3875. Construct Uniform Parity Array I

You are given an array nums1 of n distinct integers. You want to construct another array nums2 of length n such that the elements in nums2 are either all odd or all even. For each index i, you must choose exactly one of the following (in any order): nums2[i] = nums1[i] nums2[i] = nums1[i] - nums1[j], for an index j != i Return true if it is possible to construct such an array, otherwise, return false.
/**
 * @param {number[]} nums1
 * @return {boolean}
 */
var uniformArray = function(nums1) {
    // The problem allows us to transform each element by either:
    //   1) keeping nums1[i], or
    //   2) subtracting nums1[j] from it (j ≠ i)
    //
    // Because subtracting numbers can always produce either all-even
    // or all-odd results depending on what parity exists in the array,
    // it is ALWAYS possible to make the final array uniform in parity.
    //
    // Therefore the answer is

Revert last commit on remove and keep changed on another branch 

On branch `branch_a`, after accidently committing and pushing to remote, here is how to revert the last commit, while keeping changes on another branch. Assuming that nobody else has contributed.
```bash
git switch branch_a
git status

# Optional safety pointer to the accidental commit
git branch backup/accidental-commit

# Remove the commit, keeping all its changes staged
git reset --soft HEAD~1

# Correct the remote branch
git push --force-with-lease origin branch_a

# Create the intended bran

ボタン ホバーで矢印アイコンが移動 

<a href=""class="btn">
  <p>テキスト</p>
  <span class="btn__icon">
    <svg class="btn__icon--img" id="" xmlns="http://www.w3.org/2000/svg" version="1.1" viewBox="0 0 14 14">
      <path d="M1.3,6.2c-.4,0-.8.3-.8.8s.3.8.8.8v-1.5h0ZM13.2,7.5c.3-.3.3-.8,0-1.1L8.4,1.7c-.3-.3-.8-.3-1.1,0-.3.3-.3.8,0,1.1l4.2,4.2-4.2,4.2c-.3.3-.3.8,0,1.1s.8.3,1.1,0c0,0,4.8-4.8,4.8-4.8ZM1.3,6.9v.9h11.3v-1.6H1.3v.9-.2Z"/>
    </svg>
</a>

3568. Minimum Moves to Clean the Classroom

You are given an m x n grid classroom where a student volunteer is tasked with cleaning up litter scattered around the room. Each cell in the grid is one of the following: 'S': Starting position of the student 'L': Litter that must be collected (once collected, the cell becomes empty) 'R': Reset area that restores the student's energy to full capacity, regardless of their current energy level (can be used multiple times) 'X': Obstacle the student cannot pass through '.': Empty space You are also given an integer energy, representing the student's maximum energy capacity. The student starts with this energy from the starting position 'S'. Each move to an adjacent cell (up, down, left, or right) costs 1 unit of energy. If the energy reaches 0, the student can only continue if they are on a reset area 'R', which resets the energy to its maximum capacity energy. Return the minimum number of moves required to collect all litter items, or -1 if it's impossible.
/**
 * @param {string[]} classroom
 * @param {number} energy
 * @return {number}
 */
var minMoves = function(classroom, energy) {
    const m = classroom.length;
    const n = classroom[0].length;

    let start = null;
    const litters = [];

    // Find S and all L
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            const c = classroom[i][j];
            if (c === 'S') start = [i, j];
            if (c === 'L') litters.push([i, j]);
        }
    }

    const L

Extract team id and simulation id from a New Relic json extract 

# ruff: noqa: ANN401, D103, T201

"""Temporary script: group unique kelvin_simulation_ids by team_id from a JSON log export."""

import json
import sys
from collections import defaultdict
from collections.abc import Iterator
from pathlib import Path
from typing import Any

TEAM_KEY = "request_body.config.team_id"
SIMULATION_KEY = "kelvin_identifier"


def records(node: Any) -> Iterator[dict[str, Any]]:
    """Yield every dict in the structure that carries either key of interest."""
    if isinst

Django language cookies in DEV environment 

# Disable Django language cookie in development environment (localhost / 127.0.0.1)
LANGUAGE_COOKIE_ENABLED = True / False

# Alternatively, create a different cookie to each project
if DEBUG:
    LANGUAGE_COOKIE_NAME = "ktivaivrit_language"

skills 

---
name: knowledge-ops
description: Knowledge base management, ingestion, sync, and retrieval across multiple storage layers (local files, MCP memory, vector stores, Git repos). Use when the user wants to save, organize, sync, deduplicate, or search across their knowledge systems.
metadata:
  origin: ECC
---

# Knowledge Operations

Manage a multi-layered knowledge system for ingesting, organizing, syncing, and retrieving knowledge across multiple stores.

Prefer the live workspace model:
- cod

2091. Removing Minimum and Maximum From Array

You are given a 0-indexed array of distinct integers nums. There is an element in nums that has the lowest value and an element that has the highest value. We call them the minimum and maximum respectively. Your goal is to remove both these elements from the array. A deletion is defined as either removing an element from the front of the array or removing an element from the back of the array. Return the minimum number of deletions it would take to remove both the minimum and maximum element from the array.
/**
 * @param {number[]} nums
 * @return {number}
 */
var minimumDeletions = function(nums) {
    const n = nums.length;

    // Find the indices of the minimum and maximum elements
    let iMin = 0, iMax = 0;
    for (let i = 0; i < n; i++) {
        if (nums[i] < nums[iMin]) iMin = i;   // update min index
        if (nums[i] > nums[iMax]) iMax = i;   // update max index
    }

    // Normalize: L = leftmost index, R = rightmost index
    let L = Math.min(iMin, iMax);
    let R = Math.max(iMin

IELTS Listening 4-17 

C
B
A
C
A
motivation
on foot
cleaner
snack
cash
stand-by
50
energy saving bulbs

one minute
roof

hot water
one degree
mobile app
C
C
B

A
penalties

imprisonment


ten precent
health
Funding


oxidation

Corcs

格子模様を背景に指定する 

.hoge{
  --lineColor: #fff;
  --lineSize: 1px;
  --spaceSize: 20px;
  background-image: linear-gradient(to right, var(--lineColor) var(--lineSize), transparent var(--lineSize)), linear-gradient(to bottom, var(--lineColor) var(--lineSize), transparent var(--lineSize));
  background-size: var(--spaceSize) var(--spaceSize);
  background-repeat: repeat;
}

Apply green color to (.venv) in prompt in terminal 

# Add the following lines to ~/.zshrc                                                                                                                                 
# ===================================

# Apply green color to (.venv) in prompt in terminal                                        
export VIRTUAL_ENV_DISABLE_PROMPT=1                                                                                                                                          
setopt PROMPT_SUBST

#wow #macro 

macros

