in O(1). It stores data as key value pairs."See full answer

Anonymous Unicorn - "input = [ {"topic": 1, "chapter": 1, "section": 1}, {"topic": 2, "chapter": 2, "section": 1}, {"topic": 3, "chapter": 2, "section": 2}, {"topic": 4, "chapter": 1, "section": 1}, {"topic": 5, "chapter": 1, "section": 1}, {"topic": 6, "chapter": 2, "section": 2}, {"topic": 7, "chapter": 2, "section": 2}, {"topic": 8, "chapter": 2, "section": 3}, ] expected_output = [ {'chapter': 1, 'sections': [ {'section': 1, 'topics': [ {'top"See full answer

Nicholas S. - "Definitely nice to think of this without memorization, but there is a well known algorithm for this problem, which is the Levenshtein Distance. Lev(a,b) = len(a) if len(b) == 0 = len(b) if len(a) == 0 = lev(a[1:], b[1:] if a[0] == b[0] = 1 + min (lev(a, b[1:]), lev(a[1:], b), lev(a[1:], b[1:])) https://en.wikipedia.org/wiki/Levenshtein_distance I'm sure some optimizations could be made with heuristic."See full answer

Nishigandha B. - "na"See full answer

Karthik R. - "Use Dijkstra's Algorithm with priority queue"See full answer

Divya K. - "You will need to start from Browser and go all the way up to Analytic systems and methods. Everything needs to be covered"See full answer

Idan R. - " read_dir(path: str) -> list[str] returns the full path of all files and sub- directories of a given directory. is_file(path: str) -> bool: returns true if the path points to a regular file. is_dir(path: str) -> bool: returns true if the path points to a directory. read_file(path: str) -> str: reads and returns the content of the file. The algorithm: notice that storing all the file contents' is too space intensive, so we can't read all the files' contents to store and compare with each"See full answer

Lakshman B. - "As we can pass info to only one child at a time, I told that from any given node, we have to pass the info to that child(of this node) which has the largest subtree rooted at it. To calculate the subtree sizes, I used DFS. And then to calculate the minimum time to pass info to all the nodes, I used BFS picking the largest subtree child first at every node. I couldn't write the complete code in the given time and also made a mistake in telling the overall time complexity of my approach. I think t"See full answer

Matthew K. - " function diffBetweenTwoStrings(source, target) { /** @param source: string @param target: string @return: string[] */ let dp = new Array(source.length+1).fill().map(() => Array(target.length+1).fill(0)) for (let i = source.length; i>= 0; i--) { for (let j = target.length; j>= 0; j--) { if (i === source.length) { dpi = target.length - j } else if (j === target.length) { dpi = sou"See full answer

The_ A. - "The best and average of both the algorithms is same which is O(nlog(n)), but the worst time complexity of QuickSort is O(n^2){a case where all the elements are sorted in opposite to the fashion/order you want} while the worst TC for merge sort remains the same O(nlog(n)). and The SC for QS=O(logn) and MS=O(n)."See full answer

Divyani .. - "Did the code in Python"See full answer

Gabriella F. - "Table user is empy....... Problem with this problem "See full answer

Rishabh R. - "if decreasing arr, start from end and keep checking if next element increases by 1 or not. wherever not, put that value there."See full answer