Coding Interview Questions

Alvin P. - "WITH logins AS(SELECT user_id, timestamp, RANK() OVER(PARTITION BY userid ORDER BY timestamp ASC) AS loginorder FROM useractivitylog WHERE activity_type = 'LOGIN') SELECT l1.user_id, l1.timestamp AS current_login, l2.timestamp AS previous_login, (strftime('%s', l1.timestamp) - strftime('%s', l2.timestamp)) / 60 AS minutes_elapsed FROM logins AS l1 JOIN logins AS l2 ON l1.userid = l2.userid AND l1.loginorder - l2.loginorder = 1 GROUP BY l1.user_id,"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

Anonymous Armadillo - "I couldn't follow the solution offered here, but my solution seemed to pass 6/6 tests. Any feedback is welcome, thank you! def encrypt(word): en_word = "" for i in range(len(word)): if i == 0: en_word += chr(ord(word[0])+1) else: num = ord(word[i]) + ord(en_word[i-1]) while num > 122: num -= 26 en_word += chr(num) return en_word def decrypt(word): de_word = "" for i in range(len(word)): if i == 0: de_word += chr(ord(word[i]"See full answer

Ravi teja N. - "I solved it using recursion and then memoization. Used Dynamic programming approach"See full answer

Rohit B. - "class Solution: def missingNumber(self, nums: list[int]) -> int: Sorting approach n = len(nums) s = n*(n+1)//2 r = s - sum(nums) return self.r l = [3,0,1] print(missingNumber(l))"See full answer

Tiago R. - "function getCheapestCost(rootNode) { let cost = rootNode.cost; if (rootNode.children.length === 0) { return cost; } let minCost = Infinity; for (let child of rootNode.children) { minCost = Math.min(minCost, getCheapestCost(child)); } return cost + minCost; } `"See full answer

Ahmed A. - "Good Question, but I would've marked this as medium not hard difficulty, since it's just a straightforward traversal."See full answer

Anonymous Porcupine - "https://www.geeksforgeeks.org/find-local-minima-array/ I coded O(N) but after that gave a binary approach aswell. After that he also gave a varient of this problem in which, local minima means that the number is strictly less than its adjacent (we cannot do binary search there sample test case [1,1,1,1,1,1,0,1] or [1,0,1,1,1,1,1,1] using mid we cannot determine if the minima is on left or right). we have to do a linear search or find recursively."See full answer

Prashanth A. - "We are asked to calculate Sum(over value) for time in (t - window_size, t) where key in (key criteria). To develop a function to set this up. Let w be the window size. I would have an observer of some kind note the key-value, and for the first w windows just add the value to a temporary variable in memory if the key meets the key criteria. Then it would delete the oldest value and add the new value if the new key meets the criteria. At each step after "w", we would take the sum / w and store"See full answer

Anzhe M. - "Not my answer, but rather the details of this question. It should include the following functions: int insertNewCustomer(double revenue) -> returns a customer ID (assume auto-incremented & 0-based) int insertNewCustomer(double revenue, int referrerID) -> returns a customer ID (assume auto-incremented & 0-based) Set getLowestKCustomersByMinTotalRevenue(int k, double minTotalRevenue) -> returns customer IDs Note: The total revenue consists of the revenue that this customer bring"See full answer

Gowtami K. - "select DISTINCT p.product_id, p.product_name , CASE when sale_date is null then 'Not Sold' else 'Sold' END as sale_status from products p left join sales s on p.productid= s.productid `"See full answer

Aneesha K. - "Test case is wrong. It expects to sort in asc order of month_year. -- Write your query here SELECT strftime('%Y-%m', createdat) AS monthyear, COUNT(DISTINCT userid) AS numcustomers, COUNT(t.id) AS num_orders, SUM(price * quantity) AS order_amt FROM transactions t INNER JOIN products p ON t.product_id = p.id GROUP BY month_year ORDER BY month_year ; "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

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

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