How do you solve Mirror Distance of an Integer on LeetCode?

Mirror Distance of an Integer (LeetCode #3783) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(1) space complexity. Key insight: Reversing digits can be done mathematically without string manipulation.

What is the brute force solution for Mirror Distance of an Integer?

The brute force approach for Mirror Distance of an Integer is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach involves reversing the digits of the number and then calculating the absolute difference. It’s straightforward but inefficient for larger numbers. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Mirror Distance of an Integer?

Mirror Distance of an Integer uses the following concepts: Math. The recommended patterns to study are: String Manipulation, Mathematical Operations.

What are the interview tips for Mirror Distance of an Integer?

Explain your thought process clearly. Consider edge cases like single-digit numbers. Optimize your solution before coding.

What are common mistakes when solving Mirror Distance of an Integer?

Forgetting to handle leading zeros after reversing. Not using absolute value correctly.

#3783

Mirror Distance of an Integer

Easy

MathString ManipulationMathematical Operations

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n)
Space	O(1)	O(1)

💡

Intuition

Time O(log n)Space O(1)

This approach directly manipulates the number to reverse its digits without converting to a string, making it more efficient.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to store the reversed number.
2Step 2: Use a loop to extract digits from n and build the reversed number.
3Step 3: Calculate the absolute difference between n and the reversed number.

solution.py7 lines

1def mirror_distance(n):
2    rev = 0
3    temp = n
4    while temp > 0:
5        rev = rev * 10 + temp % 10
6        temp //= 10
7    return abs(n - rev)

ℹ

Complexity note: The time complexity is O(log n) because we are processing each digit of the number, and space complexity is O(1) since we are using a fixed amount of extra space.

1Reversing digits can be done mathematically without string manipulation.
2Absolute difference is always non-negative.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.