3665. Twisted Mirror Path Count

Given an m x n binary grid grid where: grid[i][j] == 0 represents an empty cell, and grid[i][j] == 1 represents a mirror. A robot starts at the top-left corner of the grid (0, 0) and wants to reach the bottom-right corner (m - 1, n - 1). It can move only right or down. If the robot attempts to move into a mirror cell, it is reflected before entering that cell: If it tries to move right into a mirror, it is turned down and moved into the cell directly below the mirror. If it tries to move down into a mirror, it is turned right and moved into the cell directly to the right of the mirror. If this reflection would cause the robot to move outside the grid boundaries, the path is considered invalid and should not be counted. Return the number of unique valid paths from (0, 0) to (m - 1, n - 1). Since the answer may be very large, return it modulo 109 + 7. Note: If a reflection moves the robot into a mirror cell, the robot is immediately reflected again based on the direction it used to enter that mirror: if it entered while moving right, it will be turned down; if it entered while moving down, it will be turned right. This process will continue until either the last cell is reached, the robot moves out of bounds or the robot moves to a non-mirror cell.
/**
 * @param {number[][]} grid
 * @return {number}
 */
var uniquePaths = function(grid) {
    const mod = 1_000_000_007;
    const down = new Array(grid[0].length).fill(0); // Ways arriving from above
    down[0] = 1; // Seed the starting cell
     
    for (const row of grid) {
        let right = 0; // Ways arriving from the left

        for (const [j, mirror] of row.entries()) {
            if (mirror === 1) {
                // A mirror reflects rightward paths down and downward paths righ

3970. Shortest Path With At Most K Consecutive Identical Characters

You are given an integer n representing the number of nodes in a directed weighted graph, numbered from 0 to n - 1. This is represented by a 2D integer array edges, where edges[i] = [ui, vi, wi] represents a directed edge from node ui to node vi with weight wi. You are also given a string labels of length n, where labels[i] is the character assigned to node i, and an integer k. Return the minimum total edge weight of a path from node 0 to node n - 1 such that the concatenation of the labels of the nodes along the path contains at most k consecutive identical characters. If no valid path exists, return -1.
/**
 * @param {number} n
 * @param {number[][]} edges
 * @param {string} labels
 * @param {number} k
 * @return {number}
 */
var shortestPath = function(n, edges, labels, k) {
    const graph = Array.from({ length: n }, () => []);
    for (const [u, v, weight] of edges) {
        graph[u].push([v, weight]);
    }

    // A node's label is fixed, so only its current run length is needed.
    const dist = Array.from({ length: n }, () => new Float64Array(k + 1).fill(Infinity));
    const startRun =

Vocab

town fuessen
Germany, State Bavaria
two years ago
mostly sightseeing
hhiking
cable card
first time high mountains


cable car
on/by foot = bei Fuß
athletic looking
very fit
do fitnesclasses

General_Listening_Practice_Set_1_test_4

Easton
30
Oktober
hotels.com
comission? * commission
13
sea view
2000
deposit
breakfast
1924
25
safety
ingridients * ingredients
wrapping
lecture
guide
10
13
factory *wheelchair
periods
specialization
essays
reading
projects
B
E *D
F
C
B
natural dip
clay
saturation level
natural spring
ecosystem?
urbanization
evaporates
*tap
tanks
food production system *balance of payments

3841. Palindromic Path Queries in a Tree

You are given an undirected tree with n nodes labeled 0 to n - 1. This is represented by a 2D array edges of length n - 1, where edges[i] = [ui, vi] indicates an undirected edge between nodes ui and vi. You are also given a string s of length n consisting of lowercase English letters, where s[i] represents the character assigned to node i. You are also given a string array queries, where each queries[i] is either: "update ui c": Change the character at node ui to c. Formally, update s[ui] = c. "query ui vi": Determine whether the string formed by the characters on the unique path from ui to vi (inclusive) can be rearranged into a palindrome. Return a boolean array answer, where answer[j] is true if the jth query of type "query ui vi"​​​​​​​ can be rearranged into a palindrome, and false otherwise.
/**
 * @param {number} n
 * @param {number[][]} edges
 * @param {string} s
 * @param {string[]} queries
 * @return {boolean[]}
 */
var palindromePath = function(n, edges, s, queries) {
    const graph = Array.from({ length: n }, () => []);
    for (const [u, v] of edges) {
        graph[u].push(v);
        graph[v].push(u);
    }

