Code Crafters

 Understanding the Problem In this LeetCode problem, we're tasked with finding the "power" of all subarrays of size K within a given array of integers. The power of a subarray is defined as: Maximum element if all elements are consecutive and sorted in ascending order. -1 otherwise. Brute Force Approach: A Simple Start A straightforward approach involves iterating over every possible subarray of size K, checking its validity, and calculating its power. Here's Java code for the same: class Solution { public int[] resultsArray(int[] nums, int k) { int n = nums.length; int[] ans = new int[n - k + 1]; for(int i = 0; i <= n-k; i++) { int f = 0; for(int j = i; j < i+k; j++) { // cur element = prev element + 1 if(j > i && nums[j] != nums[j-1] + 1) { f = 1; // Violation of the rule was found bre...

Understanding Arithmetic Progression (AP) An arithmetic progression (AP) is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant difference is known as the common difference (d).   For example, the sequence 2, 5, 8, 11, ... is an AP with a common difference of 3. Deriving the Formula Let's consider an AP with the first term a_1 and a common difference d . The nth term of this AP is given by: a_n = a_1 + (n-1)d Now, let's denote the sum of the first n terms of this AP as S_n . We can write S_n as: S_n = a_1 + (a_1 + d) + ... + (a_1 + (n-2)d) + (a_1 + (n-1)d) We can also write S_n in reverse order: S_n = (a_1 + (n-1)d) + (a_1 + (n-2)d) + ... + (a_1 + d) + a_1 Adding these two equations, we get: 2S_n = (2a_1 + (n-1)d) + (2a_1 + (n-1)d ) + ... + (2a_1 + (n-1)d) (n times) Simplifying: 2S_n = n * (2a_1 + (n-1)d) Dividing both sides by 2: S_n = (n/2) * [2a_1 + (n-1)d] This is the formula for the sum of the first n terms of ...

 Hey there, problem solvers! Today we're tackling Problem A from AtCoder Beginner Contest 43: Children and Candies. This problem might seem straightforward at first glance, but it offers an opportunity to explore two solutions with different time complexities. Let's dive in! The Problem Statement Mr. Evi at AtCoder Kindergarten is a generous soul. He's lining up N children and giving them candies based on their position. The first child receives 1 candy, the second 2 candies, and so on, until the Nth child receives N candies. Your task is to find the total number of candies Mr. Evi will need. The problem also specifies that N can range from 1 to 100. Solution 1: Brute Force Loop (Linear Time) This is a classic approach for problems where you need to iterate through a sequence and perform a calculation on each element. Here's how it works: Looping Through Children: We'll use a loop that iterates from 1 to N (number of children). Calculating Candies: Inside the lo...