Hi I did not pass the all the tests, only 9/16. so I want to know what is the problem in my code
problem link:https://www.hackerrank.com/challenges/electronics-shop/problem
function getMoneySpent(keyboards, drives, b) {
let arr = [];
let lenOfArr1 = keyboards.length;
let lenOfArr2 = drives.length;
let j = drives.length;
arr = keyboards.slice(0);
for (let number of drives) {
arr.push(number);
}
return (challenge(arr, lenOfArr1, b));
}
function challenge(arr, lenOfArr1, m) {
let result = [];
let j = lenOfArr1;
for (let i = 0; i < lenOfArr1; i++) {
if (arr[i] >= m || arr[j] >= m) return -1;
if (arr[i] + arr[j] < m) {
result.push(arr[i] + arr[j]);
}
i--;
j++;
if (j == arr.length) {
i++;
j = lenOfArr1;
}
}
if (result.length == 0) return -1;
return result.sort()[result.length - 1];
}
From the given constraints in the problem statement, you don't need such a complex solution.
Here is the Algorithm:
Initialize a variable maxValue to have value as -1.
Start two for loops over drives and keyboards and take all combinations and sum the value of each drive with each keyboard.
Inside the for loops, check if the sum of drive + keyboard and it should be less than or equal to the money Monica has and keep track of the maximum value you get from any combination in maxValue.
After the code computation, return maxValue.
Code:-
function getMoneySpent(keyboards, drives, b) {
let maxValue = -1;
for (let drive of drives) {
for(let keyboard of keyboards) {
let cost = drive + keyboard;
if(cost > maxValue && cost <= b) {
maxValue = cost;
}
}
}
return maxValue;
}
Overall Time complexity - O(n^2).
Hope this helps!
The following line causes your code to return -1 when there might be a solution:
if (arr[i] >= m || arr[j] >= m) return -1;
Even though one price may be above budget, you need to consider that there might be cheaper items available. You might even already have solutions in result, or they might still be found in a later iteration.
It is not clear to me why you first create a merged array of both input arrays. It does not seem to bring any benefit.
If you sort the input arrays, you can use a two-pointer system where you decide which one to move depending on whether the current sum is within limits or it is too great:
function getMoneySpent(keyboards, drives, b) {
keyboards = [...keyboards].sort((a, b) => b - a) // descending
drives = [...drives].sort((a, b) => a - b); // ascending
let k = keyboards.length - 1; // cheapest
let d = drives.length - 1; // most expensive
let maxSum = -1;
while (d >= 0 && k >= 0) {
let sum = keyboards[k] + drives[d];
if (sum === b) return b;
if (sum < b) {
if (sum > maxSum) maxSum = sum;
k--; // will increase sum
} else {
d--; // will decrease sum
}
}
return maxSum;
}
I have a problem I am sitting on for the past few days.
I want to write an optimal (in JS) program for verifying if a number is a Palindrome.
My current approach:
function isPalidrom2(pali){
//MOST time consuming call - I am splitting the digit into char array.
var digits = (""+pali).split("");
//To get the length of it.
var size = digits.length;
var isPali = true;
for(var i = 0; i<Math.floor(size/2); i++){
//I am comparing digits (first vs last, second vs last-1, etc.) one by one, if ANY of the pars is not correct I am breaking the loop.
if(parseInt(digits[i]) != parseInt(digits[size-i-1])){
isPali = false;
break;
}
}
return isPali;
}
It's not optimal. The biggest amount of time I am waisting is the change from INT to STRING.
And I am out of ideas
- I tried to understand the BIT operators but I can't.
- I tried to google and look for alternative approaches - but I can't find anything.
- I tried to play with different algorithms - but this one is the fastest I was able to apply.
So in short - my question is:
"how can I make it faster?"
EDIT:
So the task I want to solve:
Find all of the prime numbers within the range of all five digit numbers.
Among all of the multiplies (i*j) they are between them, find the most significant palindrome.
