RP-3

/**
 * Calculates the maximum borrowing power based on income and various parameters
 * @param incomeAfterTax - Annual income after tax
 * @param annualExpenses - Annual living expenses (default: 24000)
 * @param baseRate - Base interest rate for the loan (default: 0.0619)
 * @param bufferRate - Additional buffer rate added to base rate for stress testing (default: 0.038)
 * @param loanPeriodYears - Duration of the loan in years (default: 30)
 * @param marginPercentage - Percentage of income used as margin (default: 0.000)
 * @param marginFloor - Minimum margin amount regardless of income (default: 7000)
 * @param marginCeiling - Maximum margin amount regardless of income (default: 0)

RP-3

/*
https://leetcode.com/problems/step-by-step-directions-from-a-binary-tree-node-to-another/

IDEA:
- If we find the lowest common ancestor of the two nodes we're looking for, then we just
  need to find the path from the start to the LCA, and from the LCA to the target.
- Maintain two stacks; one for the start, and one for the end.
  - The path from the LCA to the start node will just be some number of 'U's, because
     the LCA must be equal to or above the start node (by definition). 
  - The path from the LCA to the end node will be some number of 'L's or 'R's. 

RP-3

#convert SRID to _______ outputDirectory           inputShapefileWithAdjacent .dbf, .prj, .shx files
ogr2ogr -t_srs EPSG:4326 ~/Desktop/london_boroughs greater_london_const_region.shpu

#convert to psql
          # creating tables where necessary
             # inputShapefileWithAdjacent .dbf, .prj, .shx files
                                                                       # piping the output to this postgis database
shp2pgsql -c ~/Desktop/london_boroughs/greater_london_const_region.shp | psql -d os_boundaries

# BotsonGIS cheatsheet

RP-3

/*
Pure recursive approach.
Branching factor of, at worst, a.length.
Yields O(len(a)^len(a)), which is about has bad as it gets...
*/
const stockWithKTransactionsRecursive = (k, a) => {

    const maxProfit = (i, k) => { // on the ith day, with k transactions remaining
        if(i <= 0) return 0; // if it's day 0, we can't make a profit since we need at least one previous day to buy on
        if(k === 0) return 0; // if we have no transactions remaining, just return 0;

RP-3

type void struct{}
type g = map[string](map[string]float64)
type stringSet = map[string]void

var member void

func calcEquation(equations [][]string, values []float64, queries [][]string) []float64 {
	// 1. Make a directed, weighted graph out of each equation
	graph, seen := make(g), make(stringSet)
	for i, vars := range equations {

RP-3

/*
IDEA: Iff:
  - The robot is facing North after the instructions, AND
  - The robot was displaced by any non-zero amount, THEN
the robot will hare off into the sunset. 

Otherwise, the robot will loop endlessly. 

Proof: Lots of drawing on square-ruled paper. 
*/

RP-3

func findMaximumXOR(nums []int) int {
	result := 0
	root := trieNode{}

	for _, num := range nums {
		maxAgainstNum := root.maxAgainst(num)
		if maxAgainstNum > result {
			result = maxAgainstNum
		}
		root.insert(num)

RP-3

/**
 * // Definition for a Node.
 * function Node(val) {
 *    this.val = val;
 *    this.left = null;
 *    this.right = null;
 *    this.parent = null;
 * };
 */

RP-3

/**
 * @param {number[]} nums
 * @return {number}
 */
/*
IDEA: 
- Do a prefix-multiply, i.e., a rolling product from left to
  right. 
- This strategy depends on the number of negative numbers in
  the input.

RP-3

/**
 * @param {string} secret
 * @param {string} guess
 * @return {string}
 */
var getHint = function(secret, guess) {
    let [bulls, cows]  = [0, 0];
    
    const secretDigits = new Array(10).fill(0);
    for(let i=0; i<secret.length; i++){