Another Casual Coder

One of my sons asked an interesting question: what number raised to itself equals to 25? In other words, he wanted to find X such that X^X = 25. I decided to write a simple program (no GAI for a change, just vintage coding) to do that. The key observations here are: 1/ X^X is continuous in the real domain 2/ It grows monotonically  3/ It grows very fast With that in mind, the solution is a binary search. The lower bound is 1, the upper bound can be found quickly by doubling the number, then we perform a standard binary search (some nuances there just because we're dealing with floating-point numbers). Complexity stays around Log(Log(N)), hence very fast. Code is down below, cheers. Oh... and the answer to my son is the following: 2.96321964263916 using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace Exp { internal class Program { static void Main(string[] args) { double...

Solution here is a simple sliding window approach, the caveat is to cache the mapping number -> list of unique factors which you can use across test cases. Reasonably fast prime factorization is important too, mine isn't the most optimal but fast enough to pass all test cases. Code is down below, cheers, ACC. Longest Subarray With at Most K Distinct Prime Factors - LeetCode You are given an integer array nums consisting of positive integers and an integer k . The prime factor set of a subarray is the union of the distinct prime factors of all its elements. Return the length of the longest subarray whose prime factor set contains at most k distinct prime factors. If no such subarray exists, return 0.   Example 1: Input: nums = [7,6,10,12,11], k = 3 Output: 3 Explanation: Consider the subarray [6, 10, 12] : The distinct prime factors of 6 are {2, 3} . The distinct prime factors of 10 are {2, 5} . The distinct prime factors of 12 are {2, 3} . The union of these sets is {2...