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/**
 * @param {number[]} nums
 * @return {number}
 */
var minOperations = function(nums) {
    // Step 1: Quick checks for trivial cases
    // If the array already contains a 1, spreading it is possible.
    // If the gcd of the entire array > 1, it's impossible to ever reach 1.
    
    const gcd = (a, b) => {

tatsuyax25

/**
 * @param {string[]} strs
 * @param {number} m
 * @param {number} n
 * @return {number}
 */
var findMaxForm = function(strs, m, n) {
    // Step 1: Initialize a 2D DP array
    // dp[i][j] = max number of strings we can form with i zeros and j ones
    let dp = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));

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/**
 * Calculates the minimum number of operations needed to build a strictly increasing stack
 * from the input array, ignoring zeros and removing elements that break the order.
 *
 * @param {number[]} nums - Array of integers to process.
 * @return {number} - Minimum number of operations performed.
 */
var minOperations = function(nums) {
    const stack = []; // Stack to maintain increasing sequence
    let operations = 0; // Counter for operations performed

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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