fazeelanizam13

class Node {
  // stores given value and points to nothing
  constructor (value) {
    this.value = value
    this.left = null
    this.right = null
  }
}

class Tree {

fazeelanizam13

class Node {
  // stores given value and points to nothing
  constructor (value) {
    this.value = value
    this.left = null
    this.right = null
  }
}

class Tree {

fazeelanizam13

class MaxHeap {
  // creates an array of given length
  constructor(length) {
    this.data = new Array(length)
    this.size = 0
  }

  // helpers

  rehash = () => {

fazeelanizam13

class MinHeap {
  // creates an array of given length
  constructor(length) {
    this.data = new Array(length)
    this.size = 0
  }

  // helpers

  rehash = () => {

fazeelanizam13

class Node {
  // stores given value and points to nothing
  constructor (value) {
    this.value = value
    this.left = null
    this.right = null
  }
}

class Tree {

fazeelanizam13

class Item {
  // stores given value and points to nothing
  constructor (value) {
    this.value = value
    this.next = null
  }
}

class Queue {
  // initial size is 0 and initially just has two reference items that store nothing and point to nothing

fazeelanizam13

class Item {
  // stores given value and points to nothing
  constructor (value) {
    this.value = value
    this.next = null
  }
}

class Stack {
  // initial size is 0 and initially just has a head node that stores nothing and points to nothing

fazeelanizam13

class Node {
  // stores given value and points to nothing
  constructor (value) {
    this.value = value
    this.next = null
  }
}

class LinkedList {
  // initial size is 0 and initially just has a head node that stores nothing and points to nothing

fazeelanizam13

const hashFunction = (key, tableSize) => {
  // index we are going to return
  let index = 0

  // get list of characters in key string as an array
  let numChars = String(key)
  
  // assign numChars size to index to start
  index = numChars.length
  

fazeelanizam13

// assumption - each input would have exactly one solution, and the same element may not be used twice in the input array

// brute force solution
function twoSumNaive(array, sum) {
  let indices = []

  for (let i = 0; i < array.length; i++) {
    for (let j = i + 1; j < array.length; j++) {
      if (array[i] + array[j] === sum) {
        indices.push(i)