```
/yell ¡INNER PLEASE!
/run local p=UnitPower("player")/UnitPowerMax("player")*100 SendChatMessage("MANA "..string.format("%.0f",p).."%","YELL")
```

```
/run local h=GetCombatRating(CR_HIT_MELEE) ChatFrame1:AddMessage(format("Hit:%d (%.2f%%)",h,h/32.79)) ChatFrame1:AddMessage(format("necessary 80:%d necessary 83:%d",math.max(0,164-h),math.max(0,263-h)))
```

![](https://cdn.cacher.io/attachments/u/3jyq9n8qej8bf/SsN3OYo8Y53JRNDr_RdyKKQGqk-PjsZE/wso23zliv.png)


```
/run local h=GetComb

3734. Lexicographically Smallest Palindromic Permutation Greater Than Target

You are given two strings s and target, each of length n, consisting of lowercase English letters. Return the lexicographically smallest string that is both a palindromic permutation of s and strictly greater than target. If no such permutation exists, return an empty string.
/**
 * @param {string} s
 * @param {string} target
 * @return {string}
 */
var lexPalindromicPermutation = function (s, target) {
    const n = s.length;

    // Count frequency of each character in s
    const freq = new Array(26).fill(0);
    for (const ch of s) {
        freq[ch.charCodeAt(0) - 97]++;
    }

    // Check if a palindrome is even possible:
    // at most one character may have an odd count.
    let oddCount = 0;
    for (const x of freq) {
        if (x % 2 === 1) oddCount++;

troubleshoot generic error message on bo 

com.hybris.backoffice.widgets.notificationarea.NotificationService
breakpoint

Navigating the Unpredictable World of Omegle 

In the vast and often curated landscape of the internet, there are few platforms that offer the raw, unscripted spontaneity of Omegle. For over a decade, this unique website has served as a digital crossroads, connecting strangers from across the globe for one-on-one text or video chats. It’s a place where unexpected conversations bloom, fleeting connections are forged, and the truly bizarre can unfold at a moment's notice. If you’ve ever been curious about dipping your toes into this digital oc