How do you solve Palindrome Number on LeetCode?

Palindrome Number (LeetCode #9) can be solved using Brute Force or Optimal Solution (Without String Conversion). The optimal approach is Optimal Solution (Without String Conversion) with O(log10(n)) time complexity and O(1) space complexity. Key insight: Recognizing that palindromes read the same forwards and backwards is key. This can be approached through string manipulation or mathematical reversal.

What is the brute force solution for Palindrome Number?

The brute force approach for Palindrome Number is Brute Force, with O(n²) time complexity and O(n) space complexity. The simplest way to check if a number is a palindrome is to convert it to a string and compare it with its reverse. If they are the same, the number is a palindrome. The optimal Optimal Solution (Without String Conversion) approach improves this to O(log10(n)).

What algorithm or data structure is used to solve Palindrome Number?

Palindrome Number uses the following concepts: Math. The recommended patterns to study are: Hash Map, Array, Two Pointers.

What are the interview tips for Palindrome Number?

When you first see this problem, clarify if you can use string manipulation or if you need to solve it mathematically. Communicate your thought process clearly, explaining why you chose a specific approach and how it handles edge cases. Mention edge cases like negative numbers, zero, and single-digit numbers, as these can often trip up solutions.

What are common mistakes when solving Palindrome Number?

Students often forget to handle negative numbers and numbers that end with zero, leading to incorrect results. Another common mistake is to not consider the case where the number has an odd number of digits and how to compare the two halves correctly.

Palindrome Number

Easy

MathHash MapArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution (Without String Conversion)★
Time	O(n²)	O(log10(n))
Space	O(n)	O(1)

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Intuition

Time O(log10(n))Space O(1)

Instead of converting the integer to a string, we can reverse half of the number and compare it with the other half. This avoids the overhead of string manipulation and works directly with the integer.

⚙️

Algorithm

4 steps

1Step 1: If x is negative or ends with 0 (and is not 0), return false, as it can't be a palindrome.
2Step 2: Initialize a variable to store the reversed half of the number.
3Step 3: While x is greater than the reversed half, keep extracting the last digit of x and adding it to the reversed half.
4Step 4: After the loop, compare x with the reversed half. If they are equal or if x is equal to reversed half divided by 10 (for odd-length numbers), return true; otherwise, return false.

solution.py14 lines

1# Full working Python code with comments
2
3def is_palindrome(x: int) -> bool:
4    # Step 1: Handle negative numbers and multiples of 10
5    if x < 0 or (x % 10 == 0 and x != 0):
6        return False
7    reversed_half = 0
8    # Step 2: Reverse half of the number
9    while x > reversed_half:
10        reversed_half = reversed_half * 10 + x % 10
11        x //= 10
12    # Step 3: Compare the two halves
13    return x == reversed_half or x == reversed_half // 10
14

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Complexity note: The time complexity is O(log10(n)) because we are effectively reducing the number of digits in x by half with each iteration. The space complexity is O(1) since we are using a constant amount of space.

1Recognizing that palindromes read the same forwards and backwards is key. This can be approached through string manipulation or mathematical reversal.
2Understanding how to handle edge cases, such as negative numbers and trailing zeros, is crucial for accurate solutions.

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