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Bit Manipulation by Hand: ^ means XOR, ~ is a Not (Negation) - XOR: If it's a different bit, it will be 1. EX) 1101 ^ (~1101) = 1111 - If you XOR a bit with its own negated value, you will always get 1. Therefore, the solution to a^(~a) will be a sequence of 1s. EX) 1011 & (~0

Given two integer arrays arr1 and arr2, and the integer d, return the distance value between the two arrays. The distance value is defined as the number of elements arr1[i] such that there is not any element arr2[j] where |arr1[i]-arr2[j]| d=2 |4-9|=5 > d=2 |4-1|=3 > d=2 |4-8|=4 > d=2 For arr1[1]=5 we have: |5-10|=5 > d=2 |5-9|=4 > d=2 |5-1|=4 > d=2 |5-8|=3 > d=2 For arr1[2]=8 we have: |8-10|=2

Given a positive integer num, write a function which returns True if num is a perfect square else False. Follow up: Do not use any built-in library function such as sqrt. Example 1: Input: num = 16 Output: true Example 2: Input: num = 14 Output: false Constraints: 1

Let's call an array arr a mountain if the following properties hold: arr.length >= 3 There exists some i with 0 arr[i+1] > ... > arr[arr.length - 1] Given an integer array arr that is guaranteed to be a mountain, return any i such that arr[0] arr[i + 1] > ... > arr[arr.length - 1..

Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. You must write an algorithm with O(log n) runtime complexity. Example 1: Input: nums = [1,3,5,6], target = 5 Output: 2 Example 2: Input: nums = [1,3,5,6], target = 2 Output: 1 Example 3: Input: nums = [1,3,5,6], target = 7..

We are playing the Guess Game. The game is as follows: I pick a number from 1 to n. You have to guess which number I picked. Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess. You call a pre-defined API int guess(int num), which returns three possible results: -1: Your guess is higher than the number I picked (i.e. num > pick). 1: Your gue..

class Solution { public int search(int[] nums, int target) { int high = nums.length-1; int middle, low = 0; while (low

Given a binary tree, design an algorithm that creates a linked list of all the nodes at each depth (e.g., if you have a tree with depth D, you'll have D linked lists). void createLevelLinkedList(TreeNode root, ArrayList lists, int level){ if (root == null) return; //base case LinkedList list = null; if (lists.size() == level) { list = new LinkedList(); /** * Levels are always traversed in order...

4.1 Route Between Nodes: Given a directed graph, design an algorithm to find out whether there is a route between two nodes. This problem can be solved by just simple graph traversal, such as depth-first search or breadth-first search. We start with one of the two nodes and, during traversal, check if the other node is found. We should mark any node found in the course of the algorithm as "alrea..

The two most common ways to search a graph are depth-first search and breadth-first search. In depth-first search (DFS), we start at the root (or another arbitrarily selected node) and explore each branch completely before moving on to the next branch. That is, we go deep first (hence the name depth-first search) before we go wide. In breadth-first search (BFS), we start at the root (or another ..