Stack

Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (push, top, pop, and empty). Implement the MyStack class: void push(int x) Pushes element x to the top of the stack. int pop() Removes the element on the top of the stack and returns it. int top() Returns the element on the top of the stack. boolean empty() Returns true if the stack is empty, false otherwise. Notes: You must use only standard operations of a queue, which means that only push to back, peek/pop from front, size and is empty operations are valid. Depending on your language, the queue may not be supported natively.

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty). Implement the MyQueue class: void push(int x) Pushes element x to the back of the queue. int pop() Removes the element from the front of the queue and returns it. int peek() Returns the element at the front of the queue. boolean empty() Returns true if the queue is empty, false otherwise. Notes: You must use only standard operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid. Depending on your language, the stack may not be supported natively.

Design your implementation of the circular queue. The circular queue is a linear data structure in which the operations are performed based on FIFO (First In First Out) principle, and the last position is connected back to the first position to make a circle. It is also called "Ring Buffer ". One of the benefits of the circular queue is that we can make use of the spaces in front of the queue. In a normal queue, once the queue becomes full, we cannot insert the next element even if there is a space in front of the queue. But using the circular queue, we can use the space to store new values. Implement the MyCircularQueue class: MyCircularQueue(k) Initializes the object with the size of the queue to be k. int Front() Gets the front item from the queue.

Design your implementation of the circular double-ended queue (deque). Implement the MyCircularDeque class: MyCircularDeque(int k) Initializes the deque with a maximum size of k. boolean insertFront() Adds an item at the front of Deque. Returns true if the operation is successful, or false otherwise. boolean insertLast() Adds an item at the rear of Deque. Returns true if the operation is successful, or false otherwise. boolean deleteFront() Deletes an item from the front of Deque. Returns true if the operation is successful, or false otherwise. boolean deleteLast() Deletes an item from the rear of Deque. Returns true if the operation is successful, or false otherwise. int getFront() Returns the front item from the Deque. Returns -1 if the deque is empty.

Design a max stack data structure that supports the stack operations and supports finding the stack's maximum element. Implement the MaxStack class: MaxStack() Initializes the stack object. void push(int x) Pushes element x onto the stack. int pop() Removes the element on top of the stack and returns it. int top() Gets the element on the top of the stack without removing it. int peekMax() Retrieves the maximum element in the stack without removing it. int popMax() Retrieves the maximum element in the stack and removes it. If there is more than one maximum element, only remove the top-most one. You must come up with a solution that supports O(1) for each top call and O(logn) for each other call.

We stack glasses in a pyramid, where the first row has 1 glass, the second row has 2 glasses, and so on until the 100th row. Each glass holds one cup of champagne. Then, some champagne is poured into the first glass at the top. When the topmost glass is full, any excess liquid poured will fall equally to the glass immediately to the left and right of it. When those glasses become full, any excess champagne will fall equally to the left and right of those glasses, and so on. (A glass at the bottom row has its excess champagne fall on the floor.) For example, after one cup of champagne is poured, the top most glass is full. After two cups of champagne are poured, the two glasses on the second row are half full.

Given a single positive integer x, we will write an expression of the form x (op1) x (op2) x (op3) x ... where each operator op1, op2, etc. is either addition, subtraction, multiplication, or division (+, -, , or /). For example, with x = 3, we might write 3 3 / 3 + 3 - 3 which is a value of 3. When writing such an expression, we adhere to the following conventions: The division operator (/) returns rational numbers. There are no parentheses placed anywhere. We use the usual order of operations: multiplication and division happen before addition and subtraction. It is not allowed to use the unary negation operator (-). For example, "x - x " is a valid expression as it only uses subtraction, but "-x + x " is not because it uses negation.

You have an infinite number of stacks arranged in a row and numbered (left to right) from 0, each of the stacks has the same maximum capacity. Implement the DinnerPlates class: DinnerPlates(int capacity) Initializes the object with the maximum capacity of the stacks capacity. void push(int val) Pushes the given integer val into the leftmost stack with a size less than capacity. int pop() Returns the value at the top of the rightmost non-empty stack and removes it from that stack, and returns -1 if all the stacks are empty. int popAtStack(int index) Returns the value at the top of the stack with the given index index and removes it from that stack or returns -1 if the stack with that given index is empty.

In an infinite chess board with coordinates from -infinity to +infinity, you have a knight at square [0, 0]. A knight has 8 possible moves it can make, as illustrated below. Each move is two squares in a cardinal direction, then one square in an orthogonal direction. Return the minimum number of steps needed to move the knight to the square [x, y]. It is guaranteed the answer exists.

Design a stack that supports increment operations on its elements. Implement the CustomStack class: CustomStack(int maxSize) Initializes the object with maxSize which is the maximum number of elements in the stack. void push(int x) Adds x to the top of the stack if the stack has not reached the maxSize. int pop() Pops and returns the top of the stack or -1 if the stack is empty. void inc(int k, int val) Increments the bottom k elements of the stack by val. If there are less than k elements in the stack, increment all the elements in the stack.

Design a queue that supports push and pop operations in the front, middle, and back. Implement the FrontMiddleBack class: FrontMiddleBack() Initializes the queue. void pushFront(int val) Adds val to the front of the queue. void pushMiddle(int val) Adds val to the middle of the queue. void pushBack(int val) Adds val to the back of the queue. int popFront() Removes the front element of the queue and returns it. If the queue is empty, return -1. int popMiddle() Removes the middle element of the queue and returns it. If the queue is empty, return -1. int popBack() Removes the back element of the queue and returns it. If the queue is empty, return -1.

Design a queue-like data structure that moves the most recently used element to the end of the queue. Implement the MRUQueue class: MRUQueue(int n) constructs the MRUQueue with n elements: [1,2,3,...,n]. int fetch(int k) moves the kth element (1-indexed) to the end of the queue and returns it.

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. Implement the MinStack class: MinStack() initializes the stack object. void push(int val) pushes the element val onto the stack. void pop() removes the element on the top of the stack. int top() gets the top element of the stack. int getMin() retrieves the minimum element in the stack. You must implement a solution with O(1) time complexity for each function.

You are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation. Evaluate the expression. Return an integer that represents the value of the expression. Note that: The valid operators are '+', '-', '', and '/'. Each operand may be an integer or another expression. The division between two integers always truncates toward zero. There will not be any division by zero. The input represents a valid arithmetic expression in a reverse polish notation. The answer and all the intermediate calculations can be represented in a 32-bit integer.

Given an array of integers temperatures represents the daily temperatures, return an array answer such that answer[i] is the number of days you have to wait after the ith day to get a warmer temperature. If there is no future day for which this is possible, keep answer[i] == 0 instead.

There are n cars going to the same destination along a one-lane road. The destination is target miles away. You are given two integer array position and speed, both of length n, where position[i] is the position of the ith car and speed[i] is the speed of the ith car (in miles per hour). A car can never pass another car ahead of it, but it can catch up to it and drive bumper to bumper at the same speed. The faster car will slow down to match the slower car's speed. The distance between these two cars is ignored (i.e., they are assumed to have the same position). A car fleet is some non-empty set of cars driving at the same position and same speed. Note that a single car is also a car fleet.

Given an array of integers heights representing the histogram's bar height where the width of each bar is 1, return the area of the largest rectangle in the histogram.