Add solution #1745

JoshCrozier · JoshCrozier · commit 69de7e1df688 · 2025-04-28T19:12:15.000-05:00
diff --git a/README.md b/README.md
@@ -1,4 +1,4 @@
-# 1,517 LeetCode solutions in JavaScript
+# 1,518 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1344,6 +1344,7 @@
 1742|[Maximum Number of Balls in a Box](./solutions/1742-maximum-number-of-balls-in-a-box.js)|Easy|
 1743|[Restore the Array From Adjacent Pairs](./solutions/1743-restore-the-array-from-adjacent-pairs.js)|Medium|
 1744|[Can You Eat Your Favorite Candy on Your Favorite Day?](./solutions/1744-can-you-eat-your-favorite-candy-on-your-favorite-day.js)|Medium|
+1745|[Palindrome Partitioning IV](./solutions/1745-palindrome-partitioning-iv.js)|Hard|
 1748|[Sum of Unique Elements](./solutions/1748-sum-of-unique-elements.js)|Easy|
 1749|[Maximum Absolute Sum of Any Subarray](./solutions/1749-maximum-absolute-sum-of-any-subarray.js)|Medium|
 1752|[Check if Array Is Sorted and Rotated](./solutions/1752-check-if-array-is-sorted-and-rotated.js)|Easy|
diff --git a/solutions/1745-palindrome-partitioning-iv.js b/solutions/1745-palindrome-partitioning-iv.js
@@ -0,0 +1,37 @@
+/**
+ * 1745. Palindrome Partitioning IV
+ * https://leetcode.com/problems/palindrome-partitioning-iv/
+ * Difficulty: Hard
+ *
+ * Given a string s, return true if it is possible to split the string s into three non-empty
+ * palindromic substrings. Otherwise, return false.
+ *
+ * A string is said to be palindrome if it the same string when reversed.
+ */
+
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+var checkPartitioning = function(s) {
+  const n = s.length;
+  const isPalindrome = Array.from({ length: n }, () => Array(n).fill(false));
+
+  for (let i = n - 1; i >= 0; i--) {
+    for (let j = i; j < n; j++) {
+      if (s[i] === s[j] && (j - i <= 2 || isPalindrome[i + 1][j - 1])) {
+        isPalindrome[i][j] = true;
+      }
+    }
+  }
+
+  for (let i = 1; i < n - 1; i++) {
+    for (let j = i; j < n - 1; j++) {
+      if (isPalindrome[0][i - 1] && isPalindrome[i][j] && isPalindrome[j + 1][n - 1]) {
+        return true;
+      }
+    }
+  }
+
+  return false;
+};
