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// https://en.wikipedia.org/wiki/6174_(number) | |
const kaprekar = 6174; | |
const routine = (input) => { | |
const res = new Map(); | |
let i = 0; | |
while (input !== kaprekar) { | |
input = cycle(input); | |
res.set(i++, input); |
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/* | |
You are given an array of non-negative integers that represents a two-dimensional elevation map where each element is unit-width wall and the integer is the height. Suppose it will rain and all spots between two walls get filled up. | |
Compute how many units of water remain trapped on the map in O(N) time and O(1) space. | |
For example, given the input [2, 1, 2], we can hold 1 unit of water in the middle. | |
Given the input [3, 0, 1, 3, 0, 5], we can hold 3 units in the first index, 2 in the second, and 3 in the fourth index (we cannot hold 5 since it would run off to the left), so we can trap 8 units of water. | |
*/ | |
/** | |
* Solution: | |
* |
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package aoc18 | |
import ( | |
"io/ioutil" | |
"log" | |
"strconv" | |
"strings" | |
) | |
type Maze struct { |
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let AmountCalculator = { | |
'new'(days) { | |
return days * 3; | |
}, | |
'regular'(days) { | |
let amount = 2; | |
return days > 2 ? (amount + (days - 2) * 1.5) : amount; | |
}, | |
'childrens'(days) { | |
let amount = 1.5; |
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var TrelloSolver = (function solver() { | |
var ALLOWED_LETTERS = 'acdegilmnoprstuw'; | |
function hash(str) { | |
var h = 7; | |
for ( var i = 0, iLimit = str.length; i < iLimit; i++ ) { | |
h = ( h * 37 + validate(ALLOWED_LETTERS.indexOf(str[i]))); |