Coding Interview Questions

Divya R. - "static boolean sudokuSolve(char board) { return sudokuSolve(board, 0, 0); } static boolean sudokuSolve(char board, int r, int c) { if(c>=board[0].length) { r=r+1; c=0; } if(r>=board.length) return true; if(boardr=='.') { for(int num=1; num<=9; num++) { boardr=(char)('0' + num); if(isValidPosition(board, r, c)) { if(sudokuSolve(board, r, c+1)) return true; } boardr='.'; } } else { return sudokuSolve(board, r, c+1); } return false; } static boolean isValidPosition(char b"See full answer

Michael B. - "Problem: Given a modified binary tree, where each node also has a pointer to it's parent, find the first common ancestor of two nodes. Answer: As it happens, the structure that we're looking at is actually a linked list (one pointer up), so the problem is identical to trying to find if two linked lists share a common node. How this works is by stacking the two chains of nodes together so they're the same length. chain1 = node1 chain2= node2 while True: chain1 = chain1.next chain2=chain"See full answer

Gowtami K. - "with cte as ( select user_id, timestamp as current_login, lag(timestamp,1) over(partition by userid order by timestamp) as previouslogin , round(abs(julianday(timestamp)-julianday(lag(timestamp,1) over(partition by userid order by timestamp)))2460)as minuteselapsed from useractivitylog where activity_type ='LOGIN' ) select userid, currentlogin, previouslogin, minuteselapsed from cte where currentlogin previouslogin `"See full answer

Frank A. - "1) create the experimental and control groups. 2) Then calculate the proportion (mean) of the true conversion rates for both groups using the convert column which counts True as 1 and False as 0. This is their conversion rates 3) calculate the statistic of the two groups by subtracting the proportion and standardizing. 4) get the p-value and compare with 0.05. 5) conclude the difference is statistically significant if the p-value is less than 0.05 otherwise no statistical difference"See full answer

Steven S. - "too many questions for clarification on this to start"See full answer

Jayveer S. - "WITH suspicious_transactions AS ( SELECT c.first_name, c.last_name, t.receipt_number, COUNT(t.receiptnumber) OVER (PARTITION BY c.customerid) AS noofoffences FROM customers c JOIN transactions t ON c.customerid = t.customerid WHERE t.receipt_number LIKE '%999%' OR t.receipt_number LIKE '%1234%' OR t.receipt_number LIKE '%XYZ%' ) SELECT first_name, last_name, receipt_number, noofoffences FROM suspicious_transactions WHERE noofoffences >= 2;"See full answer

B. T. - "You are given a string S and a number K as input, and your task is to print S to console output considering that, at most, you can print K characters per line. Example: S = "abracadabra sample" K = 11 Output: abracadabra sample Note that this problem requires the interviewee gather extra requirements from the interviewer (e.g. do we care about multiple white spaces? what if the length of a word is greater than K, ...)"See full answer

Bhavya V. - "we can create 2 sets for rows and columns and store the rows and column having 0 and then just check in a loop if the count of that row is greater than 0 llly for column then put the row and column to zero"See full answer

Alex M. - "A red-black tree is a self-balancing binary search tree. The motivation for this is that the benefits of O(logN) search, insertion, and deletion that a binary tree provides us will disappear if we let the tree get too "imbalanced" (e.g. there are too many nodes on one side of the tree or some branches have a depth that is way out of proportion to the average branch depth). This imbalance will occur if we don't adjust the tree after inserting or deleting nodes, hence our need for self-balancing c"See full answer

Arshad P. - "Schema is wrong - id from product is mapped to id from transactions, id from product should point to product_id in transcations table"See full answer

Md kamrul H. - "public class HashMap { public class Element { T key; V value; Element(T k, V v) { this.key = k; this.value = v; } } private static final int DEFAULT_CAPACITY = 16; private static final float LOAD_FACTOR = 0.75f; private LinkedList[] table = new LinkedList[DEFAULT_CAPACITY]; private int size = 0; private int threshold = (int) (DEFAULTCAPACITY * LOADFACTOR); public void put(T k"See full answer

Jayveer S. - "SELECT d.department_name, SUM(o.orderamount) AS totalrevenue FROM orders o JOIN departments d ON o.departmentid = d.departmentid WHERE o.order_date >= DATE('now', '-12 months') GROUP BY d.department_name ORDER BY total_revenue DESC; "See full answer

Tiago R. - "class TrieNode { constructor() { this.children = {}; this.isEndOfWord = false; } } class Trie { constructor() { this.root = new TrieNode(); } insert(word) { let node = this.root; for (const char of word) { if (!node.children[char]) { node.children[char] = new TrieNode(); } node = node.children[char]; } node.isEndOfWord = true; } search(word) { l"See full answer

Harshi B. - "SELECT DISTINCT title, ROUND(AVG(rating) over (partition by title),1) avg_rating, ROUND(AVG(rating) over (partition by genre),1) genre_rating FROM rating r JOIN movie m ON r.movieid=m.movieid ORDER by 1"See full answer

Aelaf G. - "This depends on the list of documents and the length of the documents. My implementation will use Trie with node containing the following: class TrieNode { is_end: boolean, instances: { docid → [wordpositions] }, children: array[26] } Look up for a word will give result instances{docid:wordposition...} dictionary (which can be further improved by methods like max instance on a document....you name it...) Trie space is proportional to the total characters in"See full answer