This is part of my series where I explain approaches to solving coding problems. This is to help me articulate my thought process better, and inspire new problem solving approaches for developers!

Problem Statement:

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward. Follow up: Could you solve it without converting the integer to a string?

Approach:

Now there is most definitely a super fun JavaScript one liner iterative solution, but I was also interested in the follow up approach.

This approach reverses the number mathematically--without converting it to a string, and then compares the results afterwards.

I decided to write about this approach since mathematical solutions can be much more efficient than iterative or recursive solutions. This was a good exercise for thinking about programming mathematically because math is a wonderful thing.

Solution:

/**
 * @param {number} x the number to check
 * @return {boolean} true if it's a palindrome number
 */
const isPalindrome = x => {
    if (x < 0) return false

    let reversed = 0, y = x

    while (y > 0) {
        const lastDigit = y % 10
        reversed = (reversed * 10) + lastDigit
        y = (y / 10) | 0
    }
    return x === reversed
}

Explanation:
First, we check and see if the number is negative. If it is, then we know it's not a palindrome because the numbers will read differently backwards and forwards.

if (x < 0) return false

If the number is positive, we'll create two variables. The first variable reversed will store our reversed number, and the second variable y is copy of our input number. y will be used to reverse the input number without modifying our original input.

The following steps take place inside our while loop:

Get the last digit of the number using the modulo (%) operator. This is one trick that can help you isolate the last digit for future problems. Here, we're dividing y by 10 and returning the remainder. Let's refer to the example input 121. The hundreds column 100 is divided by 10 with a remainder of 0, and the tens column 20 is divided by 10 with a remainder of 0. When we divided the ones column 1 by 10, we'll get a remainder of 1 since 1 can't be divided by 10 evenly. After, we save the remainder to lastDigit:

const lastDigit = y % 10

We append the last digit to reversed. We have to multiply reversed by 10 on the right side of the assignment to ensure that we're always appending lastDigit to the ones column.

reversed = (reversed * 10) + lastDigit

Remove the last digit from y by dividing it by 10, and truncating the last decimal. We can do this using the bitwise OR operator |. This is another trick that can help you in future JS problems. In this case, we're converting the result into an integer, and then return the new integer:

y = (y / 10) | 0

Finally, if reversed === x, then it's a palindrome!

This solution saved us from having to traverse an array of string digits, which means we didn't have to use extra storage for this problem! When trying to find a mathematical approach to a coding question, think about any patterns you notice, and ask yourself if you need to read one digit at a time. If so, you can definitely traverse a number's digits with modulo arithmetic and division.

Thanks for reading! As always questions, feedback, and ideas are always encouraged. Happy hacking!