2553. Separate the Digits in an Array

Given an array of positive integers nums, return an array answer that consists of the digits of each integer in nums after separating them in the same order they appear in nums. To separate the digits of an integer is to get all the digits it has in the same order. For example, for the integer 10921, the separation of its digits is [1,0,9,2,1].
/**
 * @param {number[]} nums
 * @return {number[]}
 */
var separateDigits = function(nums) {
    const res = [];

    for (let n of nums) {
        // If the number is a single digit, just append it
        if (n < 10) {
            res.push(n);
            continue;
        }

        const stack = [];

        // Extract digits from right to left using modulo/division
        while (n > 0) {
            stack.push(n % 10);
            n = Math.floor(n / 10);
        }

        // Digits were

how to runs localhost web into emulators 

Practical fix for local dev (Android emulator)
Use the same URL host as on iOS (e.g. http://localhost:3000/...) on the emulator as well, and make localhost on the emulator reach your Mac’s port 3000 with port reverse:

adb reverse tcp:3000 tcp:3000
Then set:

EXPO_PUBLIC_KYC_WEBVIEW_URL_FREE_ACTIVATION="http://localhost:3000/beacon-money-account/info/free-activation?beacon_flow=BMA"

(aligned with iOS). After reverse, localhost:3000 inside the emulator is forwarded to your host’s 3000, so you ge

How to use codegen simulating devices on playwright? 

npx playwright codegen --device="iPhone 13" playwright.dev

How to create an using test generator or codegen in Playwright? 

npx playwright -o codegen testname.spec.ts
# this command run codegen and write it on the file testname.spec.ts

extract order 

select {o:code} as 'Eazle Order Number', {o:bcordernumber} as 'BC Order Number', {o:sapordernumber} as 'SAP Order number', {os:code} as 'Status', {bu:companycode} as 'Company Code', {bu:uid} as 'SoldTo', {o:originalordercode} as 'Amended Order', convert(date,{o:creationtime}) as 'Order Creation date', convert(time,{o:creationtime}) as 'Order Creation time'
from {order as o join b2bunit as bu on {bu:pk} = {o:unit} join orderstatus as os on {os:pk} = {o:status}}
where {o:creationtime} >= '2026-05-

Perply: fears and 

rapport - build raport - positiv connection with
there is no rapport between us
it's all greek to me
bitch fest

this is no mean feat			das ist eine beachtliche Leistung
heights
rouns

xiaomi mimo 

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2770. Maximum Number of Jumps to Reach the Last Index

You are given a 0-indexed array nums of n integers and an integer target. You are initially positioned at index 0. In one step, you can jump from index i to any index j such that: 0 <= i < j < n -target <= nums[j] - nums[i] <= target Return the maximum number of jumps you can make to reach index n - 1. If there is no way to reach index n - 1, return -1.
/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var maximumJumps = function(nums, target) {
    const n = nums.length;

    // dp[i] = maximum number of jumps needed to reach index i
    // Initialize all as unreachable (-Infinity)
    const dp = Array(n).fill(-Infinity);

    // Starting point: index 0 requires 0 jumps
    dp[0] = 0;

    // Try all pairs (i, j) with i < j
    for (let j = 1; j < n; j++) {
        for (let i = 0; i < j; i++) {

            // Dif

1914. Cyclically Rotating a Grid

ou are given an m x n integer matrix grid​​​, where m and n are both even integers, and an integer k. The matrix is composed of several layers, which is shown in the below image, where each color is its own layer: A cyclic rotation of the matrix is done by cyclically rotating each layer in the matrix. To cyclically rotate a layer once, each element in the layer will take the place of the adjacent element in the counter-clockwise direction. An example rotation is shown below: Return the matrix after applying k cyclic rotations to it.
/**
 * @param {number[][]} grid
 * @param {number} k
 * @return {number[][]}
 */
var rotateGrid = function(grid, k) {
    const m = grid.length, n = grid[0].length;
    const layers = Math.min(m, n) / 2;

    for (let layer = 0; layer < layers; layer++) {
        let ring = [];

        let top = layer, bottom = m - layer - 1;
        let left = layer, right = n - layer - 1;

        // top row
        for (let j = left; j <= right; j++) ring.push(grid[top][j]);
        // right col
        for

newest firecrawl key 

fc-f361f97009ef4beb99e70ffa5f2a1cf6

copilots-project-manager.md 

---
description: "Use this agent when the user has a complex task that needs to be broken down, coordinated across multiple agents, and assembled into a complete solution.\n\nTrigger phrases include:\n- 'I have a big project, can you help organize it?'\n- 'Break this down and delegate it to the right people'\n- 'Can you manage this task and coordinate the team?'\n- 'I need multiple things done, can you orchestrate?'\n- 'Help me break this into pieces and assign them out'\n\nExamples:\n- User say

Oracle APEX URL session link 

oracle apex link session

https://pretius.com/blog/apex-application-links-active-session

Autoplay policy in Chrome 

https://developer.chrome.com/blog/autoplay/

3629. Minimum Jumps to Reach End via Prime Teleportation

You are given an integer array nums of length n. You start at index 0, and your goal is to reach index n - 1. From any index i, you may perform one of the following operations: Adjacent Step: Jump to index i + 1 or i - 1, if the index is within bounds. Prime Teleportation: If nums[i] is a prime number p, you may instantly jump to any index j != i such that nums[j] % p == 0. Return the minimum number of jumps required to reach index n - 1.
/**
 * @param {number[]} nums
 * @return {number}
 */
var minJumps = function(nums) {
    const n = nums.length;
    if (n === 1) return 0;

    // 1. Build SPF (smallest prime factor) up to max(nums)
    const maxVal = Math.max(...nums);
    const spf = Array(maxVal + 1).fill(0);
    for (let i = 2; i <= maxVal; i++) {
        if (spf[i] === 0) {
            spf[i] = i;
            if (i * i <= maxVal) {
                for (let j = i * i; j <= maxVal; j += i) {
                    if (spf[j] =

Git: Reset/Sync main with dev on branch-drift 

#Create Backup of Main (optional)
git fetch --all --tags
git checkout main
git pull

git branch main-backup-$(date +%Y%m%d)
git push origin main-backup-$(date +%Y%m%d)

# Reset Main to dev:
git checkout main
git reset --hard origin/dev
git push --force-with-lease origin main

glm 

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