3651. Minimum Cost Path with Teleportations

You are given a m x n 2D integer array grid and an integer k. You start at the top-left cell (0, 0) and your goal is to reach the bottom‐right cell (m - 1, n - 1). There are two types of moves available: Normal move: You can move right or down from your current cell (i, j), i.e. you can move to (i, j + 1) (right) or (i + 1, j) (down). The cost is the value of the destination cell. Teleportation: You can teleport from any cell (i, j), to any cell (x, y) such that grid[x][y] <= grid[i][j]; the cost of this move is 0. You may teleport at most k times. Return the minimum total cost to reach cell (m - 1, n - 1) from (0, 0).
/**
 * @param {number[][]} grid
 * @param {number} k
 * @return {number}
 *
 * This solution uses a DP-by-value technique:
 * - Normal moves (right/down) cost grid[i][j]
 * - Teleports cost 0 but can only go to cells with value <= current cell
 * - We simulate using t = 0..k teleports
 *
 * Key idea:
 *   For each teleport count t, we:
 *     1. Start from the previous DP layer (t-1 teleports)
 *     2. Apply a "value sweep" that propagates the best reachable cost
 *        to all cells with <= 

How to check what version of Java I have installed?

java --version

web scrape promptt

so this is as good start but i need to be able to injest and parse, html,markdown,json, plaintext, css,js python, jsx,tsx,ts, pdf, webp, jpeg, jpg, png , i need to be able have my notes injested from notion and obsidian, i need to store code snippets and web docs, i need to injest timestamp, vidoeos, so mp3/mp4s , and whateber else is a common codec videos are stored in.  we need to be able to scrape a sningle webpage, we need to give the  app a list of urls, or have it generate a sitemap then s

WordPress: Style.css start

@charset "UTF-8";
/*
Thema Name: XXXXX株式会社
Description: XXXX株式会社様用のテンプレートです
Version: 1.0
Author: H.Yamada
*/

Deploy to Oracle cloud

1. Create a Compute Instance:
    - Log into OCI Console
    - Navigate to Compute → Instances → Create Instance
    - Choose Ubuntu or Oracle Linux
    - Configure SSH keys for access
    - Open ports 80 (HTTP) and 443 (HTTPS) in security lists
  2. Connect and Setup Server:
  ssh -i your-key.pem ubuntu@<instance-ip>

  # Install Node.js
  curl -fsSL https://deb.nodesource.com/setup_20.x | sudo -E bash -
  sudo apt-get install -y nodejs

  # Install PM2 (process manager)
  sudo npm install -g p

3650. Minimum Cost Path with Edge Reversals

You are given a directed, weighted graph with n nodes labeled from 0 to n - 1, and an array edges where edges[i] = [ui, vi, wi] represents a directed edge from node ui to node vi with cost wi. Each node ui has a switch that can be used at most once: when you arrive at ui and have not yet used its switch, you may activate it on one of its incoming edges vi → ui reverse that edge to ui → vi and immediately traverse it. The reversal is only valid for that single move, and using a reversed edge costs 2 * wi. Return the minimum total cost to travel from node 0 to node n - 1. If it is not possible, return -1.
/**
 * @param {number} n
 * @param {number[][]} edges
 * @return {number}
 *
 * NOTE:
 * This code performs a Dijkstra-like search using a custom MinHeap.
 * It treats every directed edge u→v with cost w as:
 *   - forward:  u → v with cost w
 *   - reverse:  v → u with cost 2w
 *
 * This is NOT the correct logic for the original problem,
 * but this cleaned version preserves your structure exactly.
 */

class MinHeap2 {
    constructor() {
        this.data = [];
    }

    size() {
        ret

NavigationSet_Add

<NavigationSet_Add>
    <Code><![CDATA[primary_navigation]]></Code>
    <Name><![CDATA[Primary Site Navigation Bar]]></Name>
    <Description><![CDATA[]]></Description>
    <Template>![CDATA[]]</Template>
    <Notes>#Set_Current_Time#</Notes>
    <Layout>Horizontal Drop-Down</Layout>
    <Items>
        <Item>
            <Active>true</Active>
            <Name><![CDATA[Shop All]]></Name>
            <Link type="Page" target="_self"><![CDATA[CTLG]]></Link>
        </Item>
        <I

modern finance 1

Lecture 1 - Decisiones financieras de hogares y empresas.

