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The task

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121 Output: true

Example 2:

Input: -121 Output: false Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: 10 Output: false Explanation: Reads 01 from right to left. Therefore it is not a palindrome. Follow up:

Could you solve it without converting the integer to a string?

My solution with converting number to string

const isPalindrome = n => n >= 0 && Number([...`${n}`].reverse().join("")) === n;

console.log(isPalindrome(121));

My solution without converting number to string

const isPalindrome2 = n => {
  if (n < 0) { return false; }
  let num = Math.abs(n);
  const arr = [];
  let i = 1;
  while (num > 0) {
    const min = num % (10 ** i);
    num = num - min;
    i++;
    arr.push(min);
  }
  i = i - 2;
  let j = 0;
  return n === arr.reduce((res, x) => {
    const add = (x/ (10 ** j)) * (10 ** i);
    res += add;
    i--;
    j++;
    return res;
  }, 0);
};

console.log(isPalindrome2(121));
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  • 1
    \$\begingroup\$ That's an awfully convoluted way to check for your objective. Can you tell us more about why you did it this way? \$\endgroup\$ – Mast Apr 21 at 11:44
  • \$\begingroup\$ I wanted to solve it without converting the number to a string. @Mast \$\endgroup\$ – thadeuszlay Apr 21 at 11:56
  • \$\begingroup\$ That explains it in part, but it's still an odd way to do it. I hope my answer points out why I was so surprised. \$\endgroup\$ – Mast Apr 21 at 13:01
  • \$\begingroup\$ @thadeuszlay Palindrome is not defined on numbers, it's defined on strings. In this case you want decadic string representing the given number. You have to convert it to a string or a similar representation (e.g. array of decadic digits). \$\endgroup\$ – Sulthan Apr 21 at 18:57
  • 2
    \$\begingroup\$ @Sulthan well, you don't necessarily need an array or string. The answers show how you can do this without splitting up the digits. But it is true that you need to know the base, and that a fully explicit version of the question should ask whether the base-10 (or whatever) representation of the integer is a palindrome, since that's a property of the representation but not of the integer itself. \$\endgroup\$ – David Z Apr 21 at 22:00
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Count digits in positive integer

You can get the number of digits using log10

eg

Math.log10(13526); // is 4.131169383089324
const digits = Math.ceil(Math.log10(13526)); // 5

You can get the unit value eg 423 is 100 or 256378 is 100000 by raising 10 to the power of the number of digits minus one. Well not for powers of 10

eg

unit = 10 ** (Math.ceil(Math.log10(13526)) -1); // 10000
unit = 10 ** (Math.ceil(Math.log10(10000)) -1); // 1000 wrong for power of 10 number

To get the value we want we need to floor the log first

unit = 10 ** Math.floor(Math.log10(10000)); // 10000
unit = 10 ** Math.floor(Math.log10(13526)); // 10000 correct

or

unit = 10 ** (Math.log10(10000) | 0); // 10000
unit = 10 ** (Math.log10(13526) | 0); // 10000 

Get digit at position of positive integer

To get the digit at any position in a number divide it by 10 raised to the power of the digit position get the remainder of that divided by 10 and floor it.

const digitAt = (val, digit) => Math.floor(val / 10 ** digit % 10);

or

const digitAt = (val, digit) => val / 10 ** digit % 10 | 0;

// Note  brackets added only to clarify order and are not needed
// ((val / (10 ** digit)) % 10) | 0;

digitAt(567, 0);  // 7
digitAt(567, 1);  // 6
digitAt(567, 2);  // 5

Positive integer a palindrome in \$O(1)\$ space

With that info you can then build a function that does test in \$O(1)\$ space, as you do not need to store the digits in an array for later comparison.

To keep performance up we can avoid the slower versions of some operation. For floor we can | 0 (note that for large numbers > 2**31-1 you must use floor) and for ** use Math.pow

Rather than do the full calculation to get the digit we can store the unit value of the digit we want for the top and bottom and multiply by 10 to move up and divide by 10 to move down.

function isPalindrome(num) {
    var top = Math.pow(10, Math.log10(num) | 0), bot = 1;
    while (top >= bot) {
        if ((num / top % 10 | 0) !== (num / bot % 10 | 0)) { return false }
        top /= 10;
        bot *= 10;
    }
    return true;
}
  • The function will returns false for negative numbers but is not optimized for them
  • The function only works on integer values less than Number.MAX_SAFE_INTEGER which is 9007199254740991

In terms of performance the above function is 5 times faster for a 16 digit palindrome 2192123993212912 and 10-11 times faster for a non palindrome of 16 digits

