Data Structures & Algorithms Interview Questions

Casey C. - "Determine the requirements Perform basic checks on the frontend before submitting to the server Length Check for invalid or required characters Client-side validation messaging Perform more advanced checks on the server if required Does the password include the user's name, etc Handle server-side validation messaging"See full answer

Vishnu V. - "// Helper function to calculate the Euclidean distance between two points function distance(p1, p2) { return Math.sqrt(Math.pow(p1[0] - p2[0], 2) + Math.pow(p1[1] - p2[1], 2)); } // A helper function to find the closest pair in a given set of points within the strip function closestPairInStrip(strip, d) { let minDist = d; // Start with the current minimum distance strip.sort((a, b) => a[1] - b[1]); // Sort the strip by y-coordinate for (let i = 0; i < strip.length; i++) { "See full answer

Jonathan michael J. - "add two strings `"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

Flora K. - "USING RECURSION"See full answer

Vince S. - "func isMatch(text: String, pattern: String) -> Bool { // Convert strings to arrays for easier indexing let s = Array(text.characters) let p = Array(pattern.characters) guard !s.isEmpty && !p.isEmpty else { return true } // Create DP table: dpi represents if s[0...i-1] matches p[0...j-1] var dp = Array(repeating: Array(repeating: false, count: p.count + 1), count: s.count + 1) // Empty pattern matches empty string dp[0]["See full answer

Divyani .. - "Did the code in Python"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

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

Nishigandha B. - "na"See full answer

Kwaku K. - "Use a BFS"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