Palindrome Number

Problem Summary

Given an integer x, return true when its decimal representation reads the same from left to right and from right to left. The input fits in a signed 32 bit integer, so it may be negative. The natural follow up asks to solve the problem without converting the number to a string.

The key insight is that reversing the whole number is unnecessary and risky, because the reversal of a valid 32 bit integer may itself overflow. Reversing only the second half is enough: digits are stripped from the right of the original and appended to a growing reversed value until the two meet in the middle. Two early rejections simplify the loop, since a negative number carries its sign on one side only, and a number ending in zero would need a leading zero to match, which only zero itself can afford. When the digit count is odd the middle digit lands on the reversed side, where it plays no role, so dividing that side by ten discards it before the comparison.

Techniques

Math

Solution

function isPalindromeNumber(x: number): boolean {
    // negatives read as a minus sign first, and a trailing zero can only match a leading one in 0 itself
    if (x < 0 || (x % 10 === 0 && x !== 0)) {
        return false
    }

    // reverse only the second half, so the number never has to be turned into a string
    let reversedHalf = 0

    while (x > reversedHalf) {
        reversedHalf = reversedHalf * 10 + (x % 10)
        x = Math.floor(x / 10)
    }

    // an odd digit count leaves the middle digit on the reversed side, where it does not matter
    return x === reversedHalf || x === Math.floor(reversedHalf / 10)
};