877. Stone Game
Alice and Bob play a game with piles of stones. There are an even number of piles arranged in a row, and each pile has a positive integer number of stones piles[i].
The objective of the game is to end with the most stones. The total number of stones across all the piles is odd, so there are no ties.
Alice and Bob take turns, with Alice starting first. Each turn, a player takes the entire pile of stones either from the beginning or from the end of the row. This continues until there are no more piles left, at which point the person with the most stones wins.
Assuming Alice and Bob play optimally, return true if Alice wins the game, or false if Bob wins.
/**
* @param {number[]} piles
* @return {boolean}
*/
var stoneGame = function(piles) {
const n = piles.length;
// dp[i][j] = max score difference current player can achieve
const dp = Array.from({ length: n }, () => Array(n).fill(0));
// Base case: when i == j, only one pile is available
// The current player takes it, so the difference is piles[i]
for (let i = 0; i < n; i++) {
dp[i][i] = piles[i];
}
// Fill DP table for increasing lengths of subarraSimu alpha 28 rue du Foix p9syzn8r6s 2 PAC
{
"custom_renovation_plan": [
{
"category_technical_id": "heating",
"gesture_technical_id": "air_air_heat_pump_with_external_unit_complete",
"quantity": {
"default_value": null,
"error_margin": 0,
"unit": "unit",
"value": 1
},
"pricing": {
"price_per_unit": "100.0"
},
"spec": {
"service_technical_id": "air_Qwen3-Coder-Next QLoRA SFT smoke for HF Jobs
Qwen3-Coder-Next QLoRA SFT smoke for HF Jobs
# /// script
# requires-python = ">=3.10"
# dependencies = [
# "trl>=1.9.0,<2.0.0",
# "peft>=0.17.0",
# "bitsandbytes>=0.45.0",
# "trackio",
# "transformers>=4.57.0",
# "accelerate>=1.0.0",
# ]
# ///
"""LoRA SFT smoke test for Qwen3-Coder-Next on Hugging Face Jobs.
Goal: prove the hybrid `qwen3_next` architecture loads under TRL + PEFT and
survives a handful of optimizer steps before we spend on distillation / GRPO.
Defaults to 4-bit QLoRA so an 80B-total MoE can fit onsmithaery toolbox
{
"mcpServers": {
"toolbox": {
"type": "http",
"url": "https://mcp.smithery.run/pwdesignhtx",
"headers": {
"Authorization": "Bearer <api-key>"
}
}
}
}486. Predict the Winner
You are given an integer array nums. Two players are playing a game with this array: player 1 and player 2.
Player 1 and player 2 take turns, with player 1 starting first. Both players start the game with a score of 0. At each turn, the player takes one of the numbers from either end of the array (i.e., nums[0] or nums[nums.length - 1]) which reduces the size of the array by 1. The player adds the chosen number to their score. The game ends when there are no more elements in the array.
Return true if Player 1 can win the game. If the scores of both players are equal, then player 1 is still the winner, and you should also return true. You may assume that both players are playing optimally.
/**
* @param {number[]} nums
* @return {boolean}
*/
var predictTheWinner = function(nums) {
const n = nums.length;
// memo[l][r] will store the maximum score difference
// the current player can achieve over the opponent
// when playing optimally on nums[l..r].
const memo = Array.from({ length: n }, () => Array(n).fill(null));
function dp(l, r) {
// Base case: only one number left
// The current player takes it, so the difference is nums[l].
iaiml demo project
import asyncio
from sqlalchemy import text
from database import engine
async def get_question_counts(conn):
result = await conn.execute(
text(
"SELECT COUNT(*) AS total_questions, "
"SUM(CASE WHEN embedding IS NOT NULL THEN 1 ELSE 0 END) AS questions_with_embedding "
"FROM questions"
)
)
return result.fetchone()
async def delete_all_question_data():
async with engine.begin() as conn:
counts = await get_3014. Minimum Number of Pushes to Type Word I
You are given a string word containing distinct lowercase English letters.
Telephone keypads have keys mapped with distinct collections of lowercase English letters, which can be used to form words by pushing them. For example, the key 2 is mapped with ["a","b","c"], we need to push the key one time to type "a", two times to type "b", and three times to type "c" .
It is allowed to remap the keys numbered 2 to 9 to distinct collections of letters. The keys can be remapped to any amount of letters, but each letter must be mapped to exactly one key. You need to find the minimum number of times the keys will be pushed to type the string word.
Return the minimum number of pushes needed to type word after remapping the keys.
An example mapping of letters to keys on a telephone keypad is given below. Note that 1, *, #, and 0 do not map to any letters.
/**
* @param {string} word
* @return {number}
*/
var minimumPushes = function(word) {
// Each letter is distinct, so frequency doesn't matter.
// We have 8 keys (2–9), and each key can type:
// - 8 letters with 1 push
// - 8 letters with 2 pushes
// - 8 letters with 3 pushes
// ...and so on.
//
// The optimal strategy is to assign the cheapest push counts first.
