Leetcode #1960: Maximum Product of the Length of Two Palindromic Substrings
In this guide, we solve Leetcode #1960 Maximum Product of the Length of Two Palindromic Substrings in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed string s and are tasked with finding two non-intersecting palindromic substrings of odd length such that the product of their lengths is maximized. More formally, you want to choose four integers i, j, k, l such that 0 <= i <= j < k <= l < s.length and both the substrings s[i...j] and s[k...l] are palindromes and have odd lengths.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: String, Hash Function, Rolling Hash
Intuition
We need to scan characters while tracking positions or counts.
A simple state machine keeps the logic precise.
Approach
Iterate through the string once and update the state for each character.
Use a map or array if you need fast lookups.
Steps:
- Iterate through characters.
- Maintain necessary state.
- Build or validate the output.
Example
Input: s = "ababbb"
Output: 9
Explanation: Substrings "aba" and "bbb" are palindromes with odd length. product = 3 * 3 = 9.
Python Solution
class Solution:
def maxProduct(self, s: str) -> int:
n = len(s)
hlen = [0] * n
center = right = 0
for i in range(n):
if i < right:
hlen[i] = min(right - i, hlen[2 * center - i])
while (
0 <= i - 1 - hlen[i]
and i + 1 + hlen[i] < len(s)
and s[i - 1 - hlen[i]] == s[i + 1 + hlen[i]]
):
hlen[i] += 1
if right < i + hlen[i]:
center, right = i, i + hlen[i]
prefix = [0] * n
suffix = [0] * n
for i in range(n):
prefix[i + hlen[i]] = max(prefix[i + hlen[i]], 2 * hlen[i] + 1)
suffix[i - hlen[i]] = max(suffix[i - hlen[i]], 2 * hlen[i] + 1)
for i in range(1, n):
prefix[~i] = max(prefix[~i], prefix[~i + 1] - 2)
suffix[i] = max(suffix[i], suffix[i - 1] - 2)
for i in range(1, n):
prefix[i] = max(prefix[i - 1], prefix[i])
suffix[~i] = max(suffix[~i], suffix[~i + 1])
return max(prefix[i - 1] * suffix[i] for i in range(1, n))
Complexity
The time complexity is O(n). The space complexity is O(1) to O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.