My current approach:
function isPrime(number){
var prime = true;
var i
for(i = 2; i<= number/2; i++){
if((number%i)==0){
prime = false;
break;
}
}
return prime;
}
function get5DigitPrimeNr(){
var a5DigitsPrime = [];
var i;
for(i = 10000; i<100000; i++){
if(isPrime(i)){
a5DigitsPrime.push(i)
}
}
return a5DigitsPrime;
}
function isPalidrom(pali){
var digits = (""+pali).split("");
//we check if first and last are the same - if true, we can progress
size = digits.length;
return
(digits[0]==digits[size-1]) &&
(parseInt(digits.slice(1, Math.floor(size/2)).join("")) ==
parseInt(digits.reverse().slice(1, Math.floor(size/2)).join("")))
}
function isPalidrom2_48s(str) {
var str = str.toString();
const lower = str.substr(0, Math.floor(str.length / 2));
const upper = str.substr(Math.ceil(str.length / 2));
return lower.split("").reverse().join("") === upper;
}
function isPalidrom_22s(pali){
var digits = (""+pali).split("");
var size = digits.length;
for(var i = 0; i<Math.floor(size/2); i++){
//console.log("I am comparing: "+i+", and "+(size-i-1)+" elements in array")
//console.log("I am comparing digit: "+digits[i]+", and "+digits[(size-i-1)]+"")
if(digits[i] !== digits[size-i-1]){
//console.log('nie sa rowne, koniec')
return false;
}
}
return true;
}
function isPalidrom2_80s(pali){
return parseInt(pali) == parseInt((""+pali).split("").reverse().join(""))
}
function runme(){
var prime5digits = get5DigitPrimeNr();
var size = prime5digits.length;
var max = 0;
var message = "";
for(var i = 0; i<size; i++){
for(var j = 0; j<size; j++){
var nr = prime5digits[i]*prime5digits[j];
if(nr>max && isPalidrom2(nr)){
max = nr;
message = 'biggest palidrome nr: '+nr+', made from numbers: '+prime5digits[i]+' x '+prime5digits[j];
}
}
}
console.log(message)
}
function timeMe(){
var t0 = performance.now();
runme();
var t1 = performance.now();
console.log("Function took " + millisToMinutesAndSeconds(t1 - t0) + " s to find the perfect palindrom.")
}
//helper functons:
function millisToMinutesAndSeconds(millis) {
var minutes = Math.floor(millis / 60000);
var seconds = ((millis % 60000) / 1000).toFixed(0);
return minutes + ":" + (seconds < 10 ? '0' : '') + seconds;
}
To keep the spirit of you code, you could exit the loop with return instead of break and use the string directly without converting to an array. Strings have, as arrays, the possibility of an access single character with an index.
function isPalidrom2(value) {
var digits = value.toString(),
length = digits.length,
i, l;
for (i = 0, l = length >> 1; i < l; i++) {
if (digits[i] !== digits[length - i - 1]) {
return false;
}
}
return true;
}
console.log(isPalidrom2(1));
console.log(isPalidrom2(12));
console.log(isPalidrom2(1221));
console.log(isPalidrom2(123));
The fastest is probably to rely on javascripts native methods:
function isPalindrome(str) {
const lower = str.substr(0, Math.floor(str.length / 2));
const upper = str.substr(Math.ceil(str.length / 2));
return lower.split("").reverse().join("") === upper;
}
Or leave away all unneccessary conversions from your code:
function isPlaindrome(str) {
const half = str.length / 2;
for(var i = 0; i < half; i++)
if(str[i] !== str[str.length - i - 1])
return false;
return true;
}
If you're trying to speed things up, you could shave a few more seconds off by optimising your isPrime(n) function.