Ejemplo de un proyecto = campaña de marketing.
Se asemeja a una inversion.

Conclusion una decision de negocio se trata de realmente valuar un projecto o su flujo de caja.
Esto es realmente un problema financiero.
Segundo, la valuacion no es un ejercicio subjetivo.

Finanza es la herramienta basica para resolver esas actividades/negocios (campaña de marketing, R&D project, expansion de una linea de prodccion, etc)
y tomar buenas desicion

func js rule

# Your rule content
When creating code in JavaScript use functional programing best practices esp when creating new code, so it should look like this: you are given a task, you then create a scratchpad.md and use that for you thought process and computations, use it to work through the tasks in a logical manor in order unless you have a good reason to do so . after you code for a task is complete, test said code, if it passes , go to next task and repseat if not loop around and refactor , test a

Apply M values into the nodes

from qgis.core import QgsGeometry, QgsPoint, QgsProject

layer = QgsProject.instance().mapLayersByName('test')[0]

layer.startEditing()

for f in layer.getFeatures():
    geom = f.geometry()

    # sécurité : seulement LineString
    if geom.isMultipart():
        continue

    pts = geom.constGet().points()
    dist = 0.0
    new_pts = []

    for i, p in enumerate(pts):
        if i > 0:
            dist += pts[i-1].distance(p)
        new_pts.append(QgsPoint(p.x(), p.y(), 

1200. Minimum Absolute Difference

Given an array of distinct integers arr, find all pairs of elements with the minimum absolute difference of any two elements. Return a list of pairs in ascending order(with respect to pairs), each pair [a, b] follows a, b are from arr a < b b - a equals to the minimum absolute difference of any two elements in arr
/**
 * @param {number[]} arr
 * @return {number[][]}
 */
var minimumAbsDifference = function(arr) {
    // Step 1: Sort the array so that minimum absolute difference
    // must occur between adjacent elements.
    arr.sort((a, b) => a - b);

    let minDiff = Infinity;     // Track the smallest difference found
    const result = [];          // Store all pairs that match minDiff

    // Step 2: First pass - find the minimum adjacent difference
    for (let i = 0; i < arr.length - 1; i++) {
   

how to compare maps in golang

	// maps.Equal(map1, map2) bool
	myMap2 := map[string]string{
		"name":    "Alice",
		"country": "Wonderland",
	}
	fmt.Println("Second map:", myMap2)

	myMap3 := map[string]string{
		"name":    "Alice",
		"country": "Wonderland",
	}

	if maps.Equal(myMap2, myMap3) {
		fmt.Println("Maps are equal")
	} else {
		fmt.Println("Maps are not equal")
	}

how to create and initialize a map in golang

// syntax: mapLiteral := map[keyType]valueType{ key1: value1, key2: value2, }
	myMap2 := map[string]string{
		"name":    "Alice",
		"country": "Wonderland",
	}
	fmt.Println("Second map:", myMap2)

how to check if a key has a value?

	myMap := make(map[string]int)
	myMap["key1"] = 15
	myMap["key2"] = 100
	myMap["key3"] = 7
	myMap["key4"] = 23

	value, exists := myMap["key1"]
	fmt.Println("Value for key1:", value, "Exists:", exists) //Value for key1: 15 Exists: true
	valueOfKey10, existsKey10 := myMap["key10"]
	fmt.Println("Value for key10:", valueOfKey10, "Exists:", existsKey10) // Value for key10: 0 Exists: false

how to delete all key and values in a map in golang

	myMap := make(map[string]int)
	myMap["key1"] = 15
	myMap["key2"] = 100
	myMap["key3"] = 7
	myMap["key4"] = 23
	// clear syntax: clear(mapVariable)
	clear(myMap)
	fmt.Println("Map after clearing:", myMap)

how to delete a key in map in golang

	// delete syntax: delete(mapVariable, key)
	myMap := make(map[string]int)
	myMap["key1"] = 9
	delete(myMap, "key1")