Coding Interview Questions

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

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

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

Jeremy D. - "Any cycle would cause the prerequisite to be greater than the course. This passes all the tests: function canFinish(_numCourses, prerequisites) { for (const [a, b] of prerequisites) { if (b > a) return false } return true } `"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

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

Dinar M. - "I used array to append. np.array does not have append and every np.vstack recreates object again. 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=[] # I"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

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

Sammy R. - "In python def find_duplicates(arr1: List[int], arr2: List[int]) -> List[int]: result = list(set(arr1) & set(arr2)) return result "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

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

Tiago R. - "function serialize(list) { for (let i=0; i 0xFFFF) { throw new Exception(String ${list[i]} is too long!); } const prefix = String.fromCharCode(length); list[i] = ${prefix}${list[i]}; console.log(list[i]) } return list.join(''); } function deserialize(s) { let i=0; const length = s.length; const output = []; while (i < length) { "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

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

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

Victor N. - "with max_score as( select test_id, employee_id, max(score) as max_score from test_results as t1 join tests as t2 on t1.test_id=t2.id group by 2,1 ) select id as employee_id, name as employee_name, sum(maxscore) as totalscore from employees as e join max_score as m on e.id=m.employee_id group by 1,2 order by 3 desc, 1 `"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

Laksitha R. - "public static int maxProfitGreedy(int[] stockPrices) { int maxProfit = 0; for(int i = 1; i todayPrice) { maxProfit += tomorrowPrice - todayPrice; } } return maxProfit; } "See full answer