Add solution #336

Author: JoshCrozier

README.md

@@ -271,6 +271,7 @@
 332|[Reconstruct Itinerary](./0332-reconstruct-itinerary.js)|Hard|
 334|[Increasing Triplet Subsequence](./0334-increasing-triplet-subsequence.js)|Medium|
 335|[Self Crossing](./0335-self-crossing.js)|Hard|
+336|[Palindrome Pairs](./0336-palindrome-pairs.js)|Hard|
 337|[House Robber III](./0337-house-robber-iii.js)|Medium|
 338|[Counting Bits](./0338-counting-bits.js)|Easy|
 341|[Flatten Nested List Iterator](./0341-flatten-nested-list-iterator.js)|Medium|

solutions/0336-palindrome-pairs.js

@@ -0,0 +1,57 @@
+/**
+ * 336. Palindrome Pairs
+ * https://leetcode.com/problems/palindrome-pairs/
+ * Difficulty: Hard
+ *
+ * You are given a 0-indexed array of unique strings words.
+ *
+ * A palindrome pair is a pair of integers (i, j) such that:
+ * - 0 <= i, j < words.length,
+ * - i != j, and
+ * - words[i] + words[j] (the concatenation of the two strings) is a palindrome.
+ *
+ * Return an array of all the palindrome pairs of words.
+ *
+ * You must write an algorithm with O(sum of words[i].length) runtime complexity.
+ */
+
+/**
+ * @param {string[]} words
+ * @return {number[][]}
+ */
+var palindromePairs = function(words) {
+  const result = [];
+  const map = new Map();
+  const rMap = new Map();
+
+  words.forEach((word, i) => {
+    map.set(word, i);
+    rMap.set(word.split('').reverse().join(''), i);
+  });
+
+  words.forEach((word, i) => {
+    for (let j = 0; j <= word.length; j++) {
+      if (isPalindrome(word, 0, j - 1)) {
+        const right = word.slice(j);
+        if (rMap.has(right) && rMap.get(right) !== i) {
+          result.push([rMap.get(right), i]);
+        }
+      }
+      if (j < word.length && isPalindrome(word, j, word.length - 1)) {
+        const left = word.slice(0, j);
+        if (rMap.has(left) && rMap.get(left) !== i) {
+          result.push([i, rMap.get(left)]);
+        }
+      }
+    }
+  });
+
+  return result;
+
+  function isPalindrome(s, start, end) {
+    while (start < end) {
+      if (s[start++] !== s[end--]) return false;
+    }
+    return true;
+  }
+};