Develop Reference

I generate 2 arrays of equal length (15), with a choice of 4 possible letters at each position - why do they come back 53% to 87% similar?

This is inspired by a Codecadamey project, where I'm learning JavaScript.
The problem is meant to simulate a simple DNA strand. It has 15 positions in the array, and those elements are either an A, T, C, or G to represent DNA bases.
There are no limits to the amount of times a single letter (base) can show up in the array.
I create 30 arrays that are made up of at least 60% C and/or G in any of the positions, these are meant to represent strong DNA strands.
I compare the strands to each other to see what % match they are. I consider a 'match' being true when there is the same base at the same position thisDNA[i] === comparisonDNA[i]
When I test a batch of 30 of these 'strong' samples to see the best and worst match levels, I find the results very tightly grouped (I ran it 3,000 times and the highest was 87%, lowest 53%), yet it is very easy for me to concieve of two samples that will survive that are a 0% match:
const sample1 = AGACCGCGCGTGGAG
const sample2 = TCTGGCGCGCACCTC
(obviously I've cheated by building these to be a 0% match and not randomly generating them)
Here's the full code: https://gist.github.com/AidaP1/0770307979e00d4e8d3c83decc0f7771
My question is as follows: Why is the grouping of matches so tight? Why do I not see anything below a 53% match after running the test thousands of times?
Full code:
// Returns a random DNA base
const returnRandBase = () => {
const dnaBases = ['A', 'T', 'C', 'G']
return dnaBases[Math.floor(Math.random() * 4)]
}
// Returns a random single stand of DNA containing 15 bases
const mockUpStrand = () => {
const newStrand = []
for (let i = 0; i < 15; i++) {
newStrand.push(returnRandBase())
}
return newStrand
}
const pAequorFactory = (num, arr) => { //factory function for new strand specimen
return {
specimenNum: num,
dna: arr,
mutate() {
//console.log(`old dna: ${this.dna}`) //checking log
let randomBaseIndex = Math.floor(Math.random() * this.dna.length) /// chooses a location to exchange the base
let newBase = returnRandBase()
while (newBase === this.dna[randomBaseIndex]) { // Rolls a new base until newBase !== current base at that position
newBase = returnRandBase()
}
this.dna[randomBaseIndex] = newBase;
//console.log(`New dna: ${this.dna}`) //checking log
return this.dna;
},
compareDNA(pAequor) { // compare two strands and output to the console
let thisDNA = this.dna;
let compDNA = pAequor.dna;
let matchCount = 0
for (i = 0; i < this.dna.length; i++) { //cycles through each array and log position + base matches on matchCount
if (thisDNA[i] === compDNA[i]) {
matchCount += 1;
};
};
let percMatch = Math.round(matchCount / this.dna.length * 100) //multiply by 100 to make 0.25 display as 25 in console log
console.log(`specimen #${this.specimenNum} and specimen #${pAequor.specimenNum} are a ${percMatch}% DNA match.`)
return percMatch;
},
compareDNAbulk(pAequor) { //as above, but does not log to console. Used for large arrays in findMostRelated()
let thisDNA = this.dna;
let compDNA = pAequor.dna;
let matchCount = 0
for (i = 0; i < this.dna.length; i++) {
if (thisDNA[i] === compDNA[i]) {
matchCount += 1;
};
};
let percMatch = Math.round(matchCount / this.dna.length * 100) //multiply by 100 to make 0.25 display as 25 in console log
return percMatch;
},
willLikelySurvive() { // looks for >= 60% of bases as either G or C
let countCG = 0;
this.dna.forEach(base => {
if (base === 'C' || base === 'G') {
countCG += 1;
}
})
//console.log(countCG) // testing
//console.log(this.dna) // testing
return countCG / this.dna.length >= 0.6
},
complementStrand() {
return this.dna.map(base => {
switch (base) {
case 'A':
return 'T';
case 'T':
return 'A';
case 'C':
return 'G';
case 'G':
return 'C';
}
})
} //close method
} // close object
} // close factory function
function generatepAequorArray(num) { // Generatess 'num' pAequor that .willLikelySurvive() = true
let pAequorArray = []; //result array
for (i = 0; pAequorArray.length < num; i++) {
let newpAequor = pAequorFactory(i, mockUpStrand()); // runs factory until there are 30 items in the result array
if (newpAequor.willLikelySurvive() === true) {
pAequorArray.push(newpAequor)
}
}
return pAequorArray;
}
function findMostRelated(array) { // champion/challenger function to find the most related specimens
let winningScore = 0;
let winner1;
let winner2;
for (let i = 0; i < array.length; i++) {
for (let j = i; j < array.length; j++) // j = i to halve the number of loops. i = 0, j = 5 is the same as i = 5, j = 0
if (i !== j) { // Do not check specimen against itself
let currentScore = array[i].compareDNAbulk(array[j]);
if (currentScore > winningScore) { // Challenger becomes the champion if they are a closer match
winningScore = currentScore;
winner1 = array[i].specimenNum;
winner2 = array[j].specimenNum;
}
}
}
let resultArray = [winner1, winner2, winningScore] // stored together for easy return
//console.log(`The most related specimens are specimen #${winner1} and specimen #${winner2}, with a ${winningScore}% match.`)
return resultArray
}
function multiArray(loops) { //test by running finding the closest match in 30 random 'will survive' samples, repaeated 1000 times. Returns the highest and lowest match across the 1000 runs
let highScore = 0;
let lowScore = 100
for (let i = 0; i < loops; i++) {
let pAequorArray = generatepAequorArray(30);
let currentArray = findMostRelated(pAequorArray);
highScore = Math.max(highScore, currentArray[2])
lowScore = Math.min(lowScore, currentArray[2])
}
return results = {
'high score': highScore,
'low score': lowScore
}
}
console.log(multiArray(10000))

