LeetCode #2217 — MEDIUM

Find Palindrome With Fixed Length

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[1,2,3,4,5,90]
3

Example 2

[2,4,6]
4

Core Insight

What unlocks the optimal approach

For any value of queries[i] and intLength, how can you check if there exists at least queries[i] palindromes of length intLength?
Since a palindrome reads the same forwards and backwards, consider how you can efficiently find the first half (ceil(intLength/2) digits) of the palindrome.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2217: Find Palindrome With Fixed Length
class Solution {
    public long[] kthPalindrome(int[] queries, int intLength) {
        int n = queries.length;
        long[] ans = new long[n];
        int l = (intLength + 1) >> 1;
        long start = (long) Math.pow(10, l - 1);
        long end = (long) Math.pow(10, l) - 1;
        for (int i = 0; i < n; ++i) {
            long v = start + queries[i] - 1;
            if (v > end) {
                ans[i] = -1;
                continue;
            }
            String s = "" + v;
            s += new StringBuilder(s).reverse().substring(intLength % 2);
            ans[i] = Long.parseLong(s);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2217: Find Palindrome With Fixed Length
func kthPalindrome(queries []int, intLength int) []int64 {
	l := (intLength + 1) >> 1
	start, end := int(math.Pow10(l-1)), int(math.Pow10(l))-1
	var ans []int64
	for _, q := range queries {
		v := start + q - 1
		if v > end {
			ans = append(ans, -1)
			continue
		}
		t := v
		if intLength%2 == 1 {
			t /= 10
		}
		for t > 0 {
			v = v*10 + t%10
			t /= 10
		}
		ans = append(ans, int64(v))
	}
	return ans
}

# Accepted solution for LeetCode #2217: Find Palindrome With Fixed Length
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

// Accepted solution for LeetCode #2217: Find Palindrome With Fixed Length
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()
    }
}

// Accepted solution for LeetCode #2217: Find Palindrome With Fixed Length
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(''),
        );
    });
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.