// That means:
// - letters 0–7 cost 1 push
// - letters 8–15 cost 2 pushes
// - letBB LOOP Query Args > append 'featured events' to existing LOOP filters
The BB LOOP module already filters out past events.
The settings already are ordering them by EventDate.
This snippet includes ONLY Featured events.
function upcomingFeaturedEventsOnHomepage_fl_builder_loop_query_args( $args ) {
if ( $args['settings']->id == 'bbloop--upcoming-featured-public-events' ) {
$args['meta_query'][] = array(
'key' => '_tribe_featured',
'compare' => '=',
'value' => 1,
'type' => 'NUMERIC'
);
}
return $args;
}
add_filter( 'fl_builder_loop_query_args', 'upcomingFeaturedEventsOnHomepage_fl_builder_loop_query_args', 20, 1 );
WSL2: increase virtual disk
# Increase virtual disk WSL2 via `diskpart`
## 1. Shutdown WSL instance
```ps
wsl --shutdown
```
## 2. Check name of WSL distr
```ps
wsl -l -v
```
## 3. Check WSL path to .vhdx file
```ps
& {
# Collect installed WSL distribution names
$wslDistros = wsl -l -v |
Select-Object -Skip 1 |
ForEach-Object -Process {
($_ -replace "`0", "" -replace "\*", "").Trim() -split '\s+' |
Select-Object -First 1
} |
Where-Object -FilterScript {
3518. Smallest Palindromic Rearrangement II
You are given a palindromic string s and an integer k.
Return the k-th lexicographically smallest palindromic permutation of s. If there are fewer than k distinct palindromic permutations, return an empty string.
Note: Different rearrangements that yield the same palindromic string are considered identical and are counted once.
/**
* @param {string} s
* @param {number} k
* @return {string}
*/
var smallestPalindrome = function (s, k) {
const MAX = 1_000_000 + 1;
const n = s.length;
// s is already palindromic; only need counts from the left half
const halfCount = Array(26).fill(0);
for (let i = 0; i < Math.floor(n / 2); i++) {
halfCount[s.charCodeAt(i) - 97]++;
}
// nCk with early cap
function nCk(n, r) {
if (r === 0 || r === n) return 1;
r = Math.min(r, n -3517. Smallest Palindromic Rearrangement I
You are given a palindromic string s.
Return the lexicographically smallest palindromic permutation of s.
/**
* @param {string} s
* @return {string}
*/
var smallestPalindrome = function(s) {
// Count characters
const count = Array(26).fill(0);
for (const ch of s) {
count[ch.charCodeAt(0) - 97]++;
}
let left = "";
let mid = "";
// Build left half + detect middle
for (let i = 0; i < 26; i++) {
const c = String.fromCharCode(97 + i);
if (count[i] % 2 === 1) mid = c; // odd count → middle
left += c.repeat(Math.floor(count[i] / 2));
IELTS-up Listening Sample 5
32
electricity
internet
bus station
Hooper
passport
employment contract
reference from a fried or colleague
6:30 PM
At the caller's future apartment
24
18
5
19
Familiy ticket
107
online
confirmation email
by telephone
in person
7 weeks
5 people
grab people's attention
Marketing
Design of custom logos
Branding
E-commerce
Customer attraction
Ground floor
assignment
A
A?
C
A
C
rent earth?
susseptable
frequent sand storms
eye sight
architectural deeds1464. Maximum Product of Two Elements in an Array
Given the array of integers nums, you will choose two different indices i and j of that array. Return the maximum value of (nums[i]-1)*(nums[j]-1).
/**
* @param {number[]} nums
* @return {number}
*/
var maxProduct = function(nums) {
// We want the two largest numbers in the array.
// Initialize max1 as the largest, max2 as the second largest.
let max1 = 0;
let max2 = 0;
// Loop through each number in the array
for (let n of nums) {
// If n is greater than or equal to the current largest
if (n >= max1) {
// Shift max1 down to max2
max2 = max1;
// Update max1 to tRebase `--onto`
Say the 1st commit on the branch was `8f8949a` and the commit before that as `8b1ed4d` (the "tip"), then run `git rebase -onto develop 8b1ed4d`, assuming the base branch is `develop`.
git rebase --onto <base-branch> <last-commit-to-start-from>Launch Affinity Designer 2 - AutoHotKey
#Requires AutoHotkey v2.0
SetTitleMatchMode(2)
Run('"' EnvGet("LOCALAPPDATA") '\Microsoft\WindowsApps\AffinityDesigner2.exe"')
if WinWait("Affinity Designer",, 15)
{
WinRestore("Affinity Designer")
WinActivate("Affinity Designer")
}Launch Edge with URL as argument - AutoHotKey
#Requires AutoHotkey v2.0
; First command-line argument = URL
if A_Args.Length < 1
{
MsgBox "Usage:`nOpenUrl.ahk <URL>"
ExitApp
}
url := A_Args[1]
; Open the URL in the default browser
Run(url)
; Open ChatGPT in the default browser
; ** Not in use **
;Run("url)
; Wait for Microsoft Edge (remove this block if you use another browser)
if WinWait("ahk_exe msedge.exe",,10)
{
WinActivate("ahk_exe msedge.exe")
}