パーミッションの読み方

# パーミッションの読み方

## 基本構造

パーミッションは3桁の数字で表現
各桁は所有者、グループ、その他のユーザーの権限を示す

### 例: 755の場合

1桁目(7): 所有者の権限
2桁目(5): グループの権限
3桁目(5): その他のユーザーの権限


## 数字の意味
各桁は以下の数値の合計:

4 = 読み取り (read, r)
2 = 書き込み (write, w)
1 = 実行 (execute, x)
0 = 権限なし (-)

### 計算例:
7 = 4+2+1 = rwx (すべて可能)
6 = 4+2   = rw- (読み書き可能)
5 = 4+1   = r-x (読み実行可能)
4 = 4     = r-- (読み取りのみ)
0 = 0     = --- (権限なし)


## よく使うパーミッション
644 (rw-r--r--)
  所有者: 読み書き
  グループ: 読み取り
  その他: 読み取り
  → 一般的なファイル、設定ファイル、ドキュメント

755 (rwxr-xr-x)
  所有者: 読み書き実行
  グループ: 

aria-labeledbyで紐づけていればhtmlが変更されたとき読み上げも変更される

<p id="counter-text">カウント: 0</span>
<button type="button" aria-labelledby="counter-text" id="count-button">クリックするとカウントが読み上げられるが、このテキストは無視される</button>

取得予定もしくは取得済みのドメインが過去に使用されているか確認する方法

https://whois.jprs.jp/

javascriptのsetとhasのサンプル

// new Set(): 重複のない値の集合を作成する(Setオブジェクト)
// - 同じ値は一度だけ格納される
// - インデックスは存在しない
const mySet = new Set();

// Set.add(): セットに値を追加する
mySet.add("apple");
mySet.add("banana");
mySet.add("apple"); // 重複のため実際には追加されない

console.log("mySet:", mySet);

// Set.has(): セットに指定した値が存在するかを確認する
// 戻り値: boolean (true/false)
console.log('mySet.has("apple"):', mySet.has("apple")); // true
console.log('mySet.has("banana"):', mySet.has("banana")); // true
console.log('mySet.has("orange"):', mySet.has("orange")); // false

Vertical Menu Dropdown

https://www.youtube.com/watch?v=-hG2pg22Hwg Elementor's WordPress Menu set vertically has a problem with how it presents sub-items. Well - I - just - solved it - and you will be - surprised!
.elementor-nav-menu--layout-vertical .elementor-nav-menu .sub-menu {
    position: static !important;  
    display: none;                   
    padding-left: 0px;   
    margin: 0px !important;
    animation: menuSlide 0.25s ease;
    overflow: hidden;
}

/* --- Parent list items --- */
.elementor-nav-menu--layout-vertical .elementor-nav-menu li {
    position: static !important;
}

/* --- Open state (SmartMenus adds .sm-open) --- */
.elementor-nav-menu--layout-vertical .element

数量と組み合わせの購入制限ロジックサンプル

<form id="purchaseForm">
  <div>
    <label>商品A: <input type="number" name="productA" min="0" value="0" /></label>
  </div>
  <div>
    <label>商品B: <input type="number" name="productB" min="0" value="0" /></label>
  </div>
  <div>
    <label>商品C: <input type="number" name="productC" min="0" value="0" /></label>
  </div>
  <div>
    <label>商品D: <input type="number" name="productD" min="0" value="0" /></label>
  </div>
  <div>
    <label>商品E: <input type="number" name="productE" min="0" value="0" 

1523. Count Odd Numbers in an Interval Range

Given two non-negative integers low and high. Return the count of odd numbers between low and high (inclusive).
/**
 * @param {number} low
 * @param {number} high
 * @return {number}
 */
var countOdds = function(low, high) {
    // Step 1: Think about what we want:
    // We need to count how many odd numbers exist between low and high (inclusive).