    const parent = Array(n).fill(-1);
    const depth = Array(n).fill(0);
    const size = Array(n).fill(1);
    const heavy = Array(n).fill(-1);
    const order = [0]

3882. Minimum XOR Path in a Grid

You are given a 2D integer array grid of size m * n. You start at the top-left cell (0, 0) and want to reach the bottom-right cell (m - 1, n - 1). At each step, you may move either right or down. The cost of a path is defined as the bitwise XOR of all the values in the cells along that path, including the start and end cells. Return the minimum possible XOR value among all valid paths from (0, 0) to (m - 1, n - 1).
/**
 * @param {number[][]} grid
 * @return {number}
 */
var minCost = function(grid) {
    const m = grid.length;
    const n = grid[0].length;

    // possible[r][c] contains all XOR values for paths to (r, c).
    const possible = Array.from({ length: m }, () =>
        Array.from({ length: n }, () => new Set())
    );

    possible[0][0].add(grid[0][0]);

    for (let r = 0; r < m; r++) {
        for (let c = 0; c < n; c++) {
            // Extend every path arriving from above or from the le

Textarea with @# tags

<!DOCTYPE html>
<html lang="cs">
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1, viewport-fit=cover">
<title>Textarea s @zmínkami a #tagy</title>
<style>
:root{
  box-sizing:border-box;
  padding-top:env(safe-area-inset-top,0px);
  padding-bottom:env(safe-area-inset-bottom,0px);
  --bg:#f6f7f9; --surface:#fff; --text:#1c2230; --muted:#6b7385; --border:#d5d9e2;
  --accent:#3b6cf6; --hover:#eef2ff; --tag:#0f8a5f; --shadow:0 8px 24px rgb

General_Writing_set_1_test_2.pdf

TASK 1
Dear Sarah,

I'm writting to thank you for your birthday present. I wasn't expecting to have so much fun with that new video game. I've been playing the game for two weeks every evening and I still enjoy it as much as on the first day I started playing it.

As you already knew before giving me the present, my parents bought me a new gaming console for my birthday, which was exactly what I wished and dreamt of for a long time. Frank gave me an additional controller as a present so we can p

BR Cleaning Services - Cleaning Services

BR Cleaning Services - https://brcleaning.com.au/ - BR Cleaning specialises in high-quality cleaning services for homes and businesses in Rosewater and across Adelaide, SA — regular cleaning (weekly/fortnightly), one-off cleaning (move-outs, spring cleans) and deep cleaning. Licensed, police-checked cleaners, satisfaction guarantee, since 2019.
# BR Cleaning Services
[BR Cleaning Services](https://brcleaning.com.au/) provides high-quality cleaning services for homes and businesses in Rosewater and across Adelaide, SA. Services include regular cleaning (weekly/fortnightly), one-off cleaning (move-outs, spring cleans) and deep cleaning. Licensed, police-checked cleaners, satisfaction guarantee, since 2019. Address: Unit 11, 3 Henry Street, Rosewater SA 5013, Australia. Phone: 0452 514 286. Website: https://brcleaning.com.au/

ByeBye Mess Cleaning Services - House Cleaning Service in San Mateo, CA

Say goodbye to the mess — hello to a spotless home or workplace. ByeBye Mess Cleaning Services provides dependable, detail-oriented house cleaning and commercial cleaning across the San Francisco Peninsula & South Bay. Licensed and insured. Free estimates. Website: https://www.byebyemesscleaningservices.com/
# ByeBye Mess Cleaning Services

Say goodbye to the mess — hello to a spotless home or workplace. ByeBye Mess Cleaning Services provides dependable, detail-oriented house cleaning and commercial cleaning across the San Francisco Peninsula & South Bay. Licensed and insured. Free estimates.

