2119. A Number After a Double Reversal

Problem Description

This problem asks us to implement a function that performs an operation on an input integer num and checks for a certain condition. The operation involves reversing the digits of num, which we can call reversed1, and then reversing the digits of reversed1, resulting in reversed2. The condition to check is whether reversed2 is equal to the original integer num. If they are equal, the function should return true; otherwise, it should return false. It is important to note that when reversing an integer, any leading zeros that appear as a result of the reversal are not retained.

An example of the operation would be:

• If num is 123, reversed1 would be 321, and then reversed2 would be 123, which is equal to the original num.
• If num is 1200, reversed1 would be 21 (since leading zeros are not retained), and reversed2 would be 12, which is not equal to the original num.

Intuition

Upon analyzing the problem, we realize that reversing a number twice will always give us back the original number with one exception: if the original number has trailing zeros. This is because, during the first reversal, the trailing zeros are not retained, leading to a smaller number. When reversing for the second time, there is no way to get those zeros back; hence, reversed2 will not equal num if num has trailing zeros.

Therefore, the only time reversed2 is not equal to num is when num has one or more trailing zeros. The exception to this would be the number 0 itself, which remains the same even after multiple reversals.

So the solution can be reduced to a simple check:

• If num is 0, return true, because reversing 0 will always yield 0.
• Otherwise, check if num is divisible by 10 (which means it has at least one trailing zero), and if so, return false.
• In all other cases, return true, because any number without trailing zeros will stay the same after two reversals.

This is the thinking process that brings us to the concise solution provided in the code.

Learn more about Math patterns.

Solution Approach

The solution uses a straightforward logical check to determine whether the input num will remain the same after two reversals. This solution does not perform the actual reversal of the number, which would require additional operations and could be less efficient, especially for large numbers.

Algorithmically, the approach can be broken down as follows:

1. Zero Check: Directly return true if num is 0, as reversing 0 any number of times will still yield 0. This is a quick check to handle this special case.

2. Trailing Zero Check: For any number other than 0, check if it has trailing zeros, which we do by checking if num is divisible by 10 (using the modulo operator %). If it is, this means num ends with at least one zero and therefore will not be the same after two reversals. In this case, we return false.

3. Default Case: If neither of the above conditions is met, it implies that the number num does not have any trailing zeros and will remain the same after two reversals. Hence, we return true by default.

This approach is efficient because it operates in constant time (O(1)), and there is no need for additional data structures (no extra space complexity). We are using the characteristics of the problem itself to come up with a logical shortcut that bypasses the need for computation-heavy string conversions or arithmetic operations associated with actual number reversal.

The Python code for the solution is straightforward and concise:

1class Solution:
2    def isSameAfterReversals(self, num: int) -> bool:
3        # Return true if num is 0 or it does not end with a zero (not divisible by 10)
4        return num == 0 or num % 10 != 0

This code establishes the implementation of the logical checks discussed above in step-by-step fashion, resulting in an elegant and efficient solution to the problem.

Example Walkthrough

Let's consider an integer num = 1200 to illustrate the solution approach.

1. Zero Check: The function first checks if num is zero. In our example, num is 1200, so this condition fails. We do not return true here.

2. Trailing Zero Check: Next, the function checks if num ends with at least one zero. This is done by checking if num is divisible by 10. For our number, 1200 % 10 == 0 is true, which means num ends with a zero. According to our analysis, if a number ends with a zero, it will not be the same after two reversals due to loss of trailing zeros during the first reversal. Thus the function should return false.

3. Default Case: This step would only be reached if the number were not zero and did not end with a zero. But since our example number fails the Trailing Zero Check, the Default Case does not apply.

Applying the solution approach from the provided content, the function should return false for the input num = 1200. This means that after reversing the digits of 1200 to form reversed1 = 21, and then reversing reversed1 to get reversed2 = 12, reversed2 is not equal to the original num as a result of the missing trailing zeros.

Putting the example to test with the implemented code:

1solution = Solution()
2result = solution.isSameAfterReversals(1200)
3print(result)  # Output: False

The function correctly identifies that 1200, after two reversals, does not equal the original number due to the trailing zeros being dropped during the reversal process.

Solution Implementation

Time and Space Complexity

The given code checks whether a number is the same after reversing it twice. The operation involves checking if the number is zero or if the last digit of the number is not zero.

Time Complexity

The operation num % 10 is a modulo operation, which has a constant time complexity, i.e., O(1). Checking if a number is equal to zero or not (num == 0) is also a constant time operation. Since there are no loops or recursive calls, the overall time complexity of the function is O(1).

Space Complexity

The function uses a fixed amount of space, with only one input integer and no additional data structures or recursive calls that would use more space. As a result, the space complexity is constant as well, denoted as O(1). The function does not allocate any additional memory that grows with the input size, so the space used remains consistent regardless of the input.

Learn more about how to find time and space complexity quickly using problem constraints.