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/**
 * @param {number} num1
 * @param {number} num2
 * @return {number}
 */
var countOperations = function(num1, num2) {
    // initialize a counter to keep track of the number of operations
    let operations = 0;

    // Continue looping until either num1 or num2 becomes 0

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/**
 * Calculates the minimum number of operations to reduce a number `n` to 0
 * using a specific bitwise operation rule:
 * - You can flip the lowest 1-bit and all bits to its right.
 *
 * This function uses a recursive approach based on the properties of Gray code.
 *
 * @param {number} n - The input number.
 * @return {number} - Minimum number of operations to reduce `n` to 0.
 */

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/**
 * @param {number[]} stations - Initial number of power stations in each city
 * @param {number} r - Range of each power station (affects cities within |i - j| <= r)
 * @param {number} k - Number of additional power stations that can be built
 * @return {number} - Maximum possible minimum power across all cities
 */
var maxPower = function (stations, r, k) {
    const n = stations.length;

    // Step 1: Build prefix sum array to quickly compute power in a range

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/**
 * Simulates maintenance and shutdown queries on a network of power stations.
 *
 * @param {number} c - Number of power stations (1-indexed).
 * @param {number[][]} connections - Bidirectional cables between stations.
 * @param {number[][]} queries - List of queries: [1, x] for maintenance, [2, x] to shut down.
 * @return {number[]} - Results of maintenance queries.
 */
var processQueries = function(c, connections, queries) {
  // ----- Union-Find (Disjoint Set Union) Setup -----

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/**
 * @param {number[]} nums - Input array of numbers
 * @param {number} k - Size of the sliding window
 * @param {number} x - Number of top elements to sum (based on freq × value)
 * @return {number[]} - Array of sums for each window
 */
var findXSum = function(nums, k, x) {
  const n = nums.length;

  // Shortcut: if k === x, just sum the window directly

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/**
 * Computes a sliding window sum based on the x most frequent elements.
 * 
 * @param {number[]} nums - The input array of numbers.
 * @param {number} k - The size of the sliding window.
 * @param {number} x - The number of top frequent elements to include in the sum.
 * @return {number[]} - An array of sums for each window.
 */
var findXSum = function(nums, k, x) {
    const result = [];

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/**
 * Removes balloons to ensure no two adjacent balloons have the same color,
 * minimizing the total removal time.
 *
 * @param {string} colors - A string where each character represents a balloon's color.
 * @param {number[]} neededTime - An array where each element is the time to remove the corresponding balloon.
 * @return {number} - The minimum total time to remove balloons to satisfy the condition.
 */
var minCost = function(colors, neededTime) {
    let totalTime = 0; // Total time to remove balloons

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/**
 * Finds the two "sneaky" numbers added to an array that originally contained all numbers from 0 to n-1.
 * The final array has length n + 2, and includes two unknown extra numbers.
 *
 * @param {number[]} nums - Array containing numbers from 0 to n-1 plus two extra unknowns.
 * @return {number[]} - The two extra numbers that were added.
 */
var getSneakyNumbers = function (nums) {
    const n = nums.length - 2; // Original range was 0 to n-1, so n = nums.length - 2
    let xorAll = 0;

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/**
 * @param {number[]} target
 * @return {number}
 */
var minNumberOperations = function(target) {
    // Initialize the total number of operations needed
    let operations = 0;

    // Loop through the target array
    for (let i = 0; i < target.length; i++) {

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/**
 * @param {number} n
 * @return {number}
 */
var smallestNumber = function(n) {
    // Start with 1 and keep shifting left and OR-ing to build numbers like 1, 3, 7, 15, 31, etc.
    let x = 1;

    // Keep generating numbers with all bits set until we find one >= n
    while (x < n) {