greathmaster

<iframe src="https://app.box.com/embed/s/0dtgl906gz1vsrqp1w6paqxhwuarn1mv?sortColumn=date" width="500" height="400" frameborder="0" allowfullscreen webkitallowfullscreen msallowfullscreen></iframe>

greathmaster

<iframe src="https://amedmarket.app.box.com/embed/s/z5ywvjsia9qpllfj1yaq7gni6398ql6g?sortColumn=date" width="800" height="550" frameborder="0" allowfullscreen webkitallowfullscreen msallowfullscreen></iframe>

greathmaster

# Set color codes for directory and Git branch
dir_color='%F{blue}'
branch_color='%F{green}'

# Set Git branch state color codes
uncommitted_color='%F{red}'
untracked_color='%F{yellow}'
ahead_color='%F{cyan}'
behind_color='%F{magenta}'
conflicted_color='%F{red}'

greathmaster

function debounceFetch (fn, delay, onCancel = () => {}) {
  let timer = 0, isFetching = false
  
  return function (...args) {
    clearTimeout(timer)
    
    if (isFetching) {
      onCancel()
      isFetching = false
    }

greathmaster

// React: How to Correctly Build a Counter/Clock/Timer Component Using Hooks
//Taken from: https://overreacted.io/making-setinterval-declarative-with-react-hooks/

import React, { useState, useEffect, useRef } from "react";
import ReactDOM from "react-dom";

function Counter() {
  const [count, setCount] = useState(0);
  const savedCallback = useRef();

greathmaster

class UnionFind {
	constructor(size) {
		this.nodes = new Array(size).fill(-1)
	}

	find(i) {
		while(true) {
			if(i === this.nodes[i]) this.nodes[i] = -1
			if(this.nodes[i] < 0) {
				return i;

greathmaster

//Iterative solution for Leetcode Problem 114.
//Currently having out of memory issues on Leetcode, but runs flawlessly on local Node setup
var flatten = function(root) {

	if(root === null) return null;

	const DOWN = 1; //represents performing a recursive function call
	const UP = 2;   //represents returning from a recursive function call

	let stack = [];

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//Recursive Quicksort using Lomuto's partitioning method
function lomutosPartition(array, start, end) {
	let i = start;
	let pivot = array[end]

	for(let j = start; j < end; j++) {
		if(pivot > array[j]) {
			[array[j], array[i]] = [array[i], array[j]];
			i++;
		}

greathmaster

/*
	Given an array of 1's and 0's in the following format:
	[1, 1, 1, 0, 0]
	
	Use binary search to determine how many 1's are in the array.
	You can assume the 1's will always appear together and at the
	beginning of the array.
*/

function bs(array) {

greathmaster

license: gpl-3.0