Coding Interview Questions

Tiago R. - "function isValid(s) { const stack = []; for (let i=0; i < s.length; i++) { const char = s.charAt(i); if (['(', '{', '['].includes(char)) { stack.push(char); } else { const top = stack.pop(); if ((char === ')' && top !== '(') || (char === '}' && top !== '{') || (char === ']' && top !== '[')) { return false; } } } return stack.length === 0"See full answer

Gloriose H. - "Required output in the solution not the one requested from the question. only customerid, firstname, last_name and years were required. Please this needs to be very clear. Otherwise my answer is with totalorderyear as ( SELECT o.customer_id, c.first_name, c.last_name, EXTRACT(YEAR FROM o.orderdate) AS orderyear, COUNT(o.orderid) AS totalorders FROM orders o LEFT JOIN customers c ON c.customerid = o.customerid GROUP BY o.customerid, c.firstname, c.last"See full answer

Avon T. - "Initialize left pointer: Set a left pointer left to 0. Iterate through the array: Iterate through the array from left to right. If the current element is not 0, swap it with the element at the left pointer and increment left. Time complexity: O(n). The loop iterates through the entire array once, making it linear time. Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures. "See full answer

Gabriele G. - "DFS with check of an already seen node in the graph would work from collections import deque, defaultdict from typing import List def iscourseloopdfs(idcourse: int, graph: defaultdict[list]) -> bool: stack = deque([(id_course)]) seen_courses = set() while stack: print(stack) curr_course = stack.pop() if currcourse in seencourses: return True seencourses.add(currcourse) for dependency in graph[curr_course]: "See full answer

Rudra pratap S. - "select sum(orderquantity) as totalunitsorderedyesterday from orders as ord join items as it on ord.itemid=it.itemid where order_date="2023-10-14""See full answer

Michael S. - "import string from collections import defaultdict def mostcommonwords(text): d = defaultdict(int) s = text.translate(str.maketrans('', '', string.punctuation)) for w in s.lower().split(): d[w] = d[w] + 1 return sorted(sorted(d.items()), reverse=True, key=lambda x: x[1]) `"See full answer

Victor N. - "--country names are UPPERCASE but the table in the in the question showing lowercase. That's why it took me a while to figure it out until I ran the country column WITH RECURSIVE Hierarchy AS ( SELECT e.Emp_ID, CONCAT(e.FirstName, ' ', e.MiddleName, ' ', e.LastName) AS FullName, e.Manager_ID, 0 AS Level, CASE WHEN e.Country = 'IRELAND' THEN s.Salary * 1.09 WHEN e.Country = 'INDIA' THEN s.Salary * 0.012 ELSE s.Salary "See full answer

Daniel C. - "select name, stock from products p left join transactions t on p.id = t.product_id order by date desc limit 1"See full answer

Sachin R. - "If 0's aren't a concern, couldn't we just multiply all numbers. and then divide product by each number in the list ? if there's more than one zero, then we just return an array of 0s if there's one zero, then we just replace 0 with product and rest 0s. what am i missing?"See full answer

Dinesh G. - "import numpy as np class Centroid: def init(self, location, vectors): self.location = location # (D,) self.vectors = vectors # (N_i, D) class KMeans: def init(self, n_features, k): self.nfeatures = nfeatures self.centroids = [ Centroid( location=np.random.randn(n_features), vectors=np.empty((0, n_features)) ) for _ in range(k) ] def distance(self, x,"See full answer

Evan R. - "The user table no longer exists as expected - I get an error that user does not contain user_id. Note that querying the table results in only user:swuoevkivrjfta select * FROM user `"See full answer

Rick E. - " from typing import List one pass O(n) def find_duplicates(arr1: List[int], arr2: List[int]) -> List[int]: duplicates = [] i1 = i2 = 0 while i1 < len(arr1) and i2 < len(arr2): if arr1[i1] == arr2[i2]: duplicates.append(arr1[i1]) i2 += 1 i1 += 1 return duplicates debug your code below print(find_duplicates([1, 2, 3, 5, 6, 7], [3, 6, 7, 8, 20])) `"See full answer

Gabriele G. - "Without using a recursive approach, one can perform a post-order traversal by removing the parent nodes from the stack only if children were visited: def diameterOfTree(root): if root is None: return 0 diameter = 0 stack = deque([[root, False]]) # (node, visited) node_heights = {} while stack: curr_node, visited = stack[-1] if visited: heightleft = nodeheights.get(curr_node.left, 0) heightright = nodehe"See full answer

Jordan Z. - "Maybe we can use this solution: 1, connect all the strings together, and add an integer value ahead each string. 2, use Huffmans algorithm to encode the step 1 result, to make the result size smaller. 3, return the root of Huffmans tree. This solution man be slower than the common serialize method, but it can save a lot of memory, I think, at lease doing serialize is mainly for tranfering data or storing data."See full answer

Peter W. - "In the question it says: "above the overall average total posts", which to me implying a >, yet in the solution it uses >= Caused me 1 hr to find out. plz fix"See full answer

Reno S. - " ✅ Passes all test cases: Tricky: stock_prices = [1, 10, 2, 3] output: 9 (Buy on day 1 at 1 and sell on day 2 at 10) func maxProfit(_ stockPrices: [Int]) -> Int { var options: [Int] = [] // min-heap var currentProfit = 0 var maxProfit = 0 for price in stockPrices { if let cheapestOption = options.last, cheapestOption < price { if currentProfit < price { currentProfit += price // greedy profit } else { "See full answer

Matthew K. - " function climbStairs(n) { // 4 iterations of Dynamic Programming solutions: // Step 1: Recursive: // if (n <= 2) return n // return climbStairs(n-1) + climbStairs(n-2) // Step 2: Top-down Memoization // const memo = {0:0, 1:1, 2:2} // function f(x) { // if (x in memo) return memo[x] // memo[x] = f(x-1) + f(x-2) // return memo[x] // } // return f(n) // Step 3: Bottom-up Tabulation // const tab = [0,1,2] // f"See full answer

Batman X. - "def friend_distance(friends, userA, userB): step = 0 total_neighs = set() llen = len(total_neighs) total_neighs.add(userB) while len(total_neighs)!=llen: s = set() step += 1 llen = len(total_neighs) for el in total_neighs: nes = neighbours(friends, userA, el) if userA in nes: return step for p in nes: s.add(p) for el in s: total_neighs.add(el) return -1 def neighbours(A,n1, n2): out = set() for i in range(len(A[n2])): if An2: out.add(i) return out"See full answer

Christian B. - "with cte as (select ts.employee_id, e.name, t.id as test_id, max(distinct ts.score) as total from test_results as ts join tests as t on ts.test_id = t.id join employees as e on ts.employee_id = e.id group by ts.employee_id, e.name, t.id) select employee_id, name as employee_name, sum(total) as total_score from cte group by employee_id, employee_name order by total_score desc, employee_id asc ;"See full answer

Divya R. - "public static Integer[] findLargest(int[] input, int m) { if(input==null || input.length==0) return null; PriorityQueue minHeap=new PriorityQueue(); for(int i:input) { if(minHeap.size()(int)top){ minHeap.poll(); minHeap.add(i); } } } Integer[] res=minHeap.toArray(new Integer[0]); Arrays.sort(res); return res; }"See full answer