3633. Earliest Finish Time for Land and Water Rides I

You are given two categories of theme park attractions: land rides and water rides. Land rides landStartTime[i] – the earliest time the ith land ride can be boarded. landDuration[i] – how long the ith land ride lasts. Water rides waterStartTime[j] – the earliest time the jth water ride can be boarded. waterDuration[j] – how long the jth water ride lasts. A tourist must experience exactly one ride from each category, in either order. A ride may be started at its opening time or any later moment. If a ride is started at time t, it finishes at time t + duration. Immediately after finishing one ride the tourist may board the other (if it is already open) or wait until it opens. Return the earliest possible time at which the tourist can finish both rides.
/**
 * @param {number[]} landStartTime  – earliest start times for land rides
 * @param {number[]} landDuration   – durations for land rides
 * @param {number[]} waterStartTime – earliest start times for water rides
 * @param {number[]} waterDuration  – durations for water rides
 * @return {number}                 – earliest possible finish time
 */
var earliestFinishTime = function(landStartTime, landDuration, waterStartTime, waterDuration) {
    let ans = Infinity; // Track the minimum finish 

Diagrams drawing tool

https://mermaid.ai/web/

claudecode代码审查

# Code Modification & Mandatory Review Workflow

For EVERY substantial code modification in this project, you **MUST** strictly follow the independent review process below. **DO NOT SKIP THIS UNDER ANY CIRCUMSTANCES.**

1. **No Immediate Completion**: Do NOT report the task as complete immediately after writing/modifying code.
2. **Spawn a Review Subagent**: You must immediately spawn an independent Subagent.
3. **Assign the Persona & Prompt**: Pass the following exact prompt to the Subage

animate elements - js inject

//add team animations
jQuery('h1,h2').addClass('wow animate__animated animate__fadeInUp');
jQuery('h3').addClass('wow animate__animated animate__fadeInUp second-animate');
// jQuery('.block--leadership-team .team-member:nth-of-type(2)').addClass('second-animate');
// jQuery('.block--leadership-team .team-member:nth-of-type(3)').addClass('third-animate');
// jQuery('.block--leadership-team .team-member:nth-of-type(4)').addClass('fourth-animate');
// jQuery('.block--leadership-team .team-member:nt

2144. Minimum Cost of Buying Candies With Discount

A shop is selling candies at a discount. For every two candies sold, the shop gives a third candy for free. The customer can choose any candy to take away for free as long as the cost of the chosen candy is less than or equal to the minimum cost of the two candies bought. For example, if there are 4 candies with costs 1, 2, 3, and 4, and the customer buys candies with costs 2 and 3, they can take the candy with cost 1 for free, but not the candy with cost 4. Given a 0-indexed integer array cost, where cost[i] denotes the cost of the ith candy, return the minimum cost of buying all the candies.
/**
 * @param {number[]} cost
 * @return {number}
 */
var minimumCost = function(cost) {
    cost.sort((a, b) => b - a); // Sort descending

    let total = 0;

    for (let i = 0; i < cost.length; i++) {
        // Every third candy (i % 3 === 2) is free
        if (i % 3 !== 2) {
            total += cost[i];
        }
    }

    return total;
};

2126. Destroying Asteroids

You are given an integer mass, which represents the original mass of a planet. You are further given an integer array asteroids, where asteroids[i] is the mass of the ith asteroid. You can arrange for the planet to collide with the asteroids in any arbitrary order. If the mass of the planet is greater than or equal to the mass of the asteroid, the asteroid is destroyed and the planet gains the mass of the asteroid. Otherwise, the planet is destroyed. Return true if all asteroids can be destroyed. Otherwise, return false.
/**
 * @param {number} mass
 * @param {number[]} asteroids
 * @return {boolean}
 */
var asteroidsDestroyed = function(mass, asteroids) {
    // Sort asteroids so we always face the smallest one first.
    // Greedy logic: if we can't destroy the smallest, we can't destroy any larger one.
    asteroids.sort((a, b) => a - b);

    // Process each asteroid in ascending order.
    for (let a of asteroids) {
        // If our current mass is too small, we fail immediately.
        if (mass < a) retur

lmstudio-newest

sk-lm-nwY7hWqR:CsvJ6aglD21jFQvWEzI1

3161. Block Placement Queries

There exists an infinite number line, with its origin at 0 and extending towards the positive x-axis. You are given a 2D array queries, which contains two types of queries: For a query of type 1, queries[i] = [1, x]. Build an obstacle at distance x from the origin. It is guaranteed that there is no obstacle at distance x when the query is asked. For a query of type 2, queries[i] = [2, x, sz]. Check if it is possible to place a block of size sz anywhere in the range [0, x] on the line, such that the block entirely lies in the range [0, x]. A block cannot be placed if it intersects with any obstacle, but it may touch it. Note that you do not actually place the block. Queries are separate. Return a boolean array results, where results[i] is true if you can place the block specified in the ith query of type 2, and false otherwise.
/**
 * @param {number[][]} queries
 * @return {boolean[]}
 */
var getResults = function(queries) {
    const Q = queries.length;
    const MAXX = Math.min(50000, 3 * Q);

