motss

const link = 'https://example.com';
const title = 'Example.com';

const FB_APP_ID = '<YOUR_FB_APP_ID';
const facebook = `https://www.facebook.com/dialog/share?app_id=${FB_APP_ID}&display=popup&href=${link}&title=${title}`;

const twitter = `https://twitter.com/intent/tweet?text=${title}&url=${link}`;

const email = `mailto:?subject=${title}&body=${link}`;

motss

// Time complexity: O(n) where n is the number of characters.
// Space complexity: O(1) for a few variables.
function alternatingCharacters(x) {
  const len = x.length;

  if (len < 2) return 0;

  let lastChar = x[0];
  let count = 0;
  let countA = lastChar === 'a' ? 1 : 0;

motss

// Time complexity: O(n ^ 2) where is the number of levels and
//                  another n is for the number of elements in each level.
// Space complexity: O(2n + 1) to hold all elements, which is linear space.
function pascalTriangle(level) {
    if (level === 0) return [];
    if (level === 1) return [1];
    if (level === 2) return [1, 1];

    const d = Array.from(Array(level), () => []);

motss

// Given an integer n, find the minimum number of steps
// to reach integer 1.
// At each step, you can:
// * n - 1
// * n / 2, if it is divisble by 2
// * n / 3, if it is divisble by 3

// n = 0: 0 // base case 0.
// n = 1: 0 // base case 1.
// n = 2: 1 // base case 2.

motss

// i
// j
// 4 2 1 3 6 0

//   i
// j
// 4 2 1 3 6 0
// 2 4 1 3 6 0 # 2 < 4

//     i

motss

{
// Reference: https://bit.ly/2yjWDeY
// Time complexity: O(n log n) on average or O(n^2) for worst-case
// Space complexity: O(n)

function quickSort(list) {
  if (list.length < 2) return list;

  const left = [];
  const right = [];

motss

// http://www.ecma-international.org/ecma-262/5.1/#sec-11.9.3
// http://es5.github.io/#x11.9.3

// Abstract/ loose equality comparison will do type coercion
// While strict equality comparison will compare the types

null == undefined; // true

motss

function bsr(arr, m) {
  let l = 0;
  let r = arr.length - 1;

  while (l <= r) {
    const mid = (l + (r - l) / 2) | 0;
    const val = arr[mid];

    if (val === m) return mid;
    if (m > val) l = mid + 1;

motss

Attempt to recreate flight search form on Google Flights

Demo URL

motss

function bfs<T>(tree: BSTNode<T>, fn: (n: T) => void) {
  const queue = [tree];
  
  while (queue.length) {
    const cur = queue.shift();
  
    if (cur.left) queue.push(cur.left);
    if (cur.right) queue.push(cur.right);
  
    fn(cur.value);