    // Step 2: Use math instead of looping:
    // Formula: floor((high + 1) / 2) - floor(low / 2)
    // Why? Because:
    // - floor(x / 2) gives the count of even numbers up to x.
    // - floor((x + 1) / 2) gives the count of odd numbers up to x.
    // Su

km-berechnung v1

import os
import sys
import subprocess
import random
import time
import getpass
import json

# Paketprüfung
required_packages = ["googlemaps", "tqdm"]
for package in required_packages:
    try:
        __import__(package)
    except ImportError:
        subprocess.check_call([sys.executable, "-m", "pip", "install", package])

import googlemaps
from tqdm import tqdm

# Funktion: Benutzername erkennen und Pfad anpassen
def get_key_path(file_name):
    possible_users = ["Bernha

Synology_TIPS

# SYNOLOGY TIPS & TRICKS #

#### P4V:
Install P4V on synology NAS: [Link to Repo](https://github.com/FrozenStormInteractive/Perforce-Synology)

#### ENABLE SHR ( Synology Hybrid Raid )
-> Check the other snippet ( synology tag )

Initializing a PyQt5 App

This snippet show hows to jumpstart a small **PyQt5** app. PyQt6 and PySide2 also work
from PyQt5.QtWidgets import QMainWindow, QApplication
from PyQt5.QtGui import QIcon

class NameYourWindow(QMainWindow):
    def __init__(self):
        super().__init__()
        
        self.setWindowTitle("Window Title")
        
        self.show()
        
if __name__ == '__main__':
    app = QApplication([])

	app.setApplicationName("Your App Name")
	app.setOrganizationName("Your Organization Name")
	app.setApplicationVersion("0.1.0")
	
	controller = NameYourWindow()
	
	app.exec()

3578. Count Partitions With Max-Min Difference at Most K

You are given an integer array nums and an integer k. Your task is to partition nums into one or more non-empty contiguous segments such that in each segment, the difference between its maximum and minimum elements is at most k. Return the total number of ways to partition nums under this condition. Since the answer may be too large, return it modulo 109 + 7.
/**
 * Count the number of ways to partition nums into contiguous segments
 * such that in each segment, max - min <= k.
 * 
 * @param {number[]} nums - Input array of integers
 * @param {number} k - Maximum allowed difference between max and min in a segment
 * @return {number} - Number of valid partitions modulo 1e9+7
 */
var countPartitions = function (nums, k) {
    const MOD = 1e9 + 7;
    const n = nums.length;

    // dp[i] = number of ways to partition the first i elements (nums[0..i-1])

3578. Count Partitions With Max-Min Difference at Most K

You are given an integer array nums and an integer k. Your task is to partition nums into one or more non-empty contiguous segments such that in each segment, the difference between its maximum and minimum elements is at most k. Return the total number of ways to partition nums under this condition. Since the answer may be too large, return it modulo 109 + 7.
/**
 * Count the number of ways to partition nums into contiguous segments
 * such that in each segment, max - min <= k.
 * 
 * @param {number[]} nums - Input array of integers
 * @param {number} k - Maximum allowed difference between max and min in a segment
 * @return {number} - Number of valid partitions modulo 1e9+7
 */
var countPartitions = function (nums, k) {
    const MOD = 1e9 + 7;
    const n = nums.length;

    // dp[i] = number of ways to partition the first i elements (nums[0..i-1])

reisekosten

# ------------------------------------------------------------
# Automatischer Paketimport & -Installation
# ------------------------------------------------------------
import sys
import subprocess
import importlib
import importlib.util

def ensure_package(pkg_name, import_name=None):
    """
    Stellt sicher, dass ein Paket installiert und importierbar ist.
    Falls nicht installiert, wird es automatisch via pip installiert.
    """
    name_to_check = import_name or pkg_name
 

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Observer pattern

https://www.sitepoint.com/javascript-design-patterns-observer-pattern/

3432. Count Partitions with Even Sum Difference

You are given an integer array nums of length n. A partition is defined as an index i where 0 <= i < n - 1, splitting the array into two non-empty subarrays such that: Left subarray contains indices [0, i]. Right subarray contains indices [i + 1, n - 1]. Return the number of partitions where the difference between the sum of the left and right subarrays is even.
/**
 * @param {number[]} nums
 * @return {number}
 */
var countPartitions = function(nums) {
    // Step 1: Compute the total sum of the array.
    // We'll use this to quickly calculate the right subarray sum at each partition.
    let totalSum = 0;
    for (let num of nums) {
        totalSum += num;
    }

    // Step 2: Initialize variables
    let leftSum = 0;        // running sum of the left subarray
    let count = 0;          // number of valid partitions

    // Step 3: Iterate through