How do you solve A Number After a Double Reversal on LeetCode?

A Number After a Double Reversal (LeetCode #2119) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Reversing a number can lead to loss of leading zeros, which is crucial for this problem.

What is the brute force solution for A Number After a Double Reversal?

The brute force approach for A Number After a Double Reversal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves reversing the number twice and checking if the final result equals the original number. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve A Number After a Double Reversal?

A Number After a Double Reversal uses the following concepts: Math. The recommended patterns to study are: Math, String Manipulation.

What are the interview tips for A Number After a Double Reversal?

Always check edge cases like zero and numbers ending with zero. Think about how you can simplify the problem before diving into coding. Explain your thought process clearly to the interviewer; it shows your understanding.

What are common mistakes when solving A Number After a Double Reversal?

Not considering the case where the number ends with zero. Overcomplicating the solution by performing unnecessary operations.

#2119

A Number After a Double Reversal

Easy

MathMathString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(1)Space O(1)

The optimal solution recognizes that if a number ends with zero (except for zero itself), it cannot equal its double-reversed form. Hence, we can directly check this condition without performing unnecessary reversals.

⚙️

Algorithm

3 steps

1Step 1: Check if the number is zero; if yes, return true.
2Step 2: If the number ends with zero, return false.
3Step 3: Otherwise, return true since reversing twice will yield the original number.

solution.py2 lines

1def isSameAfterReversals(num):
2    return num == 0 or num % 10 != 0

ℹ

Complexity note: The time complexity is O(1) because we are only performing a constant number of operations. The space complexity is also O(1) as we are not using any additional data structures.

1Reversing a number can lead to loss of leading zeros, which is crucial for this problem.
2Any number that ends with zero (except zero itself) will not equal its double-reversed form.

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