You don't need to check every factor, only the prime factors less than sqrt(n)
If you check every number from 2 to 99999 in ascending order, you can store the results as you go, so you don't need to keep re-calculating the list of previous primes
Something like this:
var savedPrimes = [2]
function isPrime(n){
// call this method with increasing values of n (starting at 2), saving primes as we go,
// so we can safely assume that savedPrimes contains all primes less than n
for(var i=0; i<savedPrimes.length; i++)
{
var f = savedPrimes[i];
if ((n % f) == 0)
return false; // found a factor
if (f*f>=n)
break; // stop checking after f >= sqrt(n)
}
// no prime factors - we found a new prime
savedPrimes.push(n);
return true;
}
function get5DigitPrimeNr(){
var a5DigitsPrime = [];
var i;
// first find all the primes less than 10000
for(i = 3; i<10000; i++){
isPrime(i);
}
// now find (and keep) the rest of the primes up to 99999
for(i = 10000; i<100000; i++){
if(isPrime(i)){
a5DigitsPrime.push(i)
}
}
return a5DigitsPrime;
}
EDIT - when I run your code with this method, I get a time of 10 sec
Code :
There are multiple methods that you can use (dunno if they are optimal) :
Palindrom = _ => (_=''+_) === [..._].sort(_=>1).join``
Some more :
let isPalindrome = __ => (_=(__=__+'').length)==0||_==1?!0:__[0]==__.slice(-1)?isPalindrome(__.slice(1,-1)):!1
let isPalindrome = (s,i) => (i=i||0)<0||i>=(s=''+s).length>>1||s[i]==s[s.length-1-i]&&isPalindrome(s,++i);
let isPalindrome = (str) => {
var len = ~~((str=str+'').length / 2);
for (var i = 0; i < len; i++)
if (str[i] !== str[str.length - i - 1])
return false;
return true;
}
Updated now takes numbers as input
Find the least common multiple of the provided parameters using Table Method that can be evenly divided by both, as well as by all sequential numbers in the range between these parameters. There will only be two parameters. For ex [1,3], find the lcm of 1,2,3.
Note - It might create an infinite loop
function smallestCommons(arr) {
var nums = [];
var multiples = [];
if(arr[0]>arr[1]) {
var bigger = arr[0];
} else {
var bigger = arr[1];
}
for(var i=bigger;i>0;i--) {
nums.push(i);
console.log(i);
}console.log(nums + " nums");
var sums = 0;
while(sums != nums.length) {
for(var k=0;k<nums.length;k++) {
if(nums[k] % 2 === 0) {
nums[k] = nums[k]/2;
multiples.push(2);
} else if(nums[k] % 3 === 0) {
nums[k] = nums[k]/3;
multiples.push(3);
}else if(nums[k] % 5 === 0) {
nums[k] = nums[k]/5;
multiples.push(5);
}else if(nums[k] % 7 === 0) {
nums[k] = nums[k]/7;
multiples.push(7);
}else if(nums[k] === 1) {
break;
}else {
nums[k] = nums[k]/nums[k];
multiples.push(nums[k]);
}
}
for(var j = bigger; j>0;j--) {
sums = sums + nums[j];
}
}
var scm = [multiples].reduce(function(a,b){console.log(a*b)}); return scm
}
smallestCommons([1,5]);
I found this to be a simple solution, It works wonders;
Loop through all possible numbers, beginning with lower bound input (var i)
for every number, test divisibility by each number between and including input bounds (var j)
if i meets all criteria return it as answer, otherwise increment i by 1 and try again
click here for explanation of ? operator in variable initialization
function smallestCommons(arr) {
//set variables for upper and lower bounds
//incase they aren't entered in ascending order
var big = arr[0] < arr[1] ? arr[1]:arr[0],
small = arr[0] < arr[1] ? arr[0]:arr[1],
i = small;
//loop through all numbers, note the possibility of an infinite loop
while(true){
//test each number for divisibility by by both upper and lower
//bounds, as well as by all sequential numbers inbetween
for(var j = small; j <= big; j++){
if(i % j === 0){
if(j===big){
return i;
}
}else {
break;
}
}
i++;
}
}
smallestCommons([1,5]); //60
What you need is find the LCM in range (n, m) ?
Finding least common multiples by prime factorization seems better.
You can use Legendre's formula to find all prime factors of n! and m! , then just do a simple subtraction.
I am new to Javascript, I am doing a coding challenge to learn more about the language. This is not school related or anything like that, totally for my own personal growth. Here is the challenge:
Return the sum of all odd Fibonacci numbers up to and including the
passed number if it is a Fibonacci number.
I have spent the past 2 evenings working on solving this challenge. When I run my code using underscore.js it works. When I use Ramda.js it says NaN. I would think both would return NaN. I'm very surprised that I can get the correct answer from one and not the other. Any insights would be greatly appreciated!
var R = require('ramda');
function sumFibs(num) {
var fib_Arr = [];
var new_Arr = [];
var total = 0;
// I use this to tell if the fib num is greater than 2
var the_Bit = "false";
// This is used to keep track of when to stop the loop
var fib_Num = 0;
// THIS WORKS FROM HERE
// This loop generates a list of fibonacci numbers then pushes them to the fib_Arr
for(var i = 0; total < num; i++){
if (i < 1){
fib_Arr.push(0);
}
else if (i === 1){
fib_Arr.push(i);
fib_Arr.push(1);
}
else if (i === 2){
fib_Arr.push(2);
the_Bit = "true";
}
else if (the_Bit === "true"){
temp_Arr = R.last(fib_Arr,2);
temp_Arr = temp_Arr[0] + temp_Arr[1];
fib_Arr.push(temp_Arr);
total = R.last(fib_Arr);
}
// Generating the fib Array works TO HERE!!!!