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Shed Removal and Backyard Cleanups in Morgan Hill and Gilroy — Stars Junk Removal

# Shed Removal and Backyard Cleanups in Morgan Hill and Gilroy — Stars Junk Removal

Stars Junk is a family-owned, licensed and insured junk removal company in San Jose since 2008. Stars Junk helps homeowners in Morgan Hill and Gilroy clear out old sheds, backyard clutter, fencing, scrap wood and metal, patio furniture, grills, and yard debris.

## Shed Removal and Backyard Cleanups

Our team handles shed teardown and removal, backyard cleanups, fencing removal, scrap wood and metal pickup, pati

3963. Create Grid With Exactly One Path

You are given two integers m and n, representing the number of rows and columns of a grid. Construct any m x n grid consisting only of the characters '.' and '#', where: '.' represents a free cell. '#' represents an obstacle cell. A valid path is a sequence of free cells that: Starts at the top-left cell (0, 0). Ends at the bottom-right cell (m - 1, n - 1). Moves only: Right, from (i, j) to (i, j + 1), or Down, from (i, j) to (i + 1, j). Return any grid such that there is exactly one valid path from the top-left cell to the bottom-right cell.
/**
 * @param {number} m
 * @param {number} n
 * @return {string[]}
 */
var createGrid = function(m, n) {
    // Start with every cell blocked.
    const grid = Array.from({ length: m }, () => Array(n).fill("#"));

    // Open the first column for the only downward route.
    for (let i = 0; i < m; i++) {
        grid[i][0] = ".";
    }

    // Open the bottom row to reach the destination.
    for (let j = 0; j < n; j++) {
        grid[m - 1][j] = ".";
    }

    // LeetCode expects each row as 

TTA SCNJ Toggle To Scanner.js

function automateSwitchToScanner(targetScannerName) {
    // 1. Define target (Pass exact name, partial string, or leave blank to pick the first physical scanner)
    // Examples: 'KODAK Scanner: S2000w', 'S2060w', or null
    var target = targetScannerName || '';

    // 2. Open the Scan Settings dialog via DOM class
    var settingsBtn = document.querySelector('.scanSettingsButton');
    if (!settingsBtn) {
        console.warn("FAILED: Could not locate the Settings button on the toolb

TTA SCNJ Toggle To FileImport via ExJS

function automateScannerSwitchToFileImport() {
    // 1. Locate the "View or Change Scan Settings" button using the DOM class
    var settingsBtn = document.querySelector('.scanSettingsButton');

    if (!settingsBtn) {
        console.warn("FAILED: Could not locate the Settings button on the toolbar.");
        return;
    }

    // 2. Click the settings button using a standard DOM click to trigger the KTA window
    settingsBtn.click();

    // 3. Wait 400ms for the ExtJS modal win

TTA SCNJ Get browser Objects

(function() {
    var results = { WindowGlobals: [], ScanButtonProps: [] };
    
    // 1. Search global window objects
    results.WindowGlobals = Object.keys(window).filter(function(k) {
        var key = k.toLowerCase();
        return key.includes('kofax') || key.includes('capture') || key.includes('scan') || key.includes('atalasoft');
    });

    // 2. Search the specific Scan button's internal properties
    if (typeof Ext !== 'undefined' && Ext.ComponentQuery) {
        var bt

TTA SCNJ - Listen to Enter KeyPress on multi-line textbox

Ext.onReady(function () {
    setTimeout(function () {
        var controlName = 'tb_multiline';

        // 1. Locate the specific ExtJS textarea component by name
        var cmp = Ext.ComponentQuery.query('textarea[name="' + controlName + '"]')[0] || 
                  Ext.ComponentQuery.query('component[controlName="' + controlName + '"]')[0];

        if (!cmp) {
            console.warn("Could not locate textarea control: " + controlName);
            return;
        }