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Project Euler #4 - Largest Palindrome Product

Peter Kim Frank on May 28, 2019

I'm bumping a "series" that I started last year, where we collectively work on problems from Project Euler. This is Problem 4, finding the largest...

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Ali Spittel • May 28 '19

Here's my Python solution!

def palindrome_number():
    _range = xrange(100, 1000)
    palindrome = None
    for i in _range:
        for j in _range:
            prod = i * j
            if str(prod) == str(prod)[::-1]:
                if prod > palindrome:
                    palindrome = prod
    return palindrome

print(palindrome_number())

Justin • May 28 '19 • Edited

How's the run time? I think yours is Python 2?

Here's mine in Python 3 :) It's a bit more complicated because I try to write them more flexible. Like how this problem shows 2 digit numbers as an example and 3 for the actual problem, so I wrote it to take the number of digits per number as an input

import time
def largestPalindromeProduct(digits):
  min, max, temp, pal = '1', '', 0, 0
  for _ in range (1,digits):
    min += '0'
  for _ in range (0,digits):
    max += '9'
  min = int(min)-1
  max = int(max)
  start = time.time()
  for num1 in range (max,min,-1):
    for num2 in range (num1,min,-1):
      temp = num1 * num2
      if temp < pal:
        break
      if str(temp) == "".join(reversed(str(temp))):
        pal = temp
  end = time.time()
  print(pal, end=' ')
  print(end-start)

Florian Rand • May 28 '19 • Edited

My typescript solution:

function reverseInteger(n: number): number {
  return Number(String(n).split('').reverse().join(''));
}

function isPalindrome(num: number): boolean {
  if (num === reverseInteger(num)) {
    return true;
  }
  return false;
}

// I had this in separate files sorry if its confusing X)
import { max } from 'mathjs';

let number = 0;
let a = 999;

const palindromes: number[] = [];

while (a > 1) {
  for (let i = 2; i <= 999; i += 1) {
    number = a * i;
    if (isPalindrome(number)) {
      palindromes.push(number);
    }
  }
  a -= 1;
}
console.log(`Max palindrome ${max(...palindromes)}`);

Massimo Artizzu • May 29 '19

if (condition) {
  return true;
}
return false;

Please don't do this 😕 It's verbose for no reason. You can accomplish the same with just return condition;.

Florian Rand • May 29 '19

cool! yes so much simple

function isPalindrome(num: number): boolean {
  return num === reverseInteger(num);
}

thanks!

Mamoun Boussida • May 28 '19 • Edited

Here is my code on C (Brute force), the result was out after 0.172s.

#include <stdio.h>
#include <stdlib.h>

int palindrome(long x)
{
    int i,j=0;
    char ch[7];
    sprintf(ch,"%ld",x);

    char ch1[7]="";
    for(i=strlen(ch)-1;i>=0;i--,j++)
    {
        ch1[j]=ch[i];
    }

    if(strcmp(ch,ch1)==0)
        return 1;
    return 0;
}


int main()
{
    int i,j;
    long s,max=100*100;
    for(i=100;i<=999;i++)
    {
        for(j=100;j<=999;j++)
        {
            s=i*j;
            if (palindrome(s) && s>max)
                max=s;
        }
    }
    printf("Result = %ld",max);
}

Jeff Hall • May 29 '19

Here's my Ruby solution. There's probably a much more clever way to do this. Probably a more Ruby way to do it for that matter.

# Find the largest palindrome number that is the product of two three digit factors.

def check_equality(num)
  num.to_s == num.to_s.reverse
end


def find_palindrome
  r1 = (999..1)
  r2 = r1

  (r1.first).downto(r1.last).each do |i|
    (r2.first).downto(r2.last).each do |j|
      if check_equality( i * j )
        puts "#{i} * #{j} = #{i*j}"
        return
      end
    end
  end
end


find_palindrome

Massimo Artizzu • May 30 '19

If I understand it correctly (correct me if I'm wrong, I don't know Ruby), this doesn't work in general, as it prints the first palindrome product you find. But you don't know if it's the largest.

Unless you can prove it is 🤷‍♂️ (I have no idea).

Jeff Hall • May 30 '19

I'm working backwards through the range of numbers beginning with '999.' Hence the extra verbosity in the block with the call to the downto method. (999..1).each do doesn't work, and (1..999).each do really shouldn't work either because ranges are not defined by their content, just a beginning state and an end state. So counting backwards the first palindrome I find is the largest. And the outer block isn't necessary, but I include it just for the sake of being thorough I guess.

