Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root). For example: Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its bottom-up level order traversal as: [ [15,7], [9,20], [3] ] Solution # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.
# SQL Schema: Create table If Not Exists Employee (Id int, Name varchar(255), Salary int, ManagerId int) Truncate table Employee insert into Employee (Id, Name, Salary, ManagerId) values ('1', 'Joe', '70000', '3') insert into Employee (Id, Name, Salary, ManagerId) values ('2', 'Henry', '80000', '4') insert into Employee (Id, Name, Salary, ManagerId) values ('3', 'Sam', '60000', 'None') insert into Employee (Id, Name, Salary, ManagerId) values ('4', 'Max', '90000', 'None') The Employee table holds all employees including their managers.
Write a SQL query to find all duplicate emails in a table named Person. +----+---------+ | Id | Email | +----+---------+ | 1 | a@b.com | | 2 | c@d.com | | 3 | a@b.com | +----+---------+ For example, your query should return the following for the above table: +---------+ | Email | +---------+ | a@b.com | +---------+ Note: All emails are in lowercase. Solution # Write your MySQL query statement below SELECT email FROM Person WHERE email IN ( SELECT email FROM Person GROUP BY email HAVING COUNT(*) > 1 ) GROUP BY email
Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers. Note: All numbers will be positive integers. The solution set must not contain duplicate combinations. Example 1 Input: k = 3, n = 7 Output: [[1,2,4]] Example 2 Input: k = 3, n = 9 Output: [[1,2,6], [1,3,5], [2,3,4]] Solution class Solution: def combinationSum3(self, k: int, n: int) -> List[List[int]]: # Equivalent to subsets results = [] combination = [] # Start DFS self.
Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. The same repeated number may be chosen from candidates unlimited number of times. Note All numbers (including target) will be positive integers. The solution set must not contain duplicate combinations. Example 1 Input: candidates = [2,3,6,7], target = 7, A solution set is: [ [7], [2,2,3] ] Example 2 Input: candidates = [2,3,5], target = 8, A solution set is: [ [2,2,2,2], [2,3,3], [3,5] ] Solution class Solution: def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]: # Equivalent to subsets results = [] combination = [] # Edge cases if candidates == None: return results if len(candidates) == 0: results.
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n. Example Input: n = 4, k = 2 Output: [ [2,4], [3,4], [2,3], [1,2], [1,3], [1,4], ] Solution class Solution: def combine(self, n: int, k: int) -> List[List[int]]: def backtrack(start = 1, curr = []): # if the combination is done if len(curr) == k: output.append(curr[:]) return for i in range(start, n + 1): # add i into the current combination curr.
Table: Person +-------------+---------+ | Column Name | Type | +-------------+---------+ | PersonId | int | | FirstName | varchar | | LastName | varchar | +-------------+---------+ PersonId is the primary key column for this table. Table: Address +-------------+---------+ | Column Name | Type | +-------------+---------+ | AddressId | int | | PersonId | int | | City | varchar | | State | varchar | +-------------+---------+ AddressId is the primary key column for this table.
Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct. Example 1: Input: [1,2,3,1] Output: true Example 2: Input: [1,2,3,4] Output: false Example 3: Input: [1,1,1,3,3,4,3,2,4,2] Output: true Solution class Solution: def containsDuplicate(self, nums): """ :type nums: List[int] :rtype: bool """ sortedNums = sorted(nums) for i in range(1, len(nums)): if sortedNums[i] == sortedNums[i-1]: return True return False
Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST. Note next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree. Solution # Definition for a binary tree node # class TreeNode(object): # def __init__(self, x): # self.
Say you have an array for which the ith element is the price of a given stock on day i. Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times). Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).