What is the approach for finding the smallest prime palindrome?

Start by checking if the number is prime and a palindrome, and incrementally search for the smallest valid number greater than or equal to the given n.

How do you efficiently check if a number is prime?

Use trial division up to the square root of the number, which is the most efficient method for small to moderate-size numbers.

What are some common mistakes in solving the Prime Palindrome problem?

Common mistakes include inefficient prime checking, failing to correctly validate palindromes, and not optimizing the search process for large values of n.

How does the GhostInterview tool help with prime palindrome problems?

It provides targeted problem-solving scenarios, guides you through optimization steps, and helps you understand key patterns like prime checking and palindrome validation.

What are the time and space complexities for this problem?

The time complexity depends on the efficiency of prime checking and palindrome validation, while space complexity is typically O(1) if only a constant amount of space is used for processing the number.

#866

Medium

auto_awesome数学·结合·number·theory

LeetCode 题解工作台

回文质数

给你一个整数 n ，返回大于或等于 n 的最小回文质数。一个整数如果恰好有两个除数： 1 和它本身，那么它是质数。注意， 1 不是质数。例如， 2 、 3 、 5 、 7 、 11 和 13 都是质数。一个整数如果从左向右读和从右向左读是相同的，那么它是回文数。例如， 101 和…

数学数论

题目描述

给你一个整数 n ，返回大于或等于 n 的最小 回文质数。

一个整数如果恰好有两个除数：1 和它本身，那么它是质数。注意，1 不是质数。

例如，2、3、5、7、11 和 13 都是质数。

一个整数如果从左向右读和从右向左读是相同的，那么它是 回文数 。

例如，101 和 12321 都是回文数。

测试用例保证答案总是存在，并且在 [2, 2 * 10⁸] 范围内。

示例 1：

输入：n = 6
输出：7

示例 2：

输入：n = 8
输出：11

示例 3：

输入：n = 13
输出：101

提示：

1 <= n <= 10⁸

lightbulb

解题思路

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class Solution:
    def primePalindrome(self, n: int) -> int:
        def is_prime(x):
            if x < 2:
                return False
            v = 2
            while v * v <= x:
                if x % v == 0:
                    return False
                v += 1
            return True

        def reverse(x):
            res = 0
            while x:
                res = res * 10 + x % 10
                x //= 10
            return res

        while 1:
            if reverse(n) == n and is_prime(n):
                return n
            if 10**7 < n < 10**8:
                n = 10**8
            n += 1

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Look for the candidate’s understanding of prime number theory and palindrome detection.
question_mark
Pay attention to optimization strategies for both checking and searching for prime palindromes.
question_mark
Evaluate their approach to handling large inputs efficiently under the constraint of 10^8.

warning

常见陷阱

外企场景

error
Failing to optimize the prime checking process, leading to slow solutions.
error
Not validating palindrome properties correctly, resulting in incorrect outputs.
error
Using brute force without considering performance optimizations when searching for the smallest prime palindrome.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Finding the nth prime palindrome instead of just the smallest.
arrow_right_alt
Limiting the solution to prime palindromes within a certain range.
arrow_right_alt
Generalizing to find prime palindromes of even or odd length only.

help

常见问题

外企场景

继续练习

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