I need to extract every nth char of a string in Javascript

Ive been reading everything online but its not exactly what I need
var x = 'a1b2c3d4e5'
I need something to get me to
using 1 the answer should be abcde
using 2 the answer should be 12345
using 3 the answer should be b3e
the idea behind it if using 1 it grabs 1 skips 1
the idea behind it if using 2 it grabs 2 skips 2
the idea behind it if using 3 it grabs 3 skips 3
I dont want to use a for loop as it is way to long especially when your x is longer than 300000 chars.
is there a regex I can use or a function that Im not aware of?
update
I'm trying to some how implement your answers but when I use 1 that's when I face the problem. I did mention trying to stay away from for-loops the reason is resources on the server. The more clients connect the slower everything becomes. So far array.filter seem a lot quicker.
As soon as I've found it I'll accept the answer.

As others point out, it's not like regular expressions are magic; there would still be an underlying looping mechanism. Don't worry though, when it comes to loops, 300,000 is nothing -
console.time('while')
let x = 0
while (x++ < 300000)
x += 1
console.timeEnd('while')
// while: 5.135 ms
console.log(x)
// 300000
Make a big string, who cares? 300,000 is nothing -
// 10 chars repeated 30,000 times
const s =
'abcdefghij'.repeat(30000)
console.time('while string')
let x = 0
let interval = 2
let values = []
while (x < s.length)
{ values.push(s[x])
x += interval
}
let result = values.join('')
console.timeEnd('while string')
// while string: 31.990ms
console.log(result)
console.log(result.length)
// acegiacegiacegiacegiacegiacegiacegiacegia...
// 150000
Or use an interval of 3 -
const s =
'abcdefghij'.repeat(30000)
console.time('while string')
let x = 0
let interval = 3
let values = []
while (x < s.length)
{ values.push(s[x])
x += interval
}
let result = values.join('')
console.timeEnd('while string')
// while string: 25.055ms
console.log(result)
console.log(result.length)
// adgjcfibehadgjcfibehadgjcfibehadgjcfibe...
// 100000
Using a larger interval obviously results in fewer loops, so the total execution time is lower. The resulting string is shorter too.
const s =
'abcdefghij'.repeat(30000)
console.time('while string')
let x = 0
let interval = 25 // big interval
let values = []
while (x < s.length)
{ values.push(s[x])
x += interval
}
let result = values.join('')
console.timeEnd('while string')
// while string: 6.130
console.log(result)
console.log(result.length)
// afafafafafafafafafafafafafafafafafafafafafafa...
// 12000
You can achieve functional style and stack-safe speed simultaneously -
const { loop, recur } = require('./lib')
const everyNth = (s, n) =>
loop
( (acc = '', x = 0) =>
x >= s.length
? acc
: recur(acc + s[x], x + n)
)
const s = 'abcdefghij'.repeat(30000)
console.time('loop/recur')
const result = everyNth(s, 2)
console.timeEnd('loop/recur')
// loop/recur: 31.615 ms
console.log(result)
console.log(result.length)
// acegiacegiacegiacegiacegiacegiacegia ...
// 150000
The two are easily implemented -
const recur = (...values) =>
({ recur, values })
const loop = f =>
{ let acc = f()
while (acc && acc.recur === recur)
acc = f(...acc.values)
return acc
}
// ...
module.exports =
{ loop, recur, ... }
And unlike the [...str].filter(...) solutions which will always iterate through every element, our custom loop is much more flexible and receives speed benefit when a higher interval n is used -
console.time('loop/recur')
const result = everyNth(s, 25)
console.timeEnd('loop/recur')
// loop/recur: 5.770ms
console.log(result)
console.log(result.length)
// afafafafafafafafafafafafafafa...
// 12000
const recur = (...values) =>
({ recur, values })
const loop = f =>
{ let acc = f()
while (acc && acc.recur === recur)
acc = f(...acc.values)
return acc
}
const everyNth = (s, n) =>
loop
( (acc = '', x = 0) =>
x >= s.length
? acc
: recur(acc + s[x], x + n)
)
const s = 'abcdefghij'.repeat(30000)
console.time('loop/recur')
const result = everyNth(s, 2)
console.timeEnd('loop/recur')
// loop/recur: 31.615 ms
console.log(result)
console.log(result.length)
// acegiacegiacegiacegiacegiacegiacegia ...
// 150000

