See problem here.

If this problem was hard for you, you are not alone.. I hope to clear things up in this post.

Background

First, let's go over some general graph traversal algorithms.

def bfs(graph, start, end):
    # return True if there exists path from start to end
    q = [start …

See problem here.

The basic idea is to calculate the total cost for some positions to position m.

function fuel_cost(positions, m, method) {
    var cost = 0;
    for (var c of positions) {
        if (method === "constant") {
            cost += Math.abs(m - c);
        }
        else { // "step"
            cost += Math.abs(m - c) * (Math.abs(m - c …

See problem here.

My initial approach was to track all of the fish. We decrease number of days for each fish by 1 until it reaches 0. Then we reset it to 6 for existing fish and add a new fish with value 8. This works fine but it is …

See problem here.

First let's consider how to parse the input:

function parse_input(pi) {
    var arr = pi.split('\n').map(String);
    var segments = [];
    for (var line of arr) {
        var ab = line.split('->');
        var a = ab[0].split(',');
        var b = ab[1].split(',');
        segments.push({
            "x1": parseInt(a[0]), "y1": parseInt …

See problem here.

For part1, we need to do the following:

• define a function which can parse the order and boards from the puzzle input, i.e. parse_input
• define a function which marks squares equal to e in bingo board as null, i.e. mark_board
• define a function which can …

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For part1, we need to loop through the lines column by column and count the frequency of each digit in the column. If zero is more common, our first string s1 appends a zero. Otherwise, our first string s1 appends a one. Our second string is the …

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For part1, we need to keep track of horizontal position x and depth y.

function part1(pi) {
    var arr = pi.split('\n').map(String);
    var x = 0;
    var y = 0;
    for(var command of arr) {
        var comp = command.split(" ");
        var direction = comp[0];
        var magnitude = parseInt(comp …

See problem here.

For part1, we need to determine the number of entries larger than the previous entry.

function part1(pi) {
    var arr = pi.split('\n').map(Number);
    var ans = 0;
    var pe = null;
    for (var e of arr) {
        if (pe != null && e > pe) ans += 1;
        pe = e;
    }
    console.log …