Mirror Distance of an Integer

Difficulty: Easy | Acceptance: 91.50% | Paid: No Topics: Math

Problem Description

Given an integer n, find its mirror (the number formed by reversing its digits) and return the absolute difference between n and its mirror.

For example, the mirror of 123 is 321, and the mirror distance is |123 - 321| = 198.

Table of Contents

• Examples
• Constraints
• String Manipulation
• Mathematical Reversal
• Recursive Reversal

Examples

Example 1

Input:

n = 25

Output:

27

Explanation: reverse(25) = 52.

Thus, the answer is abs(25 - 52) = 27.

Example 2

Input:

n = 10

Output:

9

Explanation: reverse(10) = 01 which is 1.

Thus, the answer is abs(10 - 1) = 9.

Example 3

Input:

n = 7

Output:

0

Explanation: reverse(7) = 7.

Thus, the answer is abs(7 - 7) = 0.

Constraints

-10⁹ <= n <= 10⁹

String Manipulation

Intuition Convert the number to a string, reverse it, and convert back to an integer to find the mirror, then calculate the absolute difference.

Steps

• Extract the sign of the number
• Convert the absolute value to a string
• Reverse the string and convert back to integer
• Apply the original sign to get the mirror
• Return the absolute difference between n and its mirror

class Solution:
    def mirrorDistance(self, n: int) -> int:
        sign = -1 if n < 0 else 1
        mirror = int(str(abs(n))[::-1]) * sign
        return abs(n - mirror)

Complexity

• Time: O(log n) - Converting to string and reversing takes time proportional to the number of digits
• Space: O(log n) - String representation of the number
• Notes: Simple and readable, but uses extra space for string conversion

Mathematical Reversal

Intuition Extract digits from the number using modulo and division operations, then build the reversed number mathematically.

Steps

• Extract the sign of the number
• Use modulo 10 to get the last digit
• Build the mirror by multiplying by 10 and adding each digit
• Apply the original sign to get the mirror
• Return the absolute difference between n and its mirror

class Solution:
    def mirrorDistance(self, n: int) -> int:
        sign = -1 if n < 0 else 1
        abs_n = abs(n)
        mirror = 0
        while abs_n > 0:
            mirror = mirror * 10 + abs_n % 10
            abs_n //= 10
        mirror *= sign
        return abs(n - mirror)

Complexity

• Time: O(log n) - We process each digit exactly once
• Space: O(1) - Only using a constant amount of extra space
• Notes: Most efficient approach with constant space complexity

Recursive Reversal

Intuition Use recursion to reverse the digits by processing the last digit first and building the reversed number.

Steps

• Extract the sign of the number
• Define a recursive function that takes the remaining number and the current reversed value
• Base case: when the number is 0, return the reversed value
• Recursive case: extract the last digit and add it to the reversed value, then recurse with the remaining digits
• Apply the original sign to get the mirror
• Return the absolute difference between n and its mirror

class Solution:
    def mirrorDistance(self, n: int) -> int:
        sign = -1 if n < 0 else 1
        abs_n = abs(n)
        
        def reverse(num, rev):
            if num == 0:
                return rev
            return reverse(num // 10, rev * 10 + num % 10)
        
        mirror = reverse(abs_n, 0) * sign
        return abs(n - mirror)

Complexity

• Time: O(log n) - We process each digit exactly once
• Space: O(log n) - Recursion stack depth equals the number of digits
• Notes: Elegant solution but uses stack space due to recursion