Since I'm not an expert of regex, I'd use some fancy es6 functions to filter your chars.
var x = 'a1b2c3d4e5'
var n = 2;
var result = [...x].filter((char, index) => index % n == 0);
console.log(result);
Note that because 0 % 2 will also return 0, this will always return the first char. You can filter the first char by adding another simple check.
var result = [...x].filter((char, index) => index > 0 && index % n == 0);

As a variant:
function getNth(str, nth) {
return [...str].filter((_, i) => (i + 1) % nth === 0).join('');
}
console.log(getNth('a1b2c3d4e5', 2)); // 12345
console.log(getNth('a1b2c3d4e5', 3)); // b3e

What I'd suggest, to avoid having to iterate over the entire array, is to step straight into the known nth's.
Here's a couple of flavors:
function nthCharSubstr(str, nth) {
let res = "";
for (let i = nth - 1; i < str.length; i += nth) {
res += string[i];
}
return res;
}
More ES6-y:
const nthCharSubstr = (str, nth) =>
[...Array(parseInt(str.length / nth)).keys()] // find out the resulting number of characters and create and array with the exact length
.map(i => nth + i * nth - 1) // each item in the array now represents the resulting character's index
.reduce((res, i) => res + str[i], ""); // pull out each exact character and group them in a final string
This solution considers this comment as being valid.

Adding two numbers JS

I want to add two numbers from range 10-99,for example:
Input:16
Output:1+6=7
Input:99
Output:18
function digital_root(n) {
var z = n.toString().length;
if (z == 2) {
var x = z[0] + z[1]
return x;
}
}
console.log( digital_root(16) );
Output from this code is NaN.What should I correct?

You can try this:
function digital_root(n) {
var z = n.toString();
//use length here
if (z.length == 2) {
//convert to int
var x = parseInt(z[0]) + parseInt(z[1]);
return x;
} else {
return "not possible!";
}
}
console.log( digital_root(16) );
console.log( digital_root(99) );
console.log( digital_root(999) );

Use split to split the string in half and add the two using parseInt to convert to a number.
const sum = (s) => (''+s).split('').reduce((a,b) => parseInt(a)+parseInt(b))
↑ ↑ ↑ ↑
our coerce split sum
function to string in two both
Here a test :
const sum = (s) => (''+s).split('').reduce((a,b) => parseInt(a)+parseInt(b))
console.log(sum(12))

There are several approaches to sum digits of a number. You can convert it to a string but IDK if thats neccesary at all. You can do it with numerical operations.
var input = 2568,
sum = 0;
while (input) {
sum += input % 10;
input = Math.floor(input / 10);
}
console.log(sum);

Here's a fun short way to do it:
const number = 99
const temp = number.toString().split('')
const res = temp.reduce((a, c) => a + parseInt(c), 0) // 18
1.) Convert number to string
2.) Separate into individual numbers
3.) Use reduce to sum the numbers.

