# JavaScript Optimal verification if a number is Palindrom in JS

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;
}``````

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));``````

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 = 

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

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