Write a program that outputs the string representation of numbers from 1 to n, however: If the number is divisible by 3, output "fizz". If the number is divisible by 5, output "buzz". If the number is divisible by both 3 and 5, output "fizzbuzz". For example, for n = 15, we output: 1, 2, fizz, 4, buzz, fizz, 7, 8, fizz, buzz, 11, fizz, 13, 14, fizzbuzz.
Suppose we have a class: public class Foo { public void first() { print("first"); } public void second() { print("second"); } public void third() { print("third"); } } The same instance of Foo will be passed to three different threads. Thread A will call first(), thread B will call second(), and thread C will call third(). Design a mechanism and modify the program to ensure that second() is executed after first(), and third() is executed after second().
Five silent philosophers sit at a round table with bowls of spaghetti. Forks are placed between each pair of adjacent philosophers. Each philosopher must alternately think and eat. However, a philosopher can only eat spaghetti when they have both left and right forks. Each fork can be held by only one philosopher and so a philosopher can use the fork only if it is not being used by another philosopher. After an individual philosopher finishes eating, they need to put down both forks so that the forks become available to others.
You are given an even number of people num_people that stand around a circle and each person shakes hands with someone else, so that there are num_people / 2 handshakes total. Return the number of ways these handshakes could occur such that none of the handshakes cross. Since this number could be very big, return the answer mod 10^9 + 7 Example 1 Input: num_people = 2 Output: 1 Example 2 Input: num_people = 4 Output: 2 Explanation: There are two ways to do it, the first way is [(1,2),(3,4)] and the second one is [(2,3),(4,1)].
Given a binary tree root. Split the binary tree into two subtrees by removing 1 edge such that the product of the sums of the subtrees are maximized. Since the answer may be too large, return it modulo 10e9 + 7. Example 1 Input: root = [1,2,3,4,5,6] Output: 110 Explanation: Remove the red edge and get 2 binary trees with sum 11 and 10. Their product is 110 (11*10) Example 2 Input: root = [1,null,2,3,4,null,null,5,6] Output: 90 Explanation: Remove the red edge and get 2 binary trees with sum 15 and 6.
You are given a string s, and an array of pairs of indices in the string pairs where pairs[i] = [a, b] indicates 2 indices(0-indexed) of the string. You can swap the characters at any pair of indices in the given pairs any number of times. Return the lexicographically smallest string that s can be changed to after using the swaps. Example 1 Input: s = "dcab", pairs = [[0,3],[1,2]] Output: "bacd" Explaination: Swap s[0] and s[3], s = "bcad" Swap s[1] and s[2], s = "bacd" Example 2 Input: s = "dcab", pairs = [[0,3],[1,2],[0,2]] Output: "abcd" Explaination: Swap s[0] and s[3], s = "bcad" Swap s[0] and s[2], s = "acbd" Swap s[1] and s[2], s = "abcd" Example 3 Input: s = "cba", pairs = [[0,1],[1,2]] Output: "abc" Explaination: Swap s[0] and s[1], s = "bca" Swap s[1] and s[2], s = "bac" Swap s[0] and s[1], s = "abc" Constraints 1 <= s.
Design and implement a data structure for Least Frequently Used (LFU) cache. It should support the following operations: get and put. get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1. put(key, value) - Set or insert the value if the key is not already present. When the cache reaches its capacity, it should invalidate the least frequently used item before inserting a new item.
Given an Android 3x3 key lock screen and two integers m and n, where 1 ≤ m ≤ n ≤ 9, count the total number of unlock patterns of the Android lock screen, which consist of minimum of m keys and maximum n keys. Rules for a valid pattern: Each pattern must connect at least m keys and at most n keys. All the keys must be distinct. If the line connecting two consecutive keys in the pattern passes through any other keys, the other keys must have previously selected in the pattern.
Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive. Example 1 Input: [5,7] Output: 4 Example 2 Input: [0,1] Output: 0 Solution «««< HEAD Java class Solution { public int rangeBitwiseAnd(int m, int n) { int shift = 0; while (m < n) { m >>= 1; n >>= 1; shift++; // System.out.println("m: " + String.
You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values 0, 1 or 2, where: Each 0 marks an empty land which you can pass by freely. Each 1 marks a building which you cannot pass through. Each 2 marks an obstacle which you cannot pass through.