Stack

On a single-threaded CPU, we execute a program containing n functions. Each function has a unique ID between 0 and n-1. Function calls are stored in a call stack: when a function call starts, its ID is pushed onto the stack, and when a function call ends, its ID is popped off the stack. The function whose ID is at the top of the stack is the current function being executed. Each time a function starts or ends, we write a log with the ID, whether it started or ended, and the timestamp. ...

Given two binary search trees root1 and root2. Return a list containing all the integers from both trees sorted in ascending order. Example 1: Input: root1 = [2,1,4], root2 = [1,0,3] Output: [0,1,1,2,3,4] Example 2: Input: root1 = [0,-10,10], root2 = [5,1,7,0,2] Output: [-10,0,0,1,2,5,7,10] Example 3: Input: root1 = [], root2 = [5,1,7,0,2] Output: [0,1,2,5,7] Example 4: Input: root1 = [0,-10,10], root2 = [] Output: [-10,0,10] Example 5: Input: root1 = [1,null,8], root2 = [8,1] Output: [1,1,8,8] ...

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST): BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST. boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false. int next() Moves the pointer to the right, then returns the number at the pointer. Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST. ...

71. Simplify Path Given a string path, which is an absolute path (starting with a slash ‘/’) to a file or directory in a Unix-style file system, convert it to the simplified canonical path. In a Unix-style file system, a period ‘.’ refers to the current directory, a double period ‘..’ refers to the directory up a level, and any multiple consecutive slashes (i.e. ‘//’) are treated as a single slash ‘/’. For this problem, any other format of periods such as ‘…’ are treated as file/directory names. ...

496. Next Greater Element I The next greater element of some element x in an array is the first greater element that is to the right of x in the same array. You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2. For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1. ...

503. Next Greater Element II Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums. The next greater number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn’t exist, return -1 for this number. ...

1441. Build an Array With Stack Operations Given an array target and an integer n. In each iteration, you will read a number from list = {1,2,3…, n}. Build the target array using the following operations: Push: Read a new element from the beginning list, and push it in the array. Pop: delete the last element of the array. If the target array is already built, stop reading more elements. Return the operations to build the target array. You are guaranteed that the answer is unique. ...

199. Binary Tree Right Side View Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. Example 1: Input: root = [1,2,3,null,5,null,4] Output: [1,3,4] Example 2: Input: root = [1,null,3] Output: [1,3] Example 3: Input: root = [] Output: [] Constraints: The number of nodes in the tree is in the range [0, 100]. -100 <= Node.val <= 100 Recursive ...

![https://leetcode.com/problems/evaluate-reverse-polish-notation/] Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Note: Division between two integers should truncate toward zero. The given RPN expression is always valid. That means the expression would always evaluate to a result and there won’t be any divide by zero operation. Example 1: Input: ["2", "1", "+", "3", "*"] Output: 9 Explanation: ((2 + 1) * 3) = 9 Example 2: Input: ["4", "13", "5", "/", "+"] Output: 6 Explanation: (4 + (13 / 5)) = 6 Example 3: Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"] Output: 22 Explanation: ((10 * (6 / ((9 + 3) * -11))) + 17) + 5 = ((10 * (6 / (12 * -11))) + 17) + 5 = ((10 * (6 / -132)) + 17) + 5 = ((10 * 0) + 17) + 5 = (0 + 17) + 5 = 17 + 5 = 22 class Solution { public int evalRPN(String[] tokens) { int[] stack = new int[tokens.length]; int size = 0; for (String token: tokens) { if (token.equals("+") || token.equals("-") || token.equals("*") || token.equals("/") ) { int b = stack[--size]; int a = stack[--size]; int c; switch (token) { case "+": c = a + b; break; case "-": c = a - b; break; case "*": c = a * b; break; case "/": c = a / b; break; default: throw new RuntimeException(); } stack[size++] = c; } else { stack[size++] = Integer.valueOf(token); } } return stack[0]; } }

![https://leetcode.com/problems/crawler-log-folder/] The Leetcode file system keeps a log each time some user performs a change folder operation. The operations are described below: “../” : Move to the parent folder of the current folder. (If you are already in the main folder, remain in the same folder). “./” : Remain in the same folder. “x/” : Move to the child folder named x (This folder is guaranteed to always exist). You are given a list of strings logs where logs[i] is the operation performed by the user at the ith step. ...