Add solution #132

Author: JoshCrozier

README.md

@@ -136,6 +136,7 @@
 129|[Sum Root to Leaf Numbers](./0129-sum-root-to-leaf-numbers.js)|Medium|
 130|[Surrounded Regions](./0130-surrounded-regions.js)|Medium|
 131|[Palindrome Partitioning](./0131-palindrome-partitioning.js)|Medium|
+132|[Palindrome Partitioning II](./0132-palindrome-partitioning-ii.js)|Hard|
 133|[Clone Graph](./0133-clone-graph.js)|Medium|
 134|[Gas Station](./0134-gas-station.js)|Medium|
 135|[Candy](./0135-candy.js)|Hard|

solutions/0132-palindrome-partitioning-ii.js

@@ -0,0 +1,31 @@
+/**
+ * 132. Palindrome Partitioning II
+ * https://leetcode.com/problems/palindrome-partitioning-ii/
+ * Difficulty: Hard
+ *
+ * Given a string s, partition s such that every substring of the partition is a palindrome.
+ *
+ * Return the minimum cuts needed for a palindrome partitioning of s.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var minCut = function(s) {
+  const isPalindrome = new Array(s.length).fill().map(() => new Array(s.length).fill(false));
+  const partitions = new Array(s.length).fill(0);
+
+  for (let i = 0; i < s.length; i++) {
+    let offset = i;
+    for (let j = 0; j <= i; j++) {
+      if (s[j] === s[i] && (i - j <= 1 || isPalindrome[j + 1][i - 1])) {
+        isPalindrome[j][i] = true;
+        offset = j === 0 ? 0 : Math.min(offset, partitions[j - 1] + 1);
+      }
+    }
+    partitions[i] = offset;
+  }
+
+  return partitions[s.length - 1];
+};