What is the brute force solution for Maximum Product of the Length of Two Palindromic Substrings?

The brute force approach for Maximum Product of the Length of Two Palindromic Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible substrings, checking if they are palindromes, and then calculating the maximum product of the lengths of two non-overlapping palindromic substrings. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Product of the Length of Two Palindromic Substrings?

Maximum Product of the Length of Two Palindromic Substrings uses the following concepts: Two Pointers, String, Rolling Hash, Hash Function. The recommended patterns to study are: Two Pointers, Dynamic Programming, String Manipulation.

What are the interview tips for Maximum Product of the Length of Two Palindromic Substrings?

Explain your thought process clearly. Discuss edge cases, like very short strings. Practice Manacher's algorithm beforehand.

What are common mistakes when solving Maximum Product of the Length of Two Palindromic Substrings?

Not checking for odd-length palindromes. Overlapping substrings are mistakenly considered.

#1960

Maximum Product of the Length of Two Palindromic Substrings

Hard

Two PointersStringRolling HashHash FunctionTwo PointersDynamic ProgrammingString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Using Manacher's algorithm allows us to efficiently find all odd-length palindromic substrings in linear time. We can then use a two-pointer technique to find the maximum product of the lengths of non-overlapping palindromic substrings.

⚙️

Algorithm

3 steps

1Step 1: Use Manacher's algorithm to find the lengths of the longest odd-length palindromic substrings centered at each index.
2Step 2: Create two arrays to store the maximum lengths of palindromes to the left and right of each index.
3Step 3: Iterate through the string, and for each center, calculate the maximum product of lengths of non-overlapping palindromic substrings using the precomputed lengths.

solution.py30 lines

1def manachers_algorithm(s):
2    n = len(s)
3    p = [0] * n
4    center = right = 0
5    for i in range(n):
6        mirror = 2 * center - i
7        if i < right:
8            p[i] = min(right - i, p[mirror])
9        a, b = i + (1 + p[i]), i - (1 + p[i])
10        while a < n and b >= 0 and s[a] == s[b]:
11            p[i] += 1
12            a += 1
13            b -= 1
14        if i + p[i] > right:
15            center, right = i, i + p[i]
16    return p
17
18def max_product(s):
19    n = len(s)
20    p = manachers_algorithm(s)
21    left_max = [0] * n
22    right_max = [0] * n
23    for i in range(n):
24        left_max[i] = p[i] if i == 0 else max(left_max[i - 1], p[i])
25    for i in range(n - 1, -1, -1):
26        right_max[i] = p[i] if i == n - 1 else max(right_max[i + 1], p[i])
27    max_product = 0
28    for i in range(n - 1):
29        max_product = max(max_product, left_max[i] * right_max[i + 1])
30    return max_product

ℹ

Complexity note: The time complexity is O(n) due to the linear scan of the string for Manacher's algorithm. The space complexity is O(n) because we store the lengths of palindromes and the maximum lengths in separate arrays.

1Using Manacher's algorithm is crucial for efficiency.
2Understanding non-overlapping conditions is key to maximizing the product.

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