Your way would be the iterational way to solve this problem, but you can also use a recursive way.
Iterative solution (Imperative)
n.toString() Create String from number.
.split("") split string into chars.
.reduce(callback, startValue) reduces an array to a single value by applying the callback function to every element and updating the startValue.
(s, d) => s + parseInt(d) callback function which parses the element to an integer and adds it to s (the startValue).
0 startValue.
Recursive solution (Functional)
condition?then:else short-hand if notation.
n<10 only one digit => just return it.
n%10 the last digit of the current number (1234%10 = 4).
digital_root_recurse(...) call the function recursivly.
Math.floor(n / 10) Divide by 10 => shift dcimal point to left (1234 => 123)
... + ... add the last digit and the return value (digital root) of n/10 (1234 => 4 + root(123)).
function digital_root_string(n) {
return n.toString().split("").reduce((s, d) => s + parseInt(d), 0);
}
function digital_root_recurse(n) {
return n < 10 ? n : n % 10 + digital_root_recurse(Math.floor(n / 10));
}
console.log(digital_root_string(16));
console.log(digital_root_string(99));
console.log(digital_root_recurse(16));
console.log(digital_root_recurse(99));

The issue in your code is that you stored the length of n into z. The length is an integer, so both z[0] and [1] are undefined. The solution is to store the string into another variable and use that instead of z.
function digital_root(n) {
n = n.toString();
var l = n.length;
if (l === 2) {
return parseInt(n[0], 10) + parseInt(n[1], 10);
}
}
console.log( digital_root(16) );

Simply use var x = parseInt(n/10) + (n%10); and it will work for you.
function digital_root(n) {
var z = n.toString().length;
if (z == 2) {
var x = parseInt(n/10) + (n%10);
return x;
}
}
console.log( digital_root(16) );
console.log( digital_root(99) );
console.log( digital_root(62) );

Convert input to string, split it, convert each item back to number and sum them all:
function digital_root(n) {
return String(n).split('').map(Number).reduce((a,b) => a + b)
}
const result = digital_root(99);
console.log(result);

Look up tables and integer ranges - javascript

So I am looking to create look up tables. However I am running into a problem with integer ranges instead of just 1, 2, 3, etc. Here is what I have:
var ancient = 1;
var legendary = 19;
var epic = 251;
var rare = 1000;
var uncommon = 25000;
var common = 74629;
var poolTotal = ancient + legendary + epic + rare + uncommon + common;
var pool = general.rand(1, poolTotal);
var lootPool = {
1: function () {
return console.log("Ancient");
},
2-19: function () {
}
};
Of course I know 2-19 isn't going to work, but I've tried other things like [2-19] etc etc.
Okay, so more information:
When I call: lootPool[pool](); It will select a integer between 1 and poolTotal Depending on if it is 1 it will log it in the console as ancient. If it hits in the range of 2 through 19 it would be legendary. So on and so forth following my numbers.
EDIT: I am well aware I can easily do this with a switch, but I would like to try it this way.

Rather than making a huge lookup table (which is quite possible, but very inelegant), I'd suggest making a (small) object, choosing a random number, and then finding the first entry in the object whose value is greater than the random number:
// baseLootWeight: weights are proportional to each other
const baseLootWeight = {
ancient: 1,
legendary: 19,
epic: 251,
rare: 1000,
uncommon: 25000,
common: 74629,
};
let totalWeightSoFar = 0;
// lootWeight: weights are proportional to the total weight
const lootWeight = Object.entries(baseLootWeight).map(([rarity, weight]) => {
totalWeightSoFar += weight;
return { rarity, weight: totalWeightSoFar };
});
console.log(lootWeight);
const randomType = () => {
const rand = Math.floor(Math.random() * totalWeightSoFar);
return lootWeight
.find(({ rarity, weight }) => weight >= rand)
.rarity;
};
for (let i = 0; i < 10; i++) console.log(randomType());

Its not a lookup, but this might help you.
let loots = {
"Ancient": 1,
"Epic": 251,
"Legendary": 19
};
//We need loots sorted by value of lootType
function prepareSteps(loots) {
let steps = Object.entries(loots).map((val) => {return {"lootType": val[0], "lootVal": val[1]}});
steps.sort((a, b) => a.lootVal > b.lootVal);
return steps;
}
function getMyLoot(steps, val) {
let myLootRange;
for (var i = 0; i < steps.length; i++) {
if((i === 0 && val < steps[0].lootVal) || val === steps[i].lootVal) {
myLootRange = steps[i];
break;
}
else if( i + 1 < steps.length && val > steps[i].lootVal && val < steps[i + 1].lootVal) {
myLootRange = steps[i + 1];
break;
}
}
myLootRange && myLootRange['lootType'] ? console.log(myLootRange['lootType']) : console.log('Off Upper Limit!');
}
let steps = prepareSteps(loots);
let pool = 0;
getMyLoot(steps, pool);

JavaScript: How to reverse a number?