}
// console.log(fib_Arr); // Print out the generated fibonacci array
// if last Array element is greater than the original in
var last_Element = R.last(fib_Arr);
if (last_Element > num){
console.log("The last element of the array is bigger!");
fib_Arr.splice(-1,1); // This removes the last item from the array if it is larger than the original num input
}
// This loop removes all of the EVEN fibonacci numbers and leaves all of the ODD numbers
for (var j = 0; j < fib_Arr.length; j++){
if (fib_Arr[j] % 2 !== 0){
new_Arr.push((fib_Arr[j]));
}
}
// This checks if the original input num was a
if (num % 2 !== 0){
new_Arr.push(num);
}
else{
console.log("The original num was not a Fibonacci number!");
}
// if last Array element is the same as the original input num
var last = R.last(fib_Arr);
if (last === num){
console.log("Removing the last element of the array!");
new_Arr.splice(-1,1); // This removes the last item from the array if it is the same as the original num input
}
// Now to add all of the numbers up :-)
for (var k = 0; k < new_Arr.length; k++){
console.log("This is fib_Num: " + fib_Num);
// console.log(fib_N`);
fib_Num = fib_Num += new_Arr[k];
}
return fib_Num;
}
// TEST CASES:
// console.log(sumFibs(75025)); //.to.equal(135721);
console.log(sumFibs(75024)); //.to.equal(60696);
You have a problem on these lines :
temp_Arr = R.last(fib_Arr,2);
temp_Arr = temp_Arr[0] + temp_Arr[1];
Besides the fact that R.last does not take a second argument (that will not fail though), you are using temp_arr as an array, when it is a number. Therefore, temp_arr gets a NaN value.
You are probably looking for R.take (combined with R.reverse) or R.slice.
By changing :
temp_Arr = R.last(fib_Arr,2);
with :
temp_Arr = R.take(2, R.reverse(fib_Arr));
or with :
temp_Arr = R.slice(fib_Arr.length - 2, fib_Arr.length)(fib_Arr);
or with (bonus play with a reduce from the right) :
temp_Arr = R.reduceRight(function(arr, elem) {
return arr.length < 2 ? [elem].concat(arr) : arr;
}, [])(fib_Arr);
We get :
sumFibs(75024) === 60696
For the record, here's how you do this problem:
function fibSumTo(n) {
var f1 = 1, f2 = 1, sum = 1, t;
while (f2 <= n) {
if (f2 & 1) sum += f2;
t = f1 + f2;
f1 = f2;
f2 = t;
}
return sum;
}
There's really no need for any sort of library because there's really no need for any sort of data structure.
var _ = require('underscore');function sumUpFibs (number){
arr_of_fibs = [1,1];
current = 1; //cursor for previous location
while (true){
var num = arr_of_fibs[current] + arr_of_fibs[current - 1];
if (num <= number) {
arr_of_fibs.push(num);
current++;
} else {
break;
}
}
console.log(arr_of_fibs);
var total = 0;
_.each(arr_of_fibs, function(fib){
total += fib;
})
return total;}console.log(sumUpFibs(75025));
This may be a better implementation... Though I know you're just starting so I don't want to come off as mean : D.... Also, maybe check your test cases too.
I am just starting JS, and understand the concept of finding a factor. However, this snippet of code is what I have so far. I have the str variable that outputs nothing but the first factor which is 2. I am trying to add each (int) to the str as a list of factors. What's the wrong in below code snippet?