Massimo Artizzu • May 30 '19

The problem here is that the products you're getting aren't ordered. Which means that if you get a palindrome, you cannot be sure it's the largest.

In fact, I run your code and I got 999 * 91 = 90909, which is not correct. Even if you limit your range to 999 to 100 (we're looking for 3-digit factors after all), you'd get 995 * 583 = 580085. But the largest palindrome is 993 * 913 = 906609, which comes after the others.

Faraz Patankar • May 28 '19

Just did this super quick in JS.

function isPalindrome(num) {
  const stringifiedNum = num.toString();

  return (
    Array.from(stringifiedNum).toString() ===
    Array.from(stringifiedNum)
      .reverse()
      .toString()
  );
}

function findLargestPalindrome() {
  const start = 100;
  const end = 999;

  let largestPalindrome = 0;

  for (let i = start; i <= end; i += 1) {
    for (let j = start; j <= end; j += 1) {
      const product = i * j;
      if (isPalindrome(product) && product > largestPalindrome) {
        largestPalindrome = product;
      }
    }
  }

  return largestPalindrome;
}

findLargestPalindrome();

by • Jan 13 '20

while there are many explanations, easiest to understand was yours. the only part I struggle to understand is this one "isPalindrome(product) && product > largestPalindrome", can you please explain what it means?

Faraz Patankar • Jan 13 '20

Yes. Let's try reading it step by step:

We first check if the product is a palindrome. Because if it isn't, we don't care about it.
Then, we check if it is greater than our existing largest palindrome because our goal is to find the largest palindrome.
If both those conditions are true, the product being a palindrome and it being larger than our existing largest palindrome, we set that as our largest palindrome.

Hope that helps, let me know if you still have questions!

by • Jan 14 '20

Thank you, Faraz, I didn't know it was possible to use several conditional operators in one single line of code

Jonathan Silvestri • May 28 '19

A quadratic solution in JavaScript. I'm curious if there's a way to do this in linear time:

const array = new Array(900).fill(0).map((e, i) => i + 100);

const isPalindrome = num => {
  return (
    String(num) ===
    String(num)
      .split("")
      .reverse()
      .join("")
  );
};

const maxProduct = range => {
  let palindrome = 0;
  for (let i = 0; i < range.length; i++) {
    for (let j = i + 1; j < range.length; j++) {
      const product = array[i] * array[j];
      if (isPalindrome(product) && product > palindrome) {
        palindrome = product;
      }
    }
  }

  return palindrome;
};

console.log(maxProduct(array));

Massimo Artizzu • May 30 '19 • Edited

Not linear... but I think (without any actual proof 🤷‍♂️) (I'm a fraud 🤦‍♂️) that my solution does it in logarithmic time: dev.to/maxart2501/comment/b9m6

Edit: scratch that, no way it's not quadratic 😂 But then again, it's faster than the extensive check.

Jay • May 28 '19

Love to work on the challenges, please keep this series running.
Here's my Rust Solution: Playground

fn main() {
    let max = max_palindrome();
    println!("Max palindrome is {}, product of {} * {}", max.1, (max.0).0, (max.0).1);
}

fn max_palindrome() -> ((i32, i32), i32) {
    let range = 100..1000;
    let mut ans = ((0, 0), 0);
    for a in range.clone() {
        for b in range.clone() {
            let p = a * b;
            if is_palindrome(p) && p > ans.1 {
                ans.1 = p;
                ans.0 = (a ,b);
            }
        }
    }
    ans
}

fn is_palindrome(num: i32) -> bool {
    let str_num = num.to_string();
    str_num
        .chars()
        .zip(str_num.chars().rev())
        .all(|(c1, c2)| c1 == c2)
}

Ismail Bello • Jan 3 '20

a lazy c++ solution :)

auto larget_palindrome_product(size_t n_digits) -> size_t {

    const auto is_palidrome = [](size_t x) -> bool {
        const auto as_string = std::to_string(x);
        auto fst = as_string.begin();
        auto lst = as_string.rbegin();
        for (; fst != as_string.end(); fst++, lst++)
            if (*fst != *lst) return false;
        return true;
    };

    size_t result = 0;
    const auto lower_bound = std::pow(10, n_digits - 1);
    const auto upper_bound = std::pow(10, n_digits) - 1;

    for (size_t fst = lower_bound; fst <= upper_bound; fst++) {
        for (size_t snd = fst; snd <= upper_bound; snd++) {
            if (auto n = fst * snd;
                is_palidrome(n) && n > result) {
                result = fst * snd;
            }
        }
    }

    return result;
}

Mahmoud Bayazid • Mar 27 '20

First observation is that the number must be between 100^2 and 999^2 or in the range of [10000, 998001].

