How do you solve Largest Palindrome Product on LeetCode?

Largest Palindrome Product (LeetCode #479) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(1) space complexity. Key insight: Palindromes are symmetric, which can help reduce the number of checks needed.

What is the brute force solution for Largest Palindrome Product?

The brute force approach for Largest Palindrome Product is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible products of two n-digit numbers and checking which of these products is a palindrome. This is straightforward but inefficient for larger values of n. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Largest Palindrome Product?

Largest Palindrome Product uses the following concepts: Math, Enumeration. The recommended patterns to study are: Brute Force, Math.

What are the interview tips for Largest Palindrome Product?

Always clarify the constraints and edge cases before diving into coding. Discuss your thought process and approach with the interviewer to get feedback. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Largest Palindrome Product?

Not checking for palindromes efficiently, leading to unnecessary computations. Failing to consider the modulo operation in the final output.

#479

Largest Palindrome Product

Hard

MathEnumerationBrute ForceMath

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(1)

💡

Intuition

Time O(n²)Space O(1)

The optimal solution leverages the properties of palindromes and the structure of n-digit numbers to reduce the number of products we need to check. Instead of checking all pairs, we can start from the largest numbers and work our way down, stopping early when we find a valid palindrome.

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Algorithm

6 steps

1Step 1: Determine the range of n-digit numbers, which is from 10^(n-1) to 10^n - 1.
2Step 2: Initialize a variable to keep track of the largest palindrome found.
3Step 3: Iterate through n-digit numbers in descending order, starting from the largest.
4Step 4: For each number, generate potential palindromic products by constructing palindromes directly.
5Step 5: Check if the palindrome can be formed by multiplying two n-digit numbers.
6Step 6: Return the largest palindrome found modulo 1337.

solution.py13 lines

1# Full working Python code
2
3def largest_palindrome_product(n):
4    max_palindrome = 0
5    upper_limit = 10**n - 1
6    for i in range(upper_limit, 10**(n-1) - 1, -1):
7        for j in range(i, 10**(n-1) - 1, -1):
8            product = i * j
9            if product <= max_palindrome:
10                break
11            if str(product) == str(product)[::-1]:
12                max_palindrome = max(max_palindrome, product)
13    return max_palindrome % 1337

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we are reducing unnecessary checks by breaking early when we find a palindrome. The space complexity is O(1) as we are only using a few variables.

1Palindromes are symmetric, which can help reduce the number of checks needed.
2Starting from the largest numbers allows us to find the largest palindrome sooner.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.