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  • 2
    \$\begingroup\$ Don't suggest floating-point arithmetic when solving integer problems. \$\endgroup\$ – Roland Illig Apr 21 at 22:39
  • \$\begingroup\$ @RolandIllig Really??? Are you unfamiliar with Javascript? . Using math operators /, *, % evaluate as doubles, Meaning that what you are saying is dont use math operators to solve integer problems, \$\endgroup\$ – Blindman67 Apr 22 at 5:52
  • \$\begingroup\$ The crucial point is that IEEE 754-2008 guarantees that the +, -, *, / and sqrt operators produce results that are as close to the mathematical truth as possible. There is no such guarantee for log10 or sin, and thus using these functions may introduce rounding errors. \$\endgroup\$ – Roland Illig Apr 22 at 6:23
  • 1
    \$\begingroup\$ @RolandIllig What??? You are saying, all answers no matter what the tag must be language neutral, I think I will avoid that complication. Anyways the | operator converts to int 32 and Math.log10 in javascript will never let (Math.log10(n)|0) !== (Math.log10(n+1)|0) be true, where n is positive int (excluding 9, 99, 999, 9999, 99999... ). So prove me wrong, all you need is a number for n \$\endgroup\$ – Blindman67 Apr 22 at 6:42
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    \$\begingroup\$ @RolandIllig By the way did you read my answer "only works on integer values less than Number.MAX_SAFE_INTEGER" a clear warning regarding possible rounding errors \$\endgroup\$ – Blindman67 Apr 22 at 6:46
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Your second solution is awfully complicated for something simple.

You start out alright. No negative number can be a palindrome, so we can discount those. However, 0 is always a palindrome, so you're discounting that while you shouldn't.

You do a lot of complicated math and use an array. You could use either, but shouldn't use both. There are a couple of mathematical approaches to solve this, but you should be able to use those without iterating over the individual numbers or splitting it up at all. The most obvious solution however, is using an array and direct comparisons, without any math.

Say we got 1221. Split it up. [1, 2, 2, 1]. Iterate over the array, comparing every nth character to the last-nth character. 0th to 3rd. 1st to 2nd.

Say we got 92429. Split it up. [9, 2, 4, 2, 9]. Ignore the middle character. Handle the rest like it's an even-length number.

Based on those 2 cases, you should be able to figure out a much simpler algorithm.

Note: This answers the explicit question. Implicitly, you should wonder whether arrays should be allowed for this challenge. After all, iterating over a string or an array, it's not that different. I strongly suspect they want you to use the math-only approach.

Another approach, which is somewhat math-based and you should beware of overflows, is simply reversing the number. In pseudo-code, that would look something like this:

reverse = 0
while (number != 0) {
  reverse = reverse * 10 + number % 10;
  number /= 10;
}

Check the input versus its reversed number. If they are the same, it's a palindrome.

But this still uses extra memory to hold the additional integer we just created.

Can it be done without? Absolutely. But I'll leave that as an exercise for you.

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  • \$\begingroup\$ How would you split it up? \$\endgroup\$ – thadeuszlay Apr 21 at 16:23
  • \$\begingroup\$ You are converting the number to a string anyway, even if you don't store the string anywhere. A number is an abstract concept. To represent it, you need a string, either binary or a decadic (or any other base). When you start working with decimal representation, you are using a string representation. You cannot define palindrome without a string. \$\endgroup\$ – Sulthan Apr 21 at 19:01
  • \$\begingroup\$ @Sulthan Yes, but that's a philosophical discussion, not a review. \$\endgroup\$ – Mast Apr 21 at 20:26
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    \$\begingroup\$ You could modify the last approach to stop when reverse >= number, and then evaluate. Voila, no danger of overflow. \$\endgroup\$ – Deduplicator Apr 21 at 22:13
  • \$\begingroup\$ @Deduplicator when using your approach to test 121 and 1221, the code needs two separate tests. One for odd number of digits, one for even. Or did I miss something? \$\endgroup\$ – Roland Illig Apr 22 at 6:36
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Are you sure you should generate all those objects and arrays?
And the second code additionally looks quite complex.

Testing whether a number is palindromic is actually quite simple:

function isPalindrome(num) {
    var rev = 0
    while (rev < num) {
        rev = rev * 10 + num % 10
        num /= 10
    }
    return rev == num || rev / 10 == num
}
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  • \$\begingroup\$ This is JavaScript, therefore it's better to compare using === instead of ==. \$\endgroup\$ – Roland Illig Apr 22 at 16:36

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