LeetCode #3260 — HARD

Find the Largest Palindrome Divisible by K

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given two positive integers n and k.

An integer x is called k-palindromic if:

  • x is a palindrome.
  • x is divisible by k.

Return the largest integer having n digits (as a string) that is k-palindromic.

Note that the integer must not have leading zeros.

Example 1:

Input: n = 3, k = 5

Output: "595"

Explanation:

595 is the largest k-palindromic integer with 3 digits.

Example 2:

Input: n = 1, k = 4

Output: "8"

Explanation:

4 and 8 are the only k-palindromic integers with 1 digit.

Example 3:

Input: n = 5, k = 6

Output: "89898"

Constraints:

  • 1 <= n <= 105
  • 1 <= k <= 9
Patterns Used
📊Dynamic Programming→🎯Greedy→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given two positive integers n and k. An integer x is called k-palindromic if: x is a palindrome. x is divisible by k. Return the largest integer having n digits (as a string) that is k-palindromic. Note that the integer must not have leading zeros.

Baseline thinking

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

Pattern signal: Math · Dynamic Programming · Greedy

Example 1

3
5

Example 2

1
4

Example 3

5
6

Related Problems

  • Palindrome Number (palindrome-number)
  • Find the Closest Palindrome (find-the-closest-palindrome)
Step 02

Core Insight

What unlocks the optimal approach

  • It must have a solution since we can have all digits equal to <code>k</code>.
  • Use string dp, store modulus along with length of number currently formed.
  • Is it possible to solve greedily using divisibility rules?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #3260: Find the Largest Palindrome Divisible by K
class Solution {
  public String largestPalindrome(int n, int k) {
    StringBuilder sb = new StringBuilder();

    switch (k) {
      case 1:
        return "9".repeat(n);
      case 2:
        return n <= 2 ? "8".repeat(n)
                      : sb.append("8") //
                            .append("9".repeat(n - 2))
                            .append("8")
                            .toString();
      case 3:
      case 9:
        return "9".repeat(n);
      case 4:
        return n <= 4 ? "8".repeat(n)
                      : sb.append("88") //
                            .append("9".repeat(n - 4))
                            .append("88")
                            .toString();
      case 5:
        return n <= 2 ? "5".repeat(n)
                      : sb.append("5") //
                            .append("9".repeat(n - 2))
                            .append("5")
                            .toString();
      case 6:
        if (n <= 2) {
          return "6".repeat(n);
        } else if (n % 2 == 1) {
          final int l = n / 2 - 1;
          return sb.append("8")
              .append("9".repeat(l))
              .append("8")
              .append("9".repeat(l))
              .append("8")
              .toString();
        } else {
          final int l = n / 2 - 2;
          return sb.append("8")
              .append("9".repeat(l))
              .append("77")
              .append("9".repeat(l))
              .append("8")
              .toString();
        }
      case 8:
        return n <= 6 ? "8".repeat(n)
                      : sb.append("888") //
                            .append("9".repeat(n - 6))
                            .append("888")
                            .toString();
      default:
        String[] middle = {"",         "7",         "77",         "959",
                           "9779",     "99799",     "999999",     "9994999",
                           "99944999", "999969999", "9999449999", "99999499999"};
        final int q = n / 12;
        final int r = n % 12;
        return sb.append("999999".repeat(q))
            .append(middle[r])
            .append("999999".repeat(q))
            .toString();
    }
  }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × m)
Space
O(n × m)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.

Related Problems
#1363Largest Multiple of Threehard→#2310Sum of Numbers With Units Digit Kmedium→#2745Construct the Longest New Stringmedium→#45Jump Game IImedium→#55Jump Gamemedium→