How do you solve Find Palindrome With Fixed Length on LeetCode?

Find Palindrome With Fixed Length (LeetCode #2217) 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: Palindromes can be constructed from their first half, reducing the need to check every number.

What is the brute force solution for Find Palindrome With Fixed Length?

The brute force approach for Find Palindrome With Fixed Length is Brute Force, with O(n²) time complexity and O(n) space complexity. We can generate all possible numbers of the given length and check if they are palindromes. This is straightforward but inefficient, especially for larger lengths. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Palindrome With Fixed Length?

Find Palindrome With Fixed Length uses the following concepts: Array, Math. The recommended patterns to study are: Math, String Manipulation.

What are the interview tips for Find Palindrome With Fixed Length?

Always clarify edge cases, such as the smallest and largest possible inputs. Think about how to optimize your solution before diving into coding. Explain your thought process clearly to the interviewer; it shows your understanding.

What are common mistakes when solving Find Palindrome With Fixed Length?

Not considering leading zeros when generating palindromes. Failing to handle cases where the query exceeds the number of available palindromes.

#2217

Find Palindrome With Fixed Length

Medium

ArrayMathMathString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Instead of generating all palindromes, we can construct them directly using the first half of the number. This is efficient and avoids unnecessary checks.

⚙️

Algorithm

3 steps

1Step 1: Calculate the number of palindromes of the given length using the formula: 9 * 10^(ceil(intLength/2) - 1).
2Step 2: For each query, check if it exceeds the total number of palindromes; if so, return -1.
3Step 3: Construct the palindrome using the first half digits derived from the query.

solution.py15 lines

1def kthPalindrome(queries, intLength):
2    half_length = (intLength + 1) // 2
3    total_palindromes = 9 * (10 ** (half_length - 1))
4    answer = []
5    for q in queries:
6        if q > total_palindromes:
7            answer.append(-1)
8        else:
9            first_half = str(10**(half_length - 1) + q - 1)
10            if intLength % 2 == 0:
11                palindrome = first_half + first_half[::-1]
12            else:
13                palindrome = first_half + first_half[-2::-1]
14            answer.append(int(palindrome))
15    return answer

ℹ

Complexity note: The time complexity is O(n) because we directly compute the palindromes without generating all of them, making it efficient for large inputs.

1Palindromes can be constructed from their first half, reducing the need to check every number.
2The number of palindromes of a given length can be calculated mathematically.

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