How do you solve Prime Palindrome on LeetCode?

Prime Palindrome (LeetCode #866) 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 palindromic numbers directly reduces the search space.

What is the brute force solution for Prime Palindrome?

The brute force approach for Prime Palindrome is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each number starting from n to see if it is both prime and a palindrome. This is straightforward but can be inefficient for larger numbers. The optimal Optimal Solution approach improves this to O(n).

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

Prime Palindrome uses the following concepts: Math, Number Theory. The recommended patterns to study are: Math, Number Theory.

What are the interview tips for Prime Palindrome?

Clarify the problem requirements before jumping into coding. Discuss your thought process and potential edge cases with the interviewer. Optimize your solution iteratively — start with brute force, then refine.

What are common mistakes when solving Prime Palindrome?

Not checking edge cases like n being less than 2. Confusing the definitions of prime and palindrome.

#866

Prime Palindrome

Medium

MathNumber TheoryMathNumber Theory

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)

Instead of checking every number, we can generate palindromic numbers directly and then check if they are prime. This is more efficient because it reduces the number of candidates we need to check.

⚙️

Algorithm

3 steps

1Step 1: Generate palindromic numbers starting from n.
2Step 2: For each palindromic number, check if it is prime.
3Step 3: Return the first palindromic number that is also prime.

solution.py25 lines

1# Full working Python code
2
3def is_prime(num):
4    if num < 2:
5        return False
6    for i in range(2, int(num**0.5) + 1):
7        if num % i == 0:
8            return False
9    return True
10
11def generate_palindrome(num):
12    s = str(num)
13    return int(s + s[::-1][1:])  # Generate even-length palindrome
14
15def prime_palindrome(n):
16    if n <= 2:
17        return 2
18    for length in range(1, 9):  # Generate palindromes of increasing lengths
19        for half in range(10**(length-1), 10**length):
20            pal = generate_palindrome(half)
21            if pal >= n and is_prime(pal):
22                return pal
23            pal = generate_palindrome(half) + 1  # Odd-length palindrome
24            if pal >= n and is_prime(pal):
25                return pal

ℹ

Complexity note: The time complexity is O(n) because we generate palindromic numbers directly, significantly reducing the number of candidates we need to check for primality.

1Generating palindromic numbers directly reduces the search space.
2Checking for primality can be optimized by only checking up to the square root.

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