As the majority of numbers has 6 digits and we're looking for the largest, we ignore 5 digits numbers.

Based on this, we can construct a palindromic number as:

'abccba' =100000a+10000b+1000c+100c+10b+a

=100001a+10010b+1100c

=11(9091a+910b+100c) = p.q

This equation shows us, that either p or q, but not both must have a factor of 11!!!

def palindrome_number(x,y):

my_list =[]

for i in range(x,y):
    for j in range(x,y):
        num = i * j
        if num % 11 ==0 and str(num) == str(num)[::-1]:
            my_list.append(num)
            my_list.sort()

print(my_list[-1])

shyam1110 • Jun 21 '20

void solve()
{
fast;
ll n,largepalin=0;
cin>>n;
for(ll i=999;i>=100;i--)
{
for(ll j=999;j>=100;j--)
{
if((i*j)%11==0 && palin(i*j) && (i*j)<n) largepalin=max(i*j,largepalin);
}
}
cout<<largepalin<<endl;
return;
}

Massimo Artizzu • May 29 '19 • Edited

Sooo... there are so many good solutions, but they all kind of look the same, so I wanted a different approach 😁

What if, instead of starting with the factors and checking if their product is a palindrome, we start with palindromes and find two 3-digit factors?

We'd start with 999999, then 998899, then 997799... You get the point. So we can start with a 3-digit number (e.g. 356)... and get a palindrome out of it (e.g. 356653), and look for factors.

My solution in JavaScript:

const digits = 3;
const upper = 10 ** digits - 1;   // 999
const lower = 10 ** (digits - 1); // 100
function palindromize(num) {
  const padded = String(num).padStart(digits,'0');
  return +(padded + padded .split('').reverse().join(''));
}

let p, b;
out: for (let a = upper; a >= lower; a--) {
  p = palindromize(a);
  for (b = Math.floor(p ** .5); p/b <= upper; b--) {
    if (p % b === 0) break out;
  }
}
console.log(p / b, b);

I've tested it with up to 7 digits and it's still basically instantaneous. Over that limit you go over Number.MAX_SAFE_INTEGER and it becomes unreliable. I didn't bother using BigInts, also because you can't get the square root out of them (you can approximate it, but it's a bother).

P.S.: yes, I've used a label in JavaScript. Sue me 😂 (also useful if you have to break out of more than one cycle... which is usually rare).

Stephanie Handsteiner • May 28 '19

Am still able to just copy and paste from my Euler GitHub repo. 😂

PHP again

$largestThreeDig = 999;
$largestPalindrome = 0;
for($l = $largestThreeDig; $l > 0 && $l * $largestThreeDig > $largestPalindrome; $l--) {
    for($p = $l * $largestThreeDig; $p > $largestPalindrome; $p -= $l) {
        if((string)$p === strrev((string)$p)) {
            $largestPalindrome = $p;
        }
    }
}
 echo "Largest Palindrome: $largestPalindrome\n";

Massimo Artizzu • May 30 '19

This is another solution that I consider working in this particular case (3 digits) but not actually correct in the general case. I explained it here: dev.to/maxart2501/comment/b9me

khouloudzaiter • Aug 5 '19

Written in Java!

public class Problem4 {
    public static boolean isPalindrome(long val) {
            boolean isPalindrome = true;
            String str = Long.toString(val);
            int len = str.length();
            int i = 0;
            while (isPalindrome && i <= (len-1)/2){
                isPalindrome = str.charAt(i) == str.charAt(len-1-i);
                i++;
            }
            return isPalindrome;
        }

        public static void main(String[] args) {
            int i = 999;
            long largest = 1;
            String str = "";
            long val = 1;
        while (i>=100)
        {
            int j = i;
            while (j>=100)
            {
                val = i * j;
                if(isPalindrome(val) && largest < val)
                {
                    largest = val;
                    str = i + " x " + j;
                }
                isPalindrome (val);
                j--;
            }
            i--;
        }
        System.out.println(str+ "= "+ largest);

    }
}

vienpham2019 • Jan 9 '20

This is my solution in Javascript

console.log(largestPalindromeProduct(2));

