Coding Interview Questions

TreeOfWisdom - "Should this question be BST, not just BT? Otherwise it would not be possible to reconstruct the tree solely based on the array regardless of its order"See full answer

Astha A. - "-- Write your query here with a as (select customerid, customername, rownumber() over (order by customerid) as rindex from customers) select customerid, customername from a where rindex % 3 = 0;"See full answer

Nitin P. - "public static void sortBinaryArray(int[] array) { int len = array.length; int[] res = new int[len]; int r=len-1; for (int value : array) { if(value==1){ res[r]= 1; r--; } } System.out.println(Arrays.toString(res)); } `"See full answer

Rafał P. - "def changeString(org: str,target:str) -> bool: lOrg = len(org) lTarget = len(target) \# They have to be equal in lenght if lOrg != lTarget: return False counter1 = Counter(org) counter2 = Counter(target) \# Counter internally iterates through the input sequence, counts the number of times a given object occurs, and stores objects as keys and the counts as values. if counter1 != counter2: return False diff = sum(org[i] != target[i] for i in range(n)) return diff == 2 or (diff == 0 and any(v > 1 f"See full answer

Yash N. - "import time class Task: def init\(self, description, interval=None): self.description = description self.interval = interval self.next_run = time.time() class SimpleTaskScheduler: def init\(self): self.tasks = [] def add_task(self, description, interval=None): self.tasks.append(Task(description, interval)) def run(self, duration=60): end_time = time.time() + duration while time.time() < end_time: curr"See full answer

Arturo Z. - "function insertItem(heap, item) { heap.push(item); if (heap.length > 0) { let current = heap.length - 1; while(current > 0 && heap[Math.floor(current/2)] > heap[current]) { [heap[Math.floor(current/2)], heap[current]] = [heap[current], heap[Math.floor(current/2)]]; current = Math.floor(current/2); } } } function extractMin(heap) { let smallest = heap[0]; let len = heap.length; if (len > 2) { heap[0] = heap[len - 1]; heap.splice(len - 1); "See full answer

Robert W. - "def check_byte(octet): _""" Checks if the given string \octet\ represents a valid byte (0-255). """_ Check for empty string if not octet: return False Check if the string has non-digit characters if not octet.isdigit(): return False Check for leading zeroes in multi-digit numbers if len(octet) > 1 and octet[0] == '0': return False Check if the integer value is between 0 and 255 return 0 <= int(octet) <= 255 def va"See full answer

Jaime A. - "I might be missing something but the solution, seems to be incorrect. ... , post_pairings AS ( SELECT ps.user_id, ps.postseqid AS failpostid, ps.postseqid + 1 AS nextpostid FROM post_seq AS ps WHERE ps.issuccessfulpost IS TRUE ) -- here ps.issuccessfulpost IS TRUE the condition should be FALSE -- in that way ps.postseqid is the actual failed post(failpostid) -- Additionally, at the end the join is assumming that the sequence id is going to match the post_id, wh"See full answer

Rahul J. - "find total sum. assign that to rightsum traverse from left to right: keep updating left sum and right sum, when they match return the index. else if you reach end return -1 or not found"See full answer

Guilherme F. - "A much better solution than the one in the article, below: It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods. shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns. My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N)) class ListNode { constructor(val = 0, next = null) { th"See full answer

Richard B. - "Select interface, Count(case when issuccessfulpost then 1 end) as post_success, Count() as postattempt, ROUND((COUNT(CASE WHEN issuccessfulpost THEN 1 END) * 100 / COUNT()), 2) AS postsuccess_rate from post where interface like 'Iphone%' group by 1 order by postsuccessrate desc `"See full answer

Lin R. - "Why can not group by only on name? it gave me incorrect results when I try to do that."See full answer

Rick E. - " O(n) time, O(1) space from typing import List def maxsubarraysum(nums: List[int]) -> int: if len(nums) == 0: return 0 maxsum = currsum = nums[0] for i in range(1, len(nums)): currsum = max(currsum + nums[i], nums[i]) maxsum = max(currsum, max_sum) return max_sum debug your code below print(maxsubarraysum([-1, 2, -3, 4])) `"See full answer

Tiago R. - "function areSentencesSimilar(sentence1, sentence2, similarPairs) { if (sentence1.length !== sentence2.length) return false; for (let i=0; i (w1 === word1 && !visited.has(w2)) || (w2 === word1 && !visited.has(w1))); if (!edge) { "See full answer

Rishi G. - "Problem Statement: The Fibonacci sequence is defined as F(n) = F(n-1) + F(n-2) with F(0) = 1 and F(1) = 1. The solution is given in the problem statement itself. If the value of n = 0, return 1. If the value of n = 1, return 1. Otherwise, return the sum of data at (n - 1) and (n - 2). Explanation: The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding ones, typically starting with 0 and 1. Java Solution: public static int fib(int n"See full answer

Anonymous Unicorn - "Count items between indices within compartments compartments are delineated by by: '|' items are identified by: '*' input_inventory = "*||||" inputstartidxs = [1, 4, 6] inputendidxs = [9, 5, 8] expected_output = [3, 0, 1] Explanation: "*||||" 0123456789... indices ++ + # within compartments ^ start_idx = 1 ^ end_idx = 9 -- - # within idxs but not within compartments "*||||" 0123456789... indices "See full answer