feat: add solution 1960. Maximum Product of the Length of Two Palindromic Substrings

Author: tzuyi0817

solutions/1960. Maximum Product of the Length of Two Palindromic Substrings/README.md

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+# [1960. Maximum Product of the Length of Two Palindromic Substrings](https://leetcode.com/problems/maximum-product-of-the-length-of-two-palindromic-substrings)
+
+## Description
+
+<div class="elfjS" data-track-load="description_content"><p>You are given a <strong>0-indexed</strong> string <code>s</code> and are tasked with finding two <strong>non-intersecting palindromic </strong>substrings of <strong>odd</strong> length such that the product of their lengths is maximized.</p>
+
+<p>More formally, you want to choose four integers <code>i</code>, <code>j</code>, <code>k</code>, <code>l</code> such that <code>0 &lt;= i &lt;= j &lt; k &lt;= l &lt; s.length</code> and both the substrings <code>s[i...j]</code> and <code>s[k...l]</code> are palindromes and have odd lengths. <code>s[i...j]</code> denotes a substring from index <code>i</code> to index <code>j</code> <strong>inclusive</strong>.</p>
+
+<p>Return <em>the <strong>maximum</strong> possible product of the lengths of the two non-intersecting palindromic substrings.</em></p>
+
+<p>A <strong>palindrome</strong> is a string that is the same forward and backward. A <strong>substring</strong> is a contiguous sequence of characters in a string.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre><strong>Input:</strong> s = "ababbb"
+<strong>Output:</strong> 9
+<strong>Explanation:</strong> Substrings "aba" and "bbb" are palindromes with odd length. product = 3 * 3 = 9.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre><strong>Input:</strong> s = "zaaaxbbby"
+<strong>Output:</strong> 9
+<strong>Explanation:</strong> Substrings "aaa" and "bbb" are palindromes with odd length. product = 3 * 3 = 9.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s</code> consists of lowercase English letters.</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+## Solutions
+
+**Solution: `Manacher’s Algorithm`**
+
+- Time complexity: <em>O(n)</em>
+- Space complexity: <em>O(n)</em>
+
+<p>&nbsp;</p>
+
+### **JavaScript**
+
+```js
+/**
+ * @param {string} s
+ * @return {number}
+ */
+const maxProduct = function (s) {
+  const n = s.length;
+
+  const manacher = str => {
+    const maxExtends = Array.from({ length: n }, () => 0);
+    const leftToRight = Array.from({ length: n }, () => 1);
+    let center = 0;
+
+    for (let index = 0; index < n; index++) {
+      const r = center + maxExtends[center] - 1;
+      const mirrorIndex = center - (index - center);
+      let extend = index > r ? 1 : Math.min(maxExtends[mirrorIndex], r - index + 1);
+
+      while (index - extend >= 0 && index + extend < n && str[index - extend] === str[index + extend]) {
+        leftToRight[index + extend] = 2 * extend + 1;
+        extend += 1;
+      }
+
+      maxExtends[index] = extend;
+
+      if (index + maxExtends[index] >= r) {
+        center = index;
+      }
+    }
+
+    for (let index = 1; index < n; index++) {
+      leftToRight[index] = Math.max(leftToRight[index], leftToRight[index - 1]);
+    }
+
+    return leftToRight;
+  };
+
+  const maxLeft = manacher(s);
+  const reversed = s.split('').reverse().join('');
+  const maxRight = manacher(reversed).reverse();
+  let reuslt = 1;
+
+  for (let index = 1; index < n; index++) {
+    reuslt = Math.max(reuslt, maxLeft[index - 1] * maxRight[index]);
+  }
+
+  return reuslt;
+};
+```

solutions/1960. Maximum Product of the Length of Two Palindromic Substrings/solution.js

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+/**
+ * @param {string} s
+ * @return {number}
+ */
+const maxProduct = function (s) {
+  const n = s.length;
+
+  const manacher = str => {
+    const maxExtends = Array.from({ length: n }, () => 0);
+    const leftToRight = Array.from({ length: n }, () => 1);
+    let center = 0;
+
+    for (let index = 0; index < n; index++) {
+      const r = center + maxExtends[center] - 1;
+      const mirrorIndex = center - (index - center);
+      let extend = index > r ? 1 : Math.min(maxExtends[mirrorIndex], r - index + 1);
+
+      while (index - extend >= 0 && index + extend < n && str[index - extend] === str[index + extend]) {
+        leftToRight[index + extend] = 2 * extend + 1;
+        extend += 1;
+      }
+
+      maxExtends[index] = extend;
+
+      if (index + maxExtends[index] >= r) {
+        center = index;
+      }
+    }
+
+    for (let index = 1; index < n; index++) {
+      leftToRight[index] = Math.max(leftToRight[index], leftToRight[index - 1]);
+    }
+
+    return leftToRight;
+  };
+
+  const maxLeft = manacher(s);
+  const reversed = s.split('').reverse().join('');
+  const maxRight = manacher(reversed).reverse();
+  let reuslt = 1;
+
+  for (let index = 1; index < n; index++) {
+    reuslt = Math.max(reuslt, maxLeft[index - 1] * maxRight[index]);
+  }
+
+  return reuslt;
+};