How do you solve Find the Closest Palindrome on LeetCode?

Find the Closest Palindrome (LeetCode #564) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Generating palindromes based on the prefix of the number significantly reduces the number of candidates to check.

What is the brute force solution for Find the Closest Palindrome?

The brute force approach for Find the Closest Palindrome is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible palindromes within a reasonable range around the given number and checking which one is closest. This is straightforward but inefficient for larger numbers. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Closest Palindrome?

Find the Closest Palindrome uses the following concepts: Math, String. The recommended patterns to study are: String Manipulation, Greedy Algorithms.

What are the interview tips for Find the Closest Palindrome?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases and how they might affect your solution. Discuss your thought process and reasoning with the interviewer as you work through the problem.

What are common mistakes when solving Find the Closest Palindrome?

Not considering edge cases like single-digit numbers or very large numbers. Failing to exclude the original number from the candidates.

#564

Find the Closest Palindrome

Hard

MathStringString ManipulationGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal approach focuses on generating only a few specific palindrome candidates based on the structure of the input number, which drastically reduces the number of checks needed. By only considering the first half of the number, we can create potential palindromes efficiently.

⚙️

Algorithm

3 steps

1Step 1: Identify the length of the input number and its prefix.
2Step 2: Generate palindromes based on the prefix, adjusting it slightly to account for potential closest palindromes.
3Step 3: Compare the generated palindromes to find the closest one, ensuring to exclude the original number.

solution.py21 lines

1# Full working Python code
2
3def closest_palindrome(n):
4    num = int(n)
5    candidates = set()
6    candidates.add(num - 1)
7    candidates.add(num + 1)
8    length = len(n)
9    prefix = int(n[:(length + 1) // 2])
10    for i in [-1, 0, 1]:
11        candidate = str(prefix + i)
12        if length % 2 == 0:
13            palindrome = candidate + candidate[::-1]
14        else:
15            palindrome = candidate + candidate[-2::-1]
16        candidates.add(int(palindrome))
17    candidates.discard(num)
18    return str(min(candidates, key=lambda x: (abs(x - num), x)))
19
20# Example usage
21print(closest_palindrome('123'))

ℹ

Complexity note: The optimal solution runs in O(n) time because we only generate a few specific candidates based on the input's prefix, rather than checking all possible palindromes. The space complexity is O(n) due to the storage of candidates.

1Generating palindromes based on the prefix of the number significantly reduces the number of candidates to check.
2When comparing candidates, ensure to handle ties by choosing the smaller number.

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