    // Collect all obstacle positions
    const obs = new Set([0, MAXX]);
    for (const q of queries) {
        if (q[0] === 1) obs.add(q[1]);
    }

    const arr = Array.from(obs).sort((a, b) => a - b);
    const idx = new Map();
    arr.forEach((v, i) => idx.set(v, i));

    const n = arr.length;
    const fenw = new Array(

3300. Minimum Element After Replacement With Digit Sum

You are given an integer array nums. You replace each element in nums with the sum of its digits. Return the minimum element in nums after all replacements.
/**
 * @param {number[]} nums
 * @return {number}
 */
var minElement = function(nums) {

    // Helper function to compute the digit sum of a number.
    // Example: 999 → 27, 14 → 5
    function digitSum(x) {
        let sum = 0;

        // Extract digits one by one using modulo and integer division.
        while (x > 0) {
            sum += x % 10;          // Add the last digit
            x = Math.floor(x / 10); // Remove the last digit
        }

        return sum;
    }

    // Track th

Snow Monkey Forms フック一覧

# Snow Monkey Forms フック一覧

> **MW WP Form との違い**
>
> - SMF には **フォーム ID ごとの動的フック名**(`xxx-123` のような suffix)はありません。すべて **全フォーム共通** のフックです。
> - 問い合わせデータの DB 保存機能は標準ではなく、ファイルアップロードの一時保存のみです。

---

# 全体

| フック名 | タイプ | 説明 |
|---|---|---|
| `snow_monkey_forms_pro` | フィルター | Snow Monkey テーマ利用時の Pro 判定を上書きできる。`is_pro()` で参照。 |
| `snow_monkey_forms/enqueue/fallback_style` | フィルター | Pro 以外の環境でフォールバック CSS(sass-basis 代替)を読み込むかどうか。デフォルトは `! is_pro()`。 |

---

# イベントフック(管理画面)

| フック名 | タイプ

3093. Longest Common Suffix Queries

You are given two arrays of strings wordsContainer and wordsQuery. For each wordsQuery[i], you need to find a string from wordsContainer that has the longest common suffix with wordsQuery[i]. If there are two or more strings in wordsContainer that share the longest common suffix, find the string that is the smallest in length. If there are two or more such strings that have the same smallest length, find the one that occurred earlier in wordsContainer. Return an array of integers ans, where ans[i] is the index of the string in wordsContainer that has the longest common suffix with wordsQuery[i].
/**
 * @param {string[]} wordsContainer
 * @param {string[]} wordsQuery
 * @return {number[]}
 */
var stringIndices = function(wordsContainer, wordsQuery) {
    const n = wordsContainer.length;
    const m = wordsQuery.length;

    // Precompute lengths
    const lengths = wordsContainer.map(w => w.length);

    // Find max query length to limit trie depth
    let maxQ = 0;
    for (const q of wordsQuery) maxQ = Math.max(maxQ, q.length);

    // Trie node constructor
    function Node() {
      

skillsto check out

# Post-hoc Analyzer Agent

Analyze blind comparison results to understand WHY the winner won and generate improvement suggestions.

## Role

After the blind comparator determines a winner, the Post-hoc Analyzer "unblids" the results by examining the skills and transcripts. The goal is to extract actionable insights: what made the winner better, and how can the loser be improved?

## Inputs

You receive these parameters in your prompt:

- **winner**: "A" or "B" (from blind comparison)

gcp windows

!PWL]h]u}8moL-8

3121. Count the Number of Special Characters II

You are given a string word. A letter c is called special if it appears both in lowercase and uppercase in word, and every lowercase occurrence of c appears before the first uppercase occurrence of c. Return the number of special letters in word.
/**
 * @param {string} word
 * @return {number}
 */
var numberOfSpecialChars = function(word) {
    // first[c]  = first index (0-based) where char c appears
    // last[c]   = last index (0-based) where char c appears
    const first = Array(128).fill(-1);
    const last  = Array(128).fill(-1);

    // One pass: record first & last positions for every character
    for (let i = 0; i < word.length; i++) {
        const code = word.charCodeAt(i);
        if (first[code] === -1) first[code] = i;
 

Semantic versioning

https://docs.npmjs.com/about-semantic-versioning

how to set up specific tests in order to run in serial mode in Playwright?

// test.describe.configure({ mode: 'serial' });
// https://playwright.dev/docs/test-parallel#serial-mode
import { test, type Page } from '@playwright/test';

// Annotate entire file as serial.
test.describe.configure({ mode: 'serial' });

let page: Page;

test.beforeAll(async ({ browser }) => {
  page = await browser.newPage();
});

test.afterAll(async () => {
  await page.close();
});

test('runs first', async () => {
  await page.goto('https://playwright.dev/');
});

test('runs second', async