What is the main pattern behind the 'Find Palindrome With Fixed Length' problem?

The main pattern involves generating palindromes using mathematical manipulation of numbers and efficiently handling queries for specific palindromes based on their index.

How can I efficiently generate palindromes of a fixed length?

You can generate palindromes by constructing the first half of the number and reflecting it to form the entire palindrome. The approach differs slightly for odd and even lengths.

How do I handle large queries that ask for a non-existent palindrome?

If a query asks for a palindrome that exceeds the maximum possible palindrome for a given length, return -1 for that query.

What is the time complexity of solving the 'Find Palindrome With Fixed Length' problem?

The time complexity depends on how you calculate each palindrome. Efficient generation of palindromes can keep the complexity linear in terms of the number of queries.

Can this problem be solved without generating all the palindromes upfront?

Yes, by calculating each palindrome as needed, without pre-generating the entire set, you can optimize both time and space complexity.

#2217

Medium

auto_awesome数组·数学

LeetCode 题解工作台

找到指定长度的回文数

给你一个整数数组 queries 和一个正整数 intLength ，请你返回一个数组 answer ，其中 answer[i] 是长度为 intLength 的正回文数中第 queries[i] 小的数字，如果不存在这样的回文数，则为 -1 。回文数指的是从前往后和从后往前读一模一样的…

数组数学

题目描述

给你一个整数数组 queries 和一个正整数 intLength ，请你返回一个数组 answer ，其中 answer[i] 是长度为 intLength 的 正回文数 中第 queries[i] 小的数字，如果不存在这样的回文数，则为 -1 。

回文数 指的是从前往后和从后往前读一模一样的数字。回文数不能有前导 0 。

示例 1：

输入：queries = [1,2,3,4,5,90], intLength = 3
输出：[101,111,121,131,141,999]
解释：
长度为 3 的最小回文数依次是：
101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, ...
第 90 个长度为 3 的回文数是 999 。

示例 2：

输入：queries = [2,4,6], intLength = 4
输出：[1111,1331,1551]
解释：
长度为 4 的前 6 个回文数是：
1001, 1111, 1221, 1331, 1441 和 1551 。

提示：

1 <= queries.length <= 5 * 10⁴
1 <= queries[i] <= 10⁹
1 <= intLength <= 15

lightbulb

解题思路

方法一

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class Solution:
    def kthPalindrome(self, queries: List[int], intLength: int) -> List[int]:
        l = (intLength + 1) >> 1
        start, end = 10 ** (l - 1), 10**l - 1
        ans = []
        for q in queries:
            v = start + q - 1
            if v > end:
                ans.append(-1)
                continue
            s = str(v)
            s += s[::-1][intLength % 2 :]
            ans.append(int(s))
        return ans

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
The candidate understands how to efficiently generate palindromes based on the length and query.
question_mark
The candidate handles edge cases such as queries larger than the number of available palindromes.
question_mark
The candidate can effectively use array and mathematical techniques to minimize computation.

warning

常见陷阱

外企场景

error
Misunderstanding how to generate palindromes efficiently, leading to excessive computation or storage.
error
Failing to handle edge cases where the query exceeds the number of possible palindromes.
error
Incorrectly assuming palindromes can have leading zeros, which is not allowed in this problem.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the `intLength` were increased, making the palindromes larger?
arrow_right_alt
How would you handle queries with very large values (e.g., `queries[i]` near 10^9)?
arrow_right_alt
What if the problem allowed leading zeros in the palindromes?

help

常见问题

外企场景

继续练习

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