150 coding challenges in 25 days
What I learned from attempting Advent of Code 2017, 2016 & 2015 all at one go.
Yong Jun
24 Jan 2018
About me
by Eric Wastl
Advent of Code a series of small programming puzzles for a variety of skill levels.
They are self-contained and are just as appropriate for an expert who wants to stay sharp as they are for a beginner who is just learning to code.
Each puzzle calls upon different skills and has two parts that build on a theme.
Leaderboard
Personal Stat
Global Stat
If you need a little help
Some things I've learned
Generators
function* generator () {
// do something
yield 'someValue'
// do some more thing
yield 'anotherValue'
}
const iterator = generator()
console.log(iterator.next())
// {value: 'someValue', done: false}
console.log(iterator.next())
// {value: 'anotherValue', done: false}
console.log(iterator.next())
// {value: undefined, done: true}
AoC 2017 Day 3: Spiral Memory
Right x 1
Up x 1
Left x 2
Down x 2
Right x 3
Up x 3
Left x 4
Down x 4
function * traverse () {
let step = 1
let x = 0
let y = 0
yield [x, y]
while (true) {
for (let right = step; right > 0; right--) {
yield [++x, y]
}
for (let up = step; up > 0; up--) {
yield [x, ++y]
}
step++
for (let left = step; left > 0; left--) {
yield [--x, y]
}
for (let down = step; down > 0; down--) {
yield [x, --y]
}
step++
}
}
function manhattanDistance (target) {
let coordinates
const location = traverse()
for (let i = 0; i < target; i++) {
coordinates = location.next().value
}
return Math.abs(coordinates[0]) + Math.abs(coordinates[1])
}
Solution
Why use a generator?
-
Rather than create an object that maintain internal state by variables assignment (explicit)
- Internal state can be implicit in the position of the yield statement
- Behave sort of like the goto statement from old school imperative programming
Another example
AoC 2015 Day 14:
Reindeer Olympics
function * reindeer (speed, fly, rest) {
let distance = 0
while (true) {
for (let i = 0; i < fly; i++) {
distance += speed
yield distance
}
for (let i = 0; i < rest; i++) {
yield distance
}
}
}
RegEx
AoC 2016 Day 7:
Internet Protocol Version 7
Task:
Find ABBA
Solution:
function containsAbba (str) {
const match = str.match(/([a-z])([a-z])\2\1/)
if (!match) return false
if (match[1] !== match[2]) return true
return containsAbba(str.slice(match.index + 1))
}
Tail Call
Optimization
AoC 2015 Day 10:
Elves Look, Elves Say
Solution
// no proper tail call
function lookNsay (str) {
const match = str.match(/(.)(\1*)/)
if (match[0].length === str.length) return match[0].length + match[1]
return match[0].length + match[1] + lookNsay(str.slice(match[0].length))
}
// with proper tail call
function lookNsay (str, prefix = '') {
const match = str.match(/(.)(\1*)/)
if (match[0].length === str.length) return prefix + match[0].length + match[1]
return lookNsay(str.slice(match[0].length), prefix + match[0].length + match[1])
}
Limitations
- 'use strict'
- --harmony ⚐
- Only node 6 & 7
Workaround
- rewrite recursion into iteration
- use a stack
- eg. Depth-first-search
Enumeration
AoC 2015 Day 17:
No Such Thing as Too Much
Task:
Given containers of various sizes (eg. 20, 15, 10, 5, and 5).
Find how many different combinations total up to a given capacity.
For example if required total is 25, four ways are possible:
- 15 and 10
- 20 and 5 (the first 5)
- 20 and 5 (the second 5)
- 15, 5, and 5
Solution:
Enumerate all possible assignment combination
Eg. 000, 001, 010, 011, 100, 101, 110, 111
AoC 2015 Day 9:
All in a Single Night
Task:
Given distances between each location pair, find the
shortest route where every location is visited exactly once.
Solution:
Enumerate all possible permutation
Eg. ABC, ACB, BAC, BCA, CAB, CBA
AoC 2015 Day 13:
Knights of the Dinner Table
Task:
Find the ideal seating arrangement around a round
dining table
Solution:
Fix a start point and enumerate all possible
permutation that start and end with that point. Eg.
