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| /** | |
| * Calculates the maximum number of operations that can be performed | |
| * on a binary string `s`. An operation is defined as pairing a '1' | |
| * with the end of a block of consecutive '0's. | |
| * | |
| * @param {string} s - Input binary string consisting of '0' and '1' | |
| * @return {number} - Maximum number of operations possible | |
| */ | |
| var maxOperations = function (s) { | |
| const n = s.length; // Length of the string |
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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) => { |
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| /** | |
| * @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 = []; |