Below is my source code to reverse (as in a mirror) the given number.
I need to reverse the number using the reverse method of arrays.
<script>
var a = prompt("Enter a value");
var b, sum = 0;
var z = a;
while(a > 0)
{
b = a % 10;
sum = sum * 10 + b;
a = parseInt(a / 10);
}
alert(sum);
</script>

Low-level integer numbers reversing:
function flipInt(n){
var digit, result = 0
while( n ){
digit = n % 10 // Get right-most digit. Ex. 123/10 → 12.3 → 3
result = (result * 10) + digit // Ex. 123 → 1230 + 4 → 1234
n = n/10|0 // Remove right-most digit. Ex. 123 → 12.3 → 12
}
return result
}
// Usage:
alert(
"Reversed number: " + flipInt( +prompt("Enter a value") )
)
The above code uses bitwise operators for quick math
This method is MUCH FASTER than other methods which convert the number to an Array and then reverse it and join it again. This is a low-level blazing-fast solution.
Illustration table:
const delay = (ms = 1000) => new Promise(res => setTimeout(res, ms))
const table = document.querySelector('tbody')
async function printLine(s1, s2, op){
table.innerHTML += `<tr>
<td>${s1}</td>
<td>${s2||''}</td>
</tr>`
}
async function steps(){
printLine(123)
await delay()
printLine('12.3 →')
await delay()
printLine(12, 3)
await delay()
printLine('1.2', '3 × 10')
await delay()
printLine('1.2 →', 30)
await delay()
printLine(1, 32)
await delay()
printLine(1, '32 × 10')
await delay()
printLine('1 →', 320)
await delay()
printLine('', 321)
await delay()
}
steps()
table{ width: 200px; }
td {
border: 1px dotted #999;
}
<table>
<thead>
<tr>
<th>Current</th>
<th>Output</th>
</tr>
</thead>
<tbody>
</tbody>
</table>

Assuming #DominicTobias is correct, you can use this:
console.log(
+prompt("Enter a value").split("").reverse().join("")
)

I was recently asked how to solve this problem and this was my initial solution:
The desired output: 123 => 321, -15 => -51, 500 => 5
function revInt(num) {
// Use toString() to convert it into a String
// Use the split() method to return a new array: -123 => ['-', '1','2','3']
// Use the reverse() method to reverse the new created array: ['-', '1','2','3'] => ['3','2','1','-'];
// Use the join() method to join all elements of the array into a string
let val = num.toString().split('').reverse().join('');
// If the entered number was negative, then that '-' would be the last character in
// our newly created String, but we don't want that, instead what we want is
// for it to be the first one. So, this was the solution from the top of my head.
// The endsWith() method determines whether a string ends with the characters of a specified string
if (val.endsWith('-')) {
val = '-' + val;
return parseInt(val);
}
return parseInt(val);
}
console.log(revInt(-123));
A way better solution:
After I gave it some more thought, I came up with the following:
// Here we're converting the result of the same functions used in the above example to
// an Integer and multiplying it by the value returned from the Math.sign() function.
// NOTE: The Math.sign() function returns either a positive or negative +/- 1,
// indicating the sign of a number passed into the argument.
function reverseInt(n) {
return parseInt(n.toString().split('').reverse().join('')) * Math.sign(n)
}
console.log(reverseInt(-123));
NOTE: The 2nd solution is much more straightforward, IMHO

This is my solution, pure JS without predefined functions.
function reverseNum(number) {
var result = 0,
counter = 0;
for (i = number; i >= 1 - Number.EPSILON; i = i / 10 - (i % 10) * 0.1) {
counter = i % 10;
result = result * 10 + counter;
}
return result;
}
console.log(reverseNum(547793));

Firstly, I don't think you are using an array to store the number. You are using a java script variable.
Try out this code and see if it works.
var a = prompt("Enter a value");
var z = a;
var reverse = 0;
while(z > 0)
{
var digit = z % 10;
reverse = (reverse * 10) + digit;
z = parseInt(z / 10);
}
alert("reverse = " + reverse);

Or, as a one-liner ( x contains the integer number to be inversed):
revX=x.toFixed(0).split('').reverse().join('')-0;
The number will be separated into its individual digits, reversed and then reassembled again into a string. The -0 then converts it into a number again.

