1306. Jump Game III

Given an array of non-negative integers arr, you are initially positioned at start index of the array. When you are at index i, you can jump to i + arr[i] or i - arr[i], check if you can reach any index with value 0. Notice that you can not jump outside of the array at any time.
/**
 * @param {number[]} arr
 * @param {number} start
 * @return {boolean}
 */
var canReach = function(arr, start) {
    const n = arr.length;
    const queue = [start];
    const visited = new Array(n).fill(false);
    visited[start] = true;

    while (queue.length > 0) {
        const i = queue.shift();

        // If we reached a zero, we're done
        if (arr[i] === 0) return true;

        // Try both directions
        const next1 = i + arr[i];
        const next2 = i - arr[i];

English Reading Task: Global Affairs 

English Reading Task: Global Affairs
C2 Level – Advanced Comprehension and Analysis
Reading Text

In recent years, global affairs have become increasingly complex and
interconnected. International organisations, such as the United Nations and the
World Health Organization, play a pivotal role in addressing transnational
challenges, including climate change, pandemics, and economic instability. These
issues rarely remain confined within national borders; instead, they spread rapidly
thro

154. Find Minimum in Rotated Sorted Array II

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become: [4,5,6,7,0,1,4] if it was rotated 4 times. [0,1,4,4,5,6,7] if it was rotated 7 times. Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]]. Given the sorted rotated array nums that may contain duplicates, return the minimum element of this array. You must decrease the overall operation steps as much as possible.
/**
 * @param {number[]} nums
 * @return {number}
 */
var findMin = function(nums) {
    let left = 0, right = nums.length - 1;

    while (left < right) {
        const mid = Math.floor((left + right) / 2);

        if (nums[mid] < nums[right]) {
            right = mid;            // min is in [left, mid]
        } else if (nums[mid] > nums[right]) {
            left = mid + 1;         // min is in [mid, right]
        } else {
            right--;                // duplicates: shrink safely

153. Find Minimum in Rotated Sorted Array

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become: [4,5,6,7,0,1,2] if it was rotated 4 times. [0,1,2,4,5,6,7] if it was rotated 7 times. Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]]. Given the sorted rotated array nums of unique elements, return the minimum element of this array. You must write an algorithm that runs in O(log n) time.
/**
 * @param {number[]} nums
 * @return {number}
 */
var findMin = function(nums) {
    let left = 0;
    let right = nums.length - 1;

    while (left < right) {
        const mid = Math.floor((left + right) / 2);

        // If mid element is greater than the rightmost,
        // the minimum is in the right half.
        if (nums[mid] > nums[right]) {
            left = mid + 1;
        } else {
            // Otherwise, the minimum is in the left half (including mid)
            right = mid

PRD to html site 

https://claude.ai/public/artifacts/c314b2b5-f351-4e94-bb24-752c4d255a4b

how to record a trace a test using a command in the terminal in Playwright? 

npx playwright test --trace on
# then to see the trace
npx playwright show-report

Correlation 

Pearson Correlation
  pearson_coef, p_value = stats.pearsonr(df['']. df[''])

2784. Check if Array is Good

You are given an integer array nums. We consider an array good if it is a permutation of an array base[n]. base[n] = [1, 2, ..., n - 1, n, n] (in other words, it is an array of length n + 1 which contains 1 to n - 1 exactly once, plus two occurrences of n). For example, base[1] = [1, 1] and base[3] = [1, 2, 3, 3]. Return true if the given array is good, otherwise return false. Note: A permutation of integers represents an arrangement of these numbers.
/**
 * @param {number[]} nums
 * @return {boolean}
 */
var isGood = function(nums) {
    const n = Math.max(...nums);

    // Condition 1: length must be n + 1
    if (nums.length !== n + 1) return false;

    // Count frequencies
    const freq = new Map();
    for (let x of nums) {
        freq.set(x, (freq.get(x) || 0) + 1);
    }

    // Check 1..n-1 appear exactly once
    for (let i = 1; i < n; i++) {
        if (freq.get(i) !== 1) return false;
    }

    // Check n appears exactly twice

how to show the report in Playwright? 

npx playwright show-report

ollama api key 

ffad8d753bf24173939d651c5f284bca.j5SL1kot8x6NItLLegBysiIg

1674. Minimum Moves to Make Array Complementary

You are given an integer array nums of even length n and an integer limit. In one move, you can replace any integer from nums with another integer between 1 and limit, inclusive. The array nums is complementary if for all indices i (0-indexed), nums[i] + nums[n - 1 - i] equals the same number. For example, the array [1,2,3,4] is complementary because for all indices i, nums[i] + nums[n - 1 - i] = 5. Return the minimum number of moves required to make nums complementary.
/**
 * @param {number[]} nums
 * @param {number} limit
 * @return {number}
 */
var minMoves = function(nums, limit) {
    const n = nums.length;
    const diff = new Array(2 * limit + 2).fill(0);

    for (let i = 0; i < n / 2; i++) {
        const a = nums[i];
        const b = nums[n - 1 - i];
        const mn = Math.min(a, b);
        const mx = Math.max(a, b);
        const s = a + b;

        // Base: everything costs 2 moves
        diff[2] += 2;
        diff[2 * limit + 1] -= 2;

SHIPSTATION 

(http.request.uri.path contains "/rest/") or (http.request.uri.path contains "/graphql") or (http.request.uri.path contains "/api/")

impeccalbe commands 

## Impeccable — the “design‑fluency” agent skill

**TL;DR** – Impeccable is a portable *agent skill* that gives any code‑generation LLM (Claude Code, Cursor, Gemini‑CLI, Codex CLI, etc.) a shared design vocabulary and a set of 23 concrete commands (`/impeccable …`).  
It can *teach* the model design fundamentals, *audit* a code‑base for UI anti‑patterns, *polish* existing UI, and even *live‑edit* components directly in the browser.  
You install it once (`npx skills add pbakaus/impeccable`)

open links 

ios - xcrun simctl openurl booted "https://example.ru/"
android -  adb shell am start -a android.intent.action.VIEW -d "https://example.ru/"

resend api key 

re_MUhQG4gg_QCoTztF4heDf3rgA48ydrJYC

🪪 Authentification - Proxy Schemes 

# Authentification à un proxy HTTP — Guide pédagogique

> Comprendre les différents schémas d'authentification HTTP utilisés par les proxies d'entreprise, du plus simple au plus sophistiqué, avec leurs forces, leurs limites et leur impact sur les outils en ligne de commande (curl, pip, dbt, etc.).

---

## 1. Pourquoi un proxy demande-t-il une authentification ?

Dans une infrastructure d'entreprise, **tout trafic Internet sortant** transite par un proxy. Ce dernier joue plusieurs rôles :

- **F