function calculate(num) {
var str = "";
var int = 2;
if (num % int == 0) {
str = str + int;
int++;
} else {
int++;
}
alert(str);
}
calculate(232);
UPDATED ES6 version:
As #gengns suggested in the comments a simpler way to generate the array would be to use the spread operator and the keys method:
const factors = number => [...Array(number + 1).keys()].filter(i=>number % i === 0);
console.log(factors(36)); // [1, 2, 3, 4, 6, 9, 12, 18, 36]
ES6 version:
const factors = number => Array
.from(Array(number + 1), (_, i) => i)
.filter(i => number % i === 0)
console.log(factors(36)); // [1, 2, 3, 4, 6, 9, 12, 18, 36]
https://jsfiddle.net/1bkpq17b/
Array(number) creates an empty array of [number] places
Array.from(arr, (_, i) => i) populates the empty array with values according to position [0,1,2,3,4,5,6,7,8,9]
.filter(i => ...) filters the populated [0,1,2,3,4,5] array to the elements which satisfy the condition of number % i === 0 which leaves only the numbers that are the factors of the original number.
Note that you can go just until Math.floor(number/2) for efficiency purposes if you deal with big numbers (or small).
As an even more performant complement to #the-quodesmith's answer, once you have a factor, you know immediately what its pairing product is:
function getFactors(num) {
const isEven = num % 2 === 0;
const max = Math.sqrt(num);
const inc = isEven ? 1 : 2;
let factors = [1, num];
for (let curFactor = isEven ? 2 : 3; curFactor <= max; curFactor += inc) {
if (num % curFactor !== 0) continue;
factors.push(curFactor);
let compliment = num / curFactor;
if (compliment !== curFactor) factors.push(compliment);
}
return factors;
}
for getFactors(300) this will run the loop only 15 times, as opposed to +-150 for the original.
#Moob's answer is correct. You must use a loop. However, you can speed up the process by determining if each number is even or odd. Odd numbers don't need to be checked against every number like evens do. Odd numbers can be checked against every-other number. Also, we don't need to check past half the given number as nothing above half will work. Excluding 0 and starting with 1:
function calculate(num) {
var half = Math.floor(num / 2), // Ensures a whole number <= num.
str = '1', // 1 will be a part of every solution.
i, j;
// Determine our increment value for the loop and starting point.
num % 2 === 0 ? (i = 2, j = 1) : (i = 3, j = 2);
for (i; i <= half; i += j) {
num % i === 0 ? str += ',' + i : false;
}
str += ',' + num; // Always include the original number.
console.log(str);
}
calculate(232);
http://jsfiddle.net/r8wh715t/
While I understand in your particular case (calculating 232) computation speed isn't a factor (<-- no pun intended), it could be an issue for larger numbers or multiple calculations. I was working on Project Euler problem #12 where I needed this type of function and computation speed was crucial.
function calculate(num) {
var str = "0";
for (var i = 1; i <= num; i++) {
if (num % i == 0) {
str += ',' + i;
}
}
alert(str);
}
calculate(232);
http://jsfiddle.net/67qmt/
Below is an implementation with the time complexity O(sqrt(N)):
function(A) {
var output = [];
for (var i=1; i <= Math.sqrt(A); i++) {
if (A % i === 0) {
output.push(i);
if (i !== Math.sqrt(A)) output.push(A/i);
}
}
if (output.indexOf(A) === -1) output.push(A);
return output;
}
here is a performance friendly version with complexity O(sqrt(N)).
Output is a sorted array without using sort.
var factors = (num) => {
let fac = [], i = 1, ind = 0;
while (i <= Math.floor(Math.sqrt(num))) {
//inserting new elements in the middle using splice
if (num%i === 0) {
fac.splice(ind,0,i);
if (i != num/i) {
fac.splice(-ind,0,num/i);
}
ind++;
}
i++;
}
//swapping first and last elements
let temp = fac[fac.length - 1];
fac[fac.length - 1] = fac[0];
fac[0] = temp;
// nice sorted array of factors
return fac;
};
console.log(factors(100));
Output:
[ 1, 2, 4, 5, 10, 20, 25, 50, 100 ]
This got me an 85% on Codility (Fails on the upperlimit, over a billion).
Reducing the input by half doesn't work well on large numbers as half is still a very large loop. So I used an object to keep track of the number and it's half value, meaning that we can reduce the loop to one quarter as we work from both ends simultaneously.
