LeetCode #479 — HARD

Largest Palindrome Product

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

The Problem

Problem Statement

Given an integer n, return the largest palindromic integer that can be represented as the product of two n-digits integers. Since the answer can be very large, return it modulo 1337.

Example 1:

Input: n = 2
Output: 987
Explanation: 99 x 91 = 9009, 9009 % 1337 = 987

Example 2:

Input: n = 1
Output: 9

Constraints:

1 <= n <= 8

Step 01

Brute Force Baseline

Problem summary: Given an integer n, return the largest palindromic integer that can be represented as the product of two n-digits integers. Since the answer can be very large, return it modulo 1337.

Baseline thinking

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

Pattern signal: Math

Example 1

Example 2

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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

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 #479: Largest Palindrome Product
class Solution {
    public int largestPalindrome(int n) {
        int mx = (int) Math.pow(10, n) - 1;
        for (int a = mx; a > mx / 10; --a) {
            int b = a;
            long x = a;
            while (b != 0) {
                x = x * 10 + b % 10;
                b /= 10;
            }
            for (long t = mx; t * t >= x; --t) {
                if (x % t == 0) {
                    return (int) (x % 1337);
                }
            }
        }
        return 9;
    }
}

// Accepted solution for LeetCode #479: Largest Palindrome Product
func largestPalindrome(n int) int {
	mx := int(math.Pow10(n)) - 1
	for a := mx; a > mx/10; a-- {
		x := a
		for b := a; b != 0; b /= 10 {
			x = x*10 + b%10
		}
		for t := mx; t*t >= x; t-- {
			if x%t == 0 {
				return x % 1337
			}
		}
	}
	return 9
}

# Accepted solution for LeetCode #479: Largest Palindrome Product
class Solution:
    def largestPalindrome(self, n: int) -> int:
        mx = 10**n - 1
        for a in range(mx, mx // 10, -1):
            b = x = a
            while b:
                x = x * 10 + b % 10
                b //= 10
            t = mx
            while t * t >= x:
                if x % t == 0:
                    return x % 1337
                t -= 1
        return 9

// Accepted solution for LeetCode #479: Largest Palindrome Product
struct Solution;

impl Solution {
    fn largest_palindrome(n: i32) -> i32 {
        if n == 1 {
            return 9;
        }
        let max = 10u64.pow(n as u32) - 1;
        for i in (0..max).rev() {
            let left: String = i.to_string();
            let right: String = i.to_string().chars().rev().collect();
            let palindrome = format!("{}{}", left, right);
            if let Ok(value) = palindrome.parse::<u64>() {
                let mut j = max;
                while j * j >= value {
                    if value % j == 0 {
                        return (value % 1337) as i32;
                    }
                    j -= 1;
                }
            }
        }
        0
    }
}

#[test]
fn test() {
    let n = 2;
    let res = 987;
    assert_eq!(Solution::largest_palindrome(n), res);
    let n = 1;
    let res = 9;
    assert_eq!(Solution::largest_palindrome(n), res);
    let n = 5;
    let res = 677;
    assert_eq!(Solution::largest_palindrome(n), res);
}

// Accepted solution for LeetCode #479: Largest Palindrome Product
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #479: Largest Palindrome Product
// class Solution {
//     public int largestPalindrome(int n) {
//         int mx = (int) Math.pow(10, n) - 1;
//         for (int a = mx; a > mx / 10; --a) {
//             int b = a;
//             long x = a;
//             while (b != 0) {
//                 x = x * 10 + b % 10;
//                 b /= 10;
//             }
//             for (long t = mx; t * t >= x; --t) {
//                 if (x % t == 0) {
//                     return (int) (x % 1337);
//                 }
//             }
//         }
//         return 9;
//     }
// }

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(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

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.