Package Management & Cleanup
# Package Management & Cleanup
**Machine:** Lenovo ThinkPad T14 Gen 2
**OS:** Ubuntu 24.04 LTS (Dual Boot)
**Date:** March 30, 2026
---
## Overview
This document covers the standard commands for keeping Ubuntu packages up to date and reclaiming disk space from cached and unused files. Commands are grouped by category — run them in the order shown for a full update + cleanup cycle.
---
## 1. Update & Upgrade
### Refresh the package index
```bash
sudo apt update
```
Downloads the latest p2840. Check if Strings Can be Made Equal With Operations II
You are given two strings s1 and s2, both of length n, consisting of lowercase English letters.
You can apply the following operation on any of the two strings any number of times:
Choose any two indices i and j such that i < j and the difference j - i is even, then swap the two characters at those indices in the string.
Return true if you can make the strings s1 and s2 equal, and false otherwise.
/**
* @param {string} s1
* @param {string} s2
* @return {boolean}
*/
var checkStrings = function(s1, s2) {
// We will separate characters by index parity:
// even-indexed characters go into one bucket,
// odd-indexed characters go into another.
//
// Because the allowed operation only lets us swap
// characters whose indices differ by an EVEN number,
// we can only rearrange characters *within* the same parity group.
//
// So the question becomes:
// Tools
## Dev tools
- URL Encoder https://yoksel.github.io/url-encoder/
- SVG Optimizer https://svgomg.net/
- Online @font-face generator https://transfonter.org/
## Optimization
- Structured Data Markup Helper https://www.google.com/webmasters/markup-helper/u/0/
## Shopify
- Help Center / Support https://help.shopify.com/en#/contact
- Changelog https://shopify.dev/changelogMONGODB_URI
MONGODB_URI="mongodb+srv://weezy:weezy@cluster1.ov8pn0m.mongodb.net/"lm studio api key
sk-lm-oAmPj8OQ:ZnwH51IwH6oD64IrpVZ52839. Check if Strings Can be Made Equal With Operations I
You are given two strings s1 and s2, both of length 4, consisting of lowercase English letters.
You can apply the following operation on any of the two strings any number of times:
Choose any two indices i and j such that j - i = 2, then swap the two characters at those indices in the string.
Return true if you can make the strings s1 and s2 equal, and false otherwise.
/**
* @param {string} s1
* @param {string} s2
* @return {boolean}
*/
var canBeEqual = function(s1, s2) {
// --- Core idea ---
// We are only allowed to swap:
// - index 0 ↔ 2 (even positions)
// - index 1 ↔ 3 (odd positions)
//
// This means characters can move *within* their parity group (even/odd),
// but never cross between them.
//
// So the strings can be made equal if:
// 1. The even-index characters match as a multiset
// 2. The skills
1custom router
import { useState } from 'react'
import { useKV } from '@github/spark/hooks'
import { KnowledgeItem, ChatMessage } from './lib/types'
import { useTheme } from './hooks/use-theme'
import { Sidebar } from './components/Sidebar'
import { Dashboard } from './components/Dashboard'
import { Search } from './components/Search'
import { Upload } from './components/Upload'
import { Chat } from './components/Chat'
import { Scraper } from './components/Scraper'
import { Analytics } from './components/Analydev knowlege hub
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2573. Find the String with LCP
We define the lcp matrix of any 0-indexed string word of n lowercase English letters as an n x n grid such that:
lcp[i][j] is equal to the length of the longest common prefix between the substrings word[i,n-1] and word[j,n-1].
Given an n x n matrix lcp, return the alphabetically smallest string word that corresponds to lcp. If there is no such string, return an empty string.
A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b. For example, "aabd" is lexicographically smaller than "aaca" because the first position they differ is at the third letter, and 'b' comes before 'c'.
/**
* @param {number[][]} lcp
* @return {string}
*/
var findTheString = function(lcp) {
const n = lcp.length;
// -----------------------------
// 1. Validate the diagonal
// -----------------------------
// lcp[i][i] must equal the length of the suffix starting at i.
for (let i = 0; i < n; i++) {
if (lcp[i][i] !== n - i) {
return ""; // impossible
}
}
// -----------------------------
// 2. Union-Find to group equal positions
AWS DDoS コスト爆発防止策
# AWS DDoS コスト爆発防止策まとめ
**出典：** [DDoS攻撃でAWS請求が200万円に！S3・CloudFrontで絶対やるべきコスト爆発防止策 6選 - Qiita](https://qiita.com/miruky/items/b996e374c91923141178)
---
## 事例
2026年3月、個人開発者のS3バケットがDDoS攻撃を受け、3日間で160TBのデータ転送が発生し約$15,000（約200万円）の請求が発生。
## なぜ高額になるか
S3はデフォルトでレート制限がなく、パブリックバケットへは事実上無制限にGETリクエストを送れる。160TB転送時のデータ転送料金は約$14,140（東京リージョン）。
| データ転送量 | 単価 |
|---|---|
| 〜10TB | $0.114/GB |
| 10TB〜50TB | $0.089/GB |
| 50TB〜150TB | $0.086/GB |
| 150TB超 | $0.084/GB |
---
## 防止策（優先度順）
| 優先度 | 対策 | コスト |
|--