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Leetcode 2119. A Number After a Double Reversal

Problem Statement

Given an integer num, return true if num can be transformed into num after two reversals, else return false.

Clarifying Questions

1. What is the range of the integer num?
   • The problem constraint usually indicates that num can be any integer within typical bounds for such problems (e.g., within the range of a 32-bit signed integer: -2^31 to 2^31-1).
2. How should we handle negative numbers?
   • This problem likely only concerns non-negative integers, but it’s important to confirm any expectations regarding input limits and types.
3. Can num be zero?
   • Yes, and in that case, zero should return true because “00” after reversal is still “0”.

Strategy

The double reversal operation consists of:

1. Converting the number to a string and reversing it.
2. Reversing the string again and converting it back to an integer.

From practical observation:

• If num does not contain trailing zeroes, double reversal will return the original number.
• If num has trailing zeroes, double reversal will remove these trailing zeroes causing the number to change.

In essence:

• Any number with no trailing zero or the number 0 itself will remain the same after two reversals.

Code

public class Solution {
    public boolean isSameAfterReversals(int num) {
        // A number without trailing zeros or is zero itself remains unchanged after double reversal
        return num == 0 || num % 10 != 0;
    }
}

Time Complexity

• The computation uses basic modulus operation which is constant time operation.
• Therefore, the time complexity of the solution is O(1).

Explanation

1. num == 0: Special case handled separately, returns true since 0 remains 0 after any number of reversals.
2. num % 10 != 0: If the last digit of num is non-zero, num doesn’t have trailing zeros and remains unchanged post double reversal.

Any number having trailing zeroes will fail this condition and return false. This assures that any potential input is handled efficiently and correctly.

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