feat: add solution 1745. Palindrome Partitioning IV

Author: tzuyi0817

solutions/1745. Palindrome Partitioning IV/README.md

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+# [1745. Palindrome Partitioning IV](https://leetcode.com/problems/palindrome-partitioning-iv)
+
+## Description
+
+<div class="elfjS" data-track-load="description_content"><p>Given a string <code>s</code>, return <code>true</code> <em>if it is possible to split the string</em> <code>s</code> <em>into three <strong>non-empty</strong> palindromic substrings. Otherwise, return </em><code>false</code>.​​​​​</p>
+
+<p>A string is said to be palindrome if it the same string when reversed.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre><strong>Input:</strong> s = "abcbdd"
+<strong>Output:</strong> true
+<strong>Explanation: </strong>"abcbdd" = "a" + "bcb" + "dd", and all three substrings are palindromes.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre><strong>Input:</strong> s = "bcbddxy"
+<strong>Output:</strong> false
+<strong>Explanation: </strong>s cannot be split into 3 palindromes.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= s.length &lt;= 2000</code></li>
+	<li><code>s</code>​​​​​​ consists only of lowercase English letters.</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+## Solutions
+
+**Solution: `Dynamic Programming`**
+
+- Time complexity: <em>O(n<sup>2</sup>)</em>
+- Space complexity: <em>O(n<sup>2</sup>)</em>
+
+<p>&nbsp;</p>
+
+### **JavaScript**
+
+```js
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+const checkPartitioning = function (s) {
+  const n = s.length;
+  const dp = Array.from({ length: n }, () => new Array(n).fill(null));
+
+  const isPalindrome = (a, b) => {
+    if (a >= b) return true;
+    if (dp[a][b] !== null) return dp[a][b];
+    const result = s[a] === s[b] && isPalindrome(a + 1, b - 1);
+
+    dp[a][b] = result;
+
+    return result;
+  };
+
+  for (let a = 0; a < n - 2; a++) {
+    const segment1 = isPalindrome(0, a);
+
+    if (!segment1) continue;
+
+    for (let b = a + 1; b < n - 1; b++) {
+      const segment2 = isPalindrome(a + 1, b);
+
+      if (!segment2) continue;
+      const segment3 = isPalindrome(b + 1, n - 1);
+
+      if (segment3) return true;
+    }
+  }
+
+  return false;
+};
+```

solutions/1745. Palindrome Partitioning IV/solution.js

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+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+const checkPartitioning = function (s) {
+  const n = s.length;
+  const dp = Array.from({ length: n }, () => new Array(n).fill(null));
+
+  const isPalindrome = (a, b) => {
+    if (a >= b) return true;
+    if (dp[a][b] !== null) return dp[a][b];
+    const result = s[a] === s[b] && isPalindrome(a + 1, b - 1);
+
+    dp[a][b] = result;
+
+    return result;
+  };
+
+  for (let a = 0; a < n - 2; a++) {
+    const segment1 = isPalindrome(0, a);
+
+    if (!segment1) continue;
+
+    for (let b = a + 1; b < n - 1; b++) {
+      const segment2 = isPalindrome(a + 1, b);
+
+      if (!segment2) continue;
+      const segment3 = isPalindrome(b + 1, n - 1);
+
+      if (segment3) return true;
+    }
+  }
+
+  return false;
+};