Shahid Kamal Khan • Jan 10 '20

Here is freeCodeCamp version of it


function largestPalindromeProduct(n) {
  // Good luck!
  const isPalindrom = n => {
    return String(n).split("").reverse().join("") === String(n);
  }
  let largestPalindrome = 0;
  const lowerBound = Number(`1${Array.from({length:n}).join(0)}`);
  const upperBound= Number(`${Array.from({length: n+1}).join(9)}`);
  console.log({upperBound, lowerBound})
  for(var i=lowerBound; i<=upperBound; i++) {
    for(var j=lowerBound; j<=upperBound; j++) {
    const product = i*j;
      if(isPalindrom(product) && product > largestPalindrome ) {
        console.log(i, j, product, largestPalindrome)
        largestPalindrome = i*j;
      }
    }
  }
  return largestPalindrome;
}

largestPalindromeProduct(2);

Prabhjot Singh Rana • Feb 4 '20 • Edited

Solution in Python:

x = 0   # assuming number is Xnnnnn  1st digit
y = 0   # assuming number is nYnnnn  2nd digit
z = 0   # assuming number is nnZnnn  3rd digit

num = 999  #highest 3 digit number

palidromefound = False

highestnum = num * num

while str(highestnum) != str(highestnum)[::-1]:

    z = int(highestnum / 1000)
    z = z%10

    y = int(highestnum / 10000)
    y = y%10

    x = int(highestnum / 100000)
    x = x%10

    if z != 0 and y != 0:
        z = z-1

    # first time find the highest palidrom

    highestnum = (x*100000 + y*10000 + z*1000 + z*100 + y*10 + x)


while palidromefound == False:

    if str(highestnum) == str(highestnum)[::-1]:
        while num > 99:

            if (highestnum % num) == 0 and (highestnum / num) < 999 and (highestnum / num) > 99:
                palidromefound = True
                print(f'Highest Palidrom: {highestnum} num1: {num} and num2: {int(highestnum / num)}')
                break
            num = num -1

        else:

            num = 999

            # below are the palidromes.. the value of z is the 3rd digit, it comes from the above logic
            # when the first highest palidrom is found, but if it not a multiple of two three digit numbers
            # then the next highest palidrome is obtained:
            #
            # 992299 - 1100 = 991199  >> when z != 0
            # 991199 - 1100 = 990099  >> when z != 0
            # 990099 - 110  = 989989  >> when z == 0
            # 989989 - 1100 = 988889  >> when z != 0 


            if z == 0:
                highestnum = highestnum -110
                z = 9
            else:
                highestnum = highestnum -1100
                z = z-1

Massimo Artizzu • May 30 '19

I'm sorry if that seemed rude, that wasn't my intention at all. I did add "granted I could understand F#" in case I was wrong.

But no, I will keep on trying to understand what's going on in other people's code even out of my comfort zone, because I'll have the chance to learn something new 👌

Massimo Artizzu • May 30 '19

This is another solution that I consider working in this particolar case (3 digits) but not actually correct (granted I could understand F# 😅). I explained it here: dev.to/maxart2501/comment/b9me

Rossman • Aug 27 '22

    internal class Task4
    {
        public static string Reverse(string s)
        {
            char[] charArray = s.ToCharArray();
            Array.Reverse(charArray);
            return new string(charArray);
        }

        private bool IsPalindrome(int number) 
        {
            string strNumber = number.ToString();
            string reversedStrNumber = Reverse(strNumber);
            return strNumber == reversedStrNumber;
        }
        public Task4() 
        {
            int maxPalindrome = int.MinValue;
            for(int i = 999; i > 99; i--) 
            {
                for (int j = 999; j > 99; j--)
                {
                    int result = i * j;
                    if (result > maxPalindrome && IsPalindrome(result))
                        maxPalindrome = result;
                }
            }
            Console.WriteLine(maxPalindrome);
        }
    }

shyam1110 • Jun 21 '20

Complete Solution..!!! for Euler 004

SpyDie • Jul 3 '20 • Edited

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.

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

This is the easiest way and logical too..

list = []

for i in range(901,1000):
    for j in range(901,1000):
        prod = i*j
        if str(prod) == str(prod)[::-1]:
            list.append(prod)

    i += 1
    j += 1

list.sort()

print(list[-1])

starchcode • Jul 13 '20

here is my result using JS. avoided any use of methods. Also not a master in mathematics but here it is

Massimo Artizzu • May 30 '19

Being wrong is nothing to be ashamed of, as long as you're ready to stand corrected.

My mistake in this case is that I commented while it was very late 😴 Took note on the approach, though.

Alexey Voinov • May 28 '19

Did you try hackerrank's version? They have broader ranges for input values and pretty tight time/memory limitations. That sometimes makes even trivial project euler tasks challenging. :)