Find the largest palindrome made from the product of two 3-digit numbers - Javascript

Question

Can anyone tell me what\'s wrong with the code. Find the largest palindrome made from the product of two 3-digit numbers.

function largestPalind         


        

Answer 1

I think you can go for code given at this link http://www.mathblog.dk/project-euler-problem-4/

As this save your CPU cycle from multiplication, which is quite costly operation.

Well even in this you can make some more to make to make it more like, you can modify its while loop a bit

while (!found) {
    firstHalf--;
    palin = makePalindrome(firstHalf);
    for (int i = 999; i > 99; i--) {
        if ((palin / i) > 999 || i*i < palin) {
            break;
        }

        if ((palin % i == 0)) {
            found = true;
            factors[0] = palin / i;
            factors[1] = i;
            break;
        }
    }
}

So here instead of moving from i=999 : 100, we can write it as i=sqrt(palin):100, as you can find factorial of number within its square root. Refer link How to find Number is prime number or not!

And also you can change if(condition) to if(!(palin%i)) as comparing with zero is usually not considered a good practice also comparing takes more CPU cycle compared to your simple negating bits.

Answer 2

As a very simple solution, this one works

public class LargestPallendrome {

public static void main(String[] args) {

    int a = 999;
    int b = 999;

    long max = 0;

    while (a > 100) {
        long num = a * b;
        if (checkPallendrome(num)) {
            if (num > max)
                max = num;
        }
        if (b >= 100)
            b--;
        else {
            a--;
            b = 999;
        }
    }
    System.out.println(max);
}

public static boolean checkPallendrome(long num) {
    String a = num + "";
    String b = new StringBuffer(num + "").reverse().toString();
    if (a.equals(b))
        return true;
    return false;
}
}

Answer 3

Suggesting a solution using underscore.js. First, find all palindromes and then loop through them starting from the largest one and return the one which has two 3-digit prime factors.

function isPalindrome(num) {
    var str = num.toString();
    return str.split('').reverse().join('') === str;
}

function palindromes() {
    var max = 999 * 999;
    var min = 100 * 100;
    return _.select(_.range(max, min, -1), isPalindrome);
}

palindromes().find(function (x) {
    if (_.find(_.range(999, 100, -1), function (y) {
        return (x % y === 0 && y != x / y && x / y < 1000) ? true : false;
    })) return true;
})

Answer 4

Another Simple Solution in JavaScript

function reverseNumber(n)
{
    n = n + "";
    return n.split("").reverse().join("");
}

function palindrom(){
 var r= 1 , y =1;
var largest = 0;
 while(r <= 1000){ 
   var num1 =  r;
   var num2 = 0;
 while(num1 <= 1000 && num2 <= num1){ 
        product = num1 * num2;
   if (product == reverseNumber(product)){
     console.log(`${num1} x ${num2} = ${product}`);
          if(product > largest){
            largest = product;
          }
   }
   num1 = num1 + 1;
   num2= num2 + 1;
}
r++;
}
console.log(``)
console.log(`The largest is ${largest}`);
}
console.log(palindrom());

Answer 5

This is how I did it. I used the old fashioned way to check for a palindrome. It appears to run faster on my computer but I may be wrong. Pushing to an array, as in the above post, was definitely very slow on my computer. Noticeable lag at the console. I would recommend just checking to see if your product is greater than your current max, if it is, store that instead of pushing everything to an array. Please feel free to correct me if I'm wrong. Much appreciated.

//should find the largest palindrome made from the product of two 3 digit numbers
var largestPalindrome = function() {

    var max = 0,
        product = 0;
    for (var num1 = 999; num1 >= 100; num1--) {
        for (var num2 = 999; num2 >= 100; num2--) {
            product = num1 * num2;
            product > max && isPalindrome(product.toString()) ?  max = product : 0;
        }
    }
    return max;
};

//check to see if product is a palindrome
var isPalindrome = function(product) {
    var palindromeCheck = true;
    for (var i = 0; i < product.length / 2; i++) {
        if (product[i] != product[product.length - i - 1])
            palindromeCheck = false;
    }
    return palindromeCheck;

   //return product === product.split("").reverse().join("");
};

Answer 6

This is how I did it in Javascript. Simple & easy!

let num1 = 999;
let num2 = 999;
let arr = [];

function check(x, y)
{
    if(String(x*y) == String(x*y).split("").reverse().join(""))
    {
        return true;
    }
return false;
}

for(let i=0; i<999999; i++)
{
    if(check(num1, num2))
    {
        arr.push(num1*num2);
        num1--;
        num2 = num1+1;
    }
num2--;
}

console.log(arr.sort((x, y) => y-x)[0]);