README_EN.md

2217. Find Palindrome With Fixed Length

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Description

Given an integer array queries and a positive integer intLength, return an array answer where answer[i] is either the queries[i]th smallest positive palindrome of length intLength or -1 if no such palindrome exists.

A palindrome is a number that reads the same backwards and forwards. Palindromes cannot have leading zeros.

 

Example 1:

Input: queries = [1,2,3,4,5,90], intLength = 3
Output: [101,111,121,131,141,999]
Explanation:
The first few palindromes of length 3 are:
101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, ...
The 90th palindrome of length 3 is 999.

Example 2:

Input: queries = [2,4,6], intLength = 4
Output: [1111,1331,1551]
Explanation:
The first six palindromes of length 4 are:
1001, 1111, 1221, 1331, 1441, and 1551.

 

Constraints:

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

Solutions

Python3

Java

TypeScript

function kthPalindrome(queries: number[], intLength: number): number[] {
    const isOdd = intLength % 2 === 1;
    const bestNum = 10 ** ((intLength >> 1) + (isOdd ? 1 : 0) - 1);
    const max = bestNum * 9;
    return queries.map(v => {
        if (v > max) {
            return -1;
        }
        const num = bestNum + v - 1;
        return Number(
            num +
                (num + '')
                    .split('')
                    .reverse()
                    .slice(isOdd ? 1 : 0)
                    .join(''),
        );
    });
}

Rust

impl Solution {
    pub fn kth_palindrome(queries: Vec<i32>, int_length: i32) -> Vec<i64> {
        let is_odd = int_length & 1 == 1;
        let best_num = i32::pow(10, (int_length / 2 + if is_odd { 0 } else { -1 }) as u32);
        let max = best_num * 9;
        queries
            .iter()
            .map(|&num| {
                if num > max {
                    return -1;
                }
                let num = best_num + num - 1;
                format!(
                    "{}{}",
                    num,
                    num.to_string()
                        .chars()
                        .rev()
                        .skip(if is_odd { 1 } else { 0 })
                        .collect::<String>()
                )
                .parse()
                .unwrap()
            })
            .collect()
    }
}

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