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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2201_2300.s2217_find_palindrome_with_fixed_length;

// #Medium #Array #Math #2022_06_12_Time_37_ms_(88.60%)_Space_53.7_MB_(77.19%)

/**
 * 2217 - Find Palindrome With Fixed Length\.
 *
 * Medium
 *
 * 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 90^th 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`
**/
@SuppressWarnings("java:S2184")
public class Solution {
    public long[] kthPalindrome(int[] queries, int intLength) {
        long minHalf = (long) Math.pow(10, (intLength - 1) / 2);
        long maxIndex = (long) Math.pow(10, (intLength + 1) / 2) - minHalf;
        boolean isOdd = intLength % 2 == 1;
        long[] res = new long[queries.length];
        for (int i = 0; i < res.length; i++) {
            res[i] = queries[i] > maxIndex ? -1 : helper(queries[i], minHalf, isOdd);
        }
        return res;
    }

    private long helper(long index, long minHalf, boolean isOdd) {
        long half = minHalf + index - 1;
        long res = half;
        if (isOdd) {
            res /= 10;
        }
        while (half != 0) {
            res = res * 10 + half % 10;
            half /= 10;
        }
        return res;
    }
}