3783. Mirror Distance of an Integer

Math

TIME

O(log n)

SPACE

O(1)

Intuition

Reverse the digits of n, then compute the absolute difference.

Solution

Reverse Function

Two approaches:

String-based:

reverse = int(str(n)[::-1])

Mathematical:

def reverse(n):
    result = 0
    while n > 0:
        result = result * 10 + n % 10
        n //= 10
    return result

Both handle leading zeros correctly (e.g., 10 → 01 → 1).

Final Answer

return abs(n - reverse(n))

Complexity

Time: O(d) where d = number of digits ≈ log₁₀(n)
Space: O(1) or O(d) for string approach

Edge Cases

Palindrome numbers → distance = 0
Numbers ending in 0 (10, 100, 120) → reverse has fewer digits
Single digit → distance = 0

Code

class Solution:
    def mirrorDistance(self, n: int) -> int:
        """
        3783. Mirror Distance of an Integer
        Time: O(log n)
        Space: O(1)
        """
        reverse_n = int(str(n)[::-1])
        return abs(n - reverse_n)

/**
 * 3783. Mirror Distance of an Integer
 * Time: O(log n)
 * Space: O(1)
 */
class Solution {
    public int mirrorDistance(int n) {
        long reverseN = reverse(n);
        return (int) Math.abs(n - reverseN);
    }

    private long reverse(int n) {
        long result = 0;
        while (n > 0) {
            result = result * 10 + n % 10;
            n /= 10;
        }
        return result;
    }
}