Explanation
Using the JavaScript reverse() array method you can reverse the order of the array elements.
Code
var a = prompt("Enter a value");
var arr = [];
for (var i = 0; i < a.length; i++) {
arr[i] = a.charAt(i);
}
arr.reverse();
alert(arr);

Assuming you may want to reverse it as a true number and not a string try the following:
function reverseNumber(num){
num = num + '';
let reversedText = num.split('').reverse().join('');
let reversedNumber = parseInt(reversedText, 10);
console.log("reversed number: ", reversedNumber);
return reversedNumber;
}

Using JavaScript reverse() and Math.sign() you can reverse a number both positive and negative numbers.
var enteredNum = prompt("Enter integer");
function reverseInteger(enteredNum) {
const reveredNumber = enteredNum.toString().split('').reverse().join('');
return parseInt(reveredNumber)*Math.sign(enteredNum);
}
alert(reverseInteger(enteredNum));

function add( num:number){ //159
let d : number;
let a : number =0;
while(num > 0){ //159 15 1
d = num % 10;
a = a * 10 + d; //9 95 951
num = Math.floor(num/10); // 15 1 0
}
return a; //951
}
console.log(add(159));

Reversing a number without converting it into the string using the recursive approach.
const num = 4578;
const by10 = (num) => {
return Math.floor(num / 10);
};
const remBy10 = (num) => {
return Math.floor(num % 10);
};
const reverseNum = (num, str = "") => {
if (num.toString().length == 1) return (str += num);
return reverseNum(by10(num), (str += remBy10(num)));
};
console.log(reverseNum(num, ""));

The simplest solution is to reverse any integer in js. Doesn't work with float.
const i2a = number.toString().split("");
const a2i = parseInt(i2a.reverse().join(""));
console.log(a2i);

Apply logic of reversing number in paper and try, and you have to care about dividing because it gives float values. That's why we have to use parseInt().
function palindrome()
{
var a = document.getElementById('str').value;
var r=0 ,t=0;
while(a>0){
r=a%10;
t=t*10+r;
a=parseInt(a/10);
}
document.write(t);
}
<form>
<input type="text" id="str"/>
<input type="submit" onClick="palindrome()" />
<form>

var reverse = function(x) {
if (x > 2147483647 || x < -2147483648 || x === 0) {
return 0;
}
let isNegative = false;
if(x < 0){
isNegative = true;
x = -x;
}
const length = parseInt(Math.log10(x));
let final = 0;
let digit = x;
let mul = 0;
for(let i = length ; i >= 0; i--){
digit = parseInt(x / (10**i));
mul = 10**(length-i);
final = final + digit * mul;
x = parseInt(x % 10**i);
}
if (final > 2147483647 || final < -2147483648 ) {
return 0;
}
if(isNegative){
return -final;
}
else{
return final;
}
};
console.log(reverse(1534236469));
console.log(reverse(-123));
console.log(reverse(120));
console.log(reverse(0));
console.log(reverse(2,147,483,648));

function reverseInt(n) {
let x = n.toString();
let y = '';
for(let i of x) {
y = i + y
}
return parseInt(y) * Math.sign(n);
}

Sweet and simple:
function reverseNumber(num){
return parseInt(num.toString().split("").reverse().join(""));
}

The above code will not work for negative numbers. Instead, use the following:
/**
* #param {number} x
* #return {boolean}
*/
var isPalindrome = function(x) {
return ((x>=0) ? ((x==(x = parseInt(x.toString().split("").reverse().join("")))) ? true:false) : false);
};

The simplest way is to
Covert it into a string and apply the reverse() method
Change it back to number
Check for the value provided if negative or positive with Math.sign()
Below is my solution to that.
function reverseInt(n) {
const reversed =
n.toString().split('').reverse().join('');
return parseInt(reversed) * Math.sign(n);
}
console.log(reverseInt(12345));

My solution to reverse a string:
var text = ""
var i = 0
var array = ["1", "2", "3"]
var number = array.length
var arrayFinal = []
for (i = 0; i < array.length; i++) {
text = array[number - 1]
arrayFinal.push(text)
text = ""
number = number - 1
}
console.log(arrayFinal)