Coding Interview Questions

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

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

Anonymous Unicorn - "python: def justifywords(wordslist, width): result = [] currlinechar_count = 0 curr_words = [] for word in words_list: if curr_words: space_needed = len(word) + 1 # Space needed for the word and a preceding space else: space_needed = len(word) if currlinecharcount + spaceneeded > width: result.append(' '.join(curr_words)) curr_words = [word] currlinechar_count = len("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

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

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

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

Stanley Y. - "First of all, stack and heap memory are abstraction on top of the hardware by the compiler. The hardware is not aware of stack and heap memory. There is only a single piece of memory that a program has access to. The compiler creates the concepts of stack and heap memory to run the programs efficiently. Programs use stack memory to store local variables and a few important register values such as frame pointer and return address for program counter. This makes it easier for the compiler to gene"See full answer

Erjan G. - "1) select avg(session) from table where session> 180 2) select round(sessiontime/300)*300 as sessionbin, count() as sessioncount from table group by round(sessiontime/300)300 order by session_bin 3) SELECT t1.country AS country_a, t2.country AS country_b FROM ( SELECT country, COUNT(*) AS session_count FROM yourtablename GROUP BY country ) AS t1 JOIN ( SELECT country, COUNT(*) AS session_count FROM yourtablename `GROUP BY countr"See full answer

Dinar M. - "Even more faster and vectorized version, using np.linalg.norm - to avoid loop and np.argpartition to select lowest k. We dont need to sort whole array - we need to be sure that first k elements are lower than the rest. import numpy as np def knn(Xtrain, ytrain, X_new, k): distances = np.linalg.norm(Xtrain - Xnew, axis=1) k_indices = np.argpartition(distances, k)[:k] # O(N) selection instead of O(N log N) sort return int(np.sum(ytrain[kindices]) > k / 2.0) `"See full answer

K.nithish K. - "MOD = 10**9 + 7 def max_stability(reliability, availability): max_stability = 1 for r, a in zip(reliability, availability): Compute stability of the current server stability = r * a if stability != 0: Multiply into max_stability and take modulo maxstability = (maxstability * stability) % MOD return max_stability reliability = [1, 2, 2] availability = [1, 1, 3] print(max_stability(reliability, availability)) # Output the result mo"See full answer

Zacharias E. - "SELECT pro.id, pro.title, pro.budget, COUNT(employeeid) AS numemployees, SUM(e.salary) as total_salaries FROM projects pro JOIN employeesprojects ep ON ep.projectid = pro.id JOIN employees e ON e.id = ep.employee_id GROUP BY project_id; `"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

Nathan C. - "def mostefficientseqscore(parentheses, efficiencyratings): mes = [] for i in range(len(parentheses)): mes.append((parentheses[i], max(efficiency_ratings[i])) return sum([m[1] for m in mes]) `"See full answer

Vidhyadhar V. - "public class CircularBuffer { private T[] buffer; private int head; private int tail; private int size; private final int capacity; public CircularBuffer(int capacity) { this.capacity = capacity; this.buffer = (T[]) new Object[capacity]; this.head = 0; this.tail = 0; this.size = 0; } public void enqueue(T item) { if (isFull()) { throw new IllegalStateException("Buffer is full"); } buf"See full answer