Data Structures & Algorithms Interview Questions

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

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

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

Nishigandha B. - "na"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

Divyani .. - "Did the code in Python"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

Kunal kumar S. - "simply check its size if the size if the size is greater than n then yes it has duplicate"See full answer

Shivani N. - "did well but messed up dequeue"See full answer

Nathan B. - " This is mostly correct and fairly fast. My code has a bug somewhere where it fails on cases like the last case, where there are negative number on both ends of the array and the sums . from collections import deque debug = True # False def prdbg(*x): global debug debug = True # False if debug: print(x) else: return def max_sum(arr, start, end): if type(arr) == type(''' "See full answer

Ravi C. - "boolean isMatch(String s, String p) { int i=0; int j=0; int sID=-1; int prevM=-1; while(i<s.length()){ if(j<p.length() && (s.charAt(i)==p.charAt(j) || p.charAt(j)=='?')){ i++; j++; }else if(j<p.length() && p.charAt(j)=='*'){ sID=j; prevM=i; j++; }else if(sID!=-1){ j=sID+1; prevM++; i=prevM; }else{ return false; } } while(j<p.length() && p.charAt(j)=='*') j++; if(i!=s.length() || j!=p.leng"See full answer

Vaibhav D. - "Make current as root. 2 while current is not null, if p and q are less than current, go left. If p and q are greater than current, go right. else return current. return null"See full answer