- A > B > C > D > A
- A > B > D > C > A
- A > C > B > D > A
- A > C > D > B > A
AoC 2015 Day 15:
Science for Hungry People
Task:
With exactly 100 teaspoons of ingredients. Find
the ideal recipe
Solution:
Enumeration all possible split combinations. Eg.
0/0/100 | |||
0/1/99 | 1/0/99 | ||
0/2/98 | 1/2/98 | 2/1/98 | |
0/3/97 | 1/2/97 | 2/1/97 | 3/0/97 |
Enumerate Assignment Combinations
- Count up from 0 to
- Convert number to binary representation using Number.prototype.toString(2)
- Left pad string with zeros
- Convert to array using String.prototype.split('')
function getAssignments (items, groups = 2) {
const combinations = []
const nCombinations = Math.pow(groups, items)
const zeroPad = '0'.repeat(items)
for (let i = 0; i < nCombinations; i++) {
const combiString = (zeroPad + i.toString(groups)).slice(-items)
const combination = combiString.split('')
if (groups === 2) combinations.push(combination.map(v => +v))
else combinations.push(combination)
}
return combinations
}
Implementation
Enumerate Split Combination
- Instead of enumerating units, enumerate cuts. Eg.
- If you have k different ingredients, the number of cuts to enumerate will be k - 1
- Pick any position that is between last cut's position and N inclusive
- Cuts can overlap
- Subtract adjacent cut index to get units of ingredient
- i.e.
1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
---|
First cut is always at 0
Last cut is always at N
Second cut at 2
Third cut also at 2
Fourth cut at 5
Implementation
function getSplitCombinations (total, items) {
const combinations = []
function recurse (division, i) {
if (division.length < items - 1) {
while (i <= total) {
recurse(division.concat(i), i)
i++
}
} else {
division = [0, ...division, total]
const combination = []
for (let i = 1; i < division.length; i++) {
combination.push(division[i] - division[i - 1])
}
combinations.push(combination)
}
}
recurse([], 0)
return combinations
}
More enumeration helpers
How about lazy enumeration?
AoC 2015 Day 24:
It Hangs in the Balance
Task:
Split cargo (28 pieces) into 3 groups of equal weight.
Pick combination with least items in the first group.
Analysis:
Total possible combinations is !!!
Takes too long to enumerate every possible combination.
Need a way to exit early once a match is found.
Iterables
const iterable = {
[Symbol.iterator]: function* () {
let n = 0
while (true) {
yield n++
}
}
}
for (let n of iterable) {
if (n >= 10) break
}
Example
// lazy version of assignment enumerator
function getAssignments (items, groups = 2) {
const nCombinations = Math.pow(groups, items)
const zeroPad = '0'.repeat(items)
return {
[Symbol.iterator]: function* () {
for (let i = 0; i < nCombinations; i++) {
const combiString = (zeroPad + i.toString(groups)).slice(-items)
const combination = combiString.split('')
if (groups === 2) yield combination.map(v => +v)
else yield combination
}
}
}
}
// wrong use
const iterable = getAssignment(n)
Array.from(iterable)
[...iterable]
// correct use
const iterable = getAssignment(n)
for (let item of iterable) {
if (/* condition met */) return item
}
Graph search
Breadth-First-Search
Depth-First-Search
function bfs (root) {
const visited = {}
const unvisited = []
unvisited.push([root, 0])
while (unvisited.length > 0) {
const [next, steps] = unvisited.shift()
if (next in visited) continue
visited[next] = 1
if (found) return steps
next.children
.forEach(child => {
unvisited.push([child, steps + 1]
})
}
}
function dfs (root) {
const visited = {}
const unvisited = []
unvisited.push([root, 0])
while (unvisited.length > 0) {
const [next, steps] = unvisited.pop()
if (next in visited) continue
visited[next] = 1
if (found) return steps
next.children
.sort(sortFunc)
.reverse()
.forEach(child => {
unvisited.push([child, steps + 1]
})
}
}
Dealing with coding task of the assembly language kind
- Use Chrome inspector
- node --inspect-brk script.js
- Add debugger statement
Wrapping up
- Work it out on paper
- Premature optimization
- Readability & Extensibility > Efficiency & Speed
- Some knowledge of computation complexity is helpful
All solutions
150 coding challenges in 25 days
By yongjun21
150 coding challenges in 25 days
What I've learned from attempting Advent of Code 2017, 2016 & 2015 all at one go.
- 1,677