N=24 becomes: (1&24),(2&12),(3&8),(4&6)
function solution(N) {
const factors = {};
let num = 1;
let finished = false;
while(!finished)
{
if(factors[num] !== undefined)
{
finished = true;
}
else if(Number.isInteger(N/num))
{
factors[num] = 0;
factors[N/num]= 0;
}
num++
}
return Object.keys(factors).length;
}
Using generators in typescript in 2021
function* numberFactorGenerator(number: number): Generator<number> {
let i: number = 0;
while (i <= number) {
if (number % i === 0) {
yield i;
}
i++;
}
}
console.log([...numberFactorGenerator(12)]); // [ 1, 2, 3, 4, 6, 12 ]
function factorialize(num) {
var result = '';
if( num === 0){
return 1;
}else{
var myNum = [];
for(i = 1; i <= num; i++){
myNum.push(i);
result = myNum.reduce(function(pre,cur){
return pre * cur;
});
}
return result;
}
}
factorialize(9);
I came looking for an algorithm for this for use in factoring quadratic equations, meaning I need to consider both positive and negative numbers and factors. The below function does that and returns a list of factor pairs. Fiddle.
function getFactors(n) {
if (n === 0) {return "∞";} // Deal with 0
if (n % 1 !== 0) {return "The input must be an integer.";} // Deal with non-integers
// Check only up to the square root of the absolute value of n
// All factors above that will pair with factors below that
var absval_of_n = Math.abs(n),
sqrt_of_n = Math.sqrt(absval_of_n),
numbers_to_check = [];
for (var i=1; i <= sqrt_of_n; i++) {
numbers_to_check.push(i);
}
// Create an array of factor pairs
var factors = [];
for (var i=0; i <= numbers_to_check.length; i++) {
if (absval_of_n % i === 0) {
// Include both positive and negative factors
if (n>0) {
factors.push([i, absval_of_n/i]);
factors.push([-i, -absval_of_n/i]);
} else {
factors.push([-i, absval_of_n/i]);
factors.push([i, -absval_of_n/i]);
}
}
}
// Test for the console
console.log("FACTORS OF "+n+":\n"+
"There are "+factors.length+" factor pairs.");
for (var i=0; i<factors.length; i++) {
console.log(factors[i]);
}
return factors;
}
getFactors(-26);
function calculate(num){
var str = "0" // initializes a place holder for var str
for(i=2;i<num;i++){
var num2 = num%i;
if(num2 ==0){
str = str +i; // this line joins the factors to the var str
}
}
str1 = str.substr(1) //This removes the initial --var str = "0" at line 2
console.log(str1)
}
calculate(232);
//Output 2482958116
Here's an optimized solution using best practices, proper code style/readability, and returns the results in an ordered array.
function getFactors(num) {
const maxFactorNum = Math.floor(Math.sqrt(num));
const factorArr = [];
let count = 0; //count of factors found < maxFactorNum.
for (let i = 1; i <= maxFactorNum; i++) {
//inserting new elements in the middle using splice
if (num % i === 0) {
factorArr.splice(count, 0, i);
let otherFactor = num / i; //the other factor
if (i != otherFactor) {
//insert these factors in the front of the array
factorArr.splice(-count, 0, otherFactor);
}
count++;
}
}
//swapping first and last elements
let lastIndex = factorArr.length - 1;
let temp = factorArr[lastIndex];
factorArr[lastIndex] = factorArr[0];
factorArr[0] = temp;
return factorArr;
}
console.log(getFactors(100));
console.log(getFactors(240));
console.log(getFactors(600851475143)); //large number used in Project Euler.
I based my answer on the answer written by #Harman
We don't have to loop till end of the given number to find out all the factors. We just have to loop till reaching the given number's squareroot. After that point we, can figure out the rest of the factors by dividing the given number with the already found factors.
There is one special case with this logic. When the given number has a perfect square, then the middle factor is duplicated. The special case is also handled properly in the below code.
const findFactors = function (num) {
const startingFactors = []
const latterFactors = []
const sqrt = Math.sqrt(num)
for (let i = 1; i <= sqrt; i++) {
if (num % i == 0) {
startingFactors.push(i)
latterFactors.push(num / i)
}
}
// edge case (if number has perfect square, then the middle factor is replicated, so remove it)
if (sqrt % 1 == 0) startingFactors.pop()
return startingFactors.concat(latterFactors.reverse())
}
function factorialize(num) {
if(num === 0)
return 1;
var arr = [];
for(var i=1; i<= num; i++){
arr.push(i);
}
num = arr.reduce(function(preVal, curVal){
return preVal * curVal;
});
return num;
}
factorialize(5);