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Merge pull request #3898 from revanth1718/main
Add Solution to LC problem 1960 (Hard)
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‎dsa-solutions/lc-solutions/1900-1999/1960 - Maximum Product of the Length of Two Palindromic Substrings.md

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---
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id: maximum-product-of-length-of-two-palindromic-substrings
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title: Maximum Product of the Length of Two Palindromic Substrings
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sidebar_label: Maximum Product of the Length of Two Palindromic Substrings
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tags: [String, Palindrome, C++, Python, Java]
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description: Solve the problem of finding the maximum product of the lengths of two non-intersecting palindromic substrings in a given string.
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---
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## Problem Statement
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### Problem Description
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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.
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More formally, you want to choose four integers `i, j, k, l` such that $0 \leq i \leq j < k \leq l \lt s.length$ and both the substrings `s[i...j]` and `s[k...l]` are palindromes and have odd lengths. `s[i...j]` denotes a substring from index `i` to index `j` inclusive.
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Return the maximum possible product of the lengths of the two non-intersecting palindromic substrings.
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A palindrome is a string that is the same forward and backward. A substring is a contiguous sequence of characters in a string.
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### Example
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**Example 1:**
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```
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Input: s = "ababbb"
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Output: 9
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```
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**Explanation:** Substrings "aba" and "bbb" are palindromes with odd length. product = 3 * 3 = 9.
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### Constraints
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- $2 \leq s.length \leq 10^5$
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- `s` consists of lowercase English letters.
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## Solution
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### Intuition
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To solve this problem, we can use dynamic programming and a sliding window approach to find all possible palindromic substrings of odd length. By iterating through the string, we can determine the longest palindromic substring ending at each position and the longest palindromic substring starting at each position. Then, we calculate the maximum product of the lengths of two non-intersecting palindromic substrings.
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### Time Complexity and Space Complexity Analysis
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- **Time Complexity**: The solution involves a linear scan of the string and dynamic programming updates, making the time complexity $O(n)$.
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- **Space Complexity**: The space complexity is $O(n)$ due to the storage required for dynamic programming arrays.
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### Code
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#### C++
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```java
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class Solution {
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public int maxProduct(String s) {
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int n = s.length();
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int[] left = new int[n];
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int[] right = new int[n];
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// Calculate longest palindromic substring length ending at each index
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int l = 0, r = -1;
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for (int i = 0; i < n; i++) {
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if (i > r) {
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l = r = i;
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while (l >= 0 && r < n && s.charAt(l) == s.charAt(r)) {
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l--;
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r++;
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}
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left[i] = r - l - 1;
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} else {
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int k = (r - i) / 2;
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if (left[i - k] / 2 + i < r) {
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left[i] = left[i - k];
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} else {
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l = i - (r - i);
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while (l >= 0 && r < n && s.charAt(l) == s.charAt(r)) {
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l--;
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r++;
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}
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left[i] = r - l - 1;
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}
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}
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}
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// Calculate longest palindromic substring length starting at each index
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l = n - 1;
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r = n;
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for (int i = n - 1; i >= 0; i--) {
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if (i < l) {
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l = r = i;
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while (l >= 0 && r < n && s.charAt(l) == s.charAt(r)) {
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l--;
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r++;
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}
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right[i] = r - l - 1;
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} else {
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int k = (i - l) / 2;
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if (right[i + k] / 2 + i > l) {
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right[i] = right[i + k];
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} else {
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l = i + (i - l);
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while (l >= 0 && r < n && s.charAt(l) == s.charAt(r)) {
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l--;
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r++;
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}
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right[i] = r - l - 1;
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}
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}
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}
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// Calculate the maximum product
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```
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```cpp
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class Solution {
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public:
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int maxProduct(string s) {
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int n = s.size();
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vector<int> left(n, 0), right(n, 0);
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// Calculate longest palindromic substring length ending at each index
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for (int i = 0, l = 0, r = -1; i < n; i++) {
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if (i > r) {
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l = r = i;
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while (l >= 0 && r < n && s[l] == s[r]) l--, r++;
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left[i] = r - l - 1;
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} else {
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int k = (r - i) / 2;
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if (left[i - k] / 2 + i < r) {
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left[i] = left[i - k];
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} else {
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l = i - (r - i);
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while (l >= 0 && r < n && s[l] == s[r]) l--, r++;
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left[i] = r - l - 1;
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}
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}
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}
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// Calculate longest palindromic substring length starting at each index
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for (int i = n - 1, l = n - 1, r = n; i >= 0; i--) {
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if (i < l) {
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l = r = i;
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while (l >= 0 && r < n && s[l] == s[r]) l--, r++;
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right[i] = r - l - 1;
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} else {
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int k = (i - l) / 2;
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if (right[i + k] / 2 + i > l) {
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right[i] = right[i + k];
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} else {
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l = i + (i - l);
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while (l >= 0 && r < n && s[l] == s[r]) l--, r++;
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right[i] = r - l - 1;
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}
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}
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}
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// Calculate the maximum product
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int maxProduct = 0;
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for (int i = 0; i < n - 1; i++) {
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maxProduct = max(maxProduct, (left[i] / 2) * (right[i + 1] / 2));
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}
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return maxProduct;
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}
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};
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```
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#### Python
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```python
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class Solution:
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def maxProduct(self, s: str) -> int:
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n = len(s)
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left = [0] * n
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right = [0] * n
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# Calculate longest palindromic substring length ending at each index
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l, r = 0, -1
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for i in range(n):
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if i > r:
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l, r = i, i
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while l >= 0 and r < n and s[l] == s[r]:
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l -= 1
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r += 1
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left[i] = r - l - 1
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else:
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k = (r - i) // 2
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if left[i - k] // 2 + i < r:
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left[i] = left[i - k]
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else:
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l = i - (r - i)
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while l >= 0 and r < n and s[l] == s[r]:
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l -= 1
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r += 1
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left[i] = r - l - 1
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# Calculate longest palindromic substring length starting at each index
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l, r = n - 1, n
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for i in range(n - 1, -1, -1):
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if i < l:
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l, r = i, i
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while l >= 0 and r < n and s[l] == s[r]:
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l -= 1
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r += 1
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right[i] = r - l - 1
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else:
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k = (i - l) // 2
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if right[i + k] // 2 + i > l:
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right[i] = right[i + k]
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else:
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l = i + (i - l)
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while l >= 0 and r < n and s[l] == s[r]:
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l -= 1
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r += 1
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right[i] = r - l - 1
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# Calculate the maximum product
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maxProduct = 0
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for i in range(n - 1):
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maxProduct = max(maxProduct, (left[i] // 2) * (right[i + 1] // 2))
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return maxProduct
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```

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