Add solution #1048

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,119 LeetCode solutions in JavaScript
+# 1,120 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -854,6 +854,7 @@
 1044|[Longest Duplicate Substring](./solutions/1044-longest-duplicate-substring.js)|Hard|
 1046|[Last Stone Weight](./solutions/1046-last-stone-weight.js)|Easy|
 1047|[Remove All Adjacent Duplicates In String](./solutions/1047-remove-all-adjacent-duplicates-in-string.js)|Easy|
+1048|[Longest String Chain](./solutions/1048-longest-string-chain.js)|Medium|
 1051|[Height Checker](./solutions/1051-height-checker.js)|Easy|
 1071|[Greatest Common Divisor of Strings](./solutions/1071-greatest-common-divisor-of-strings.js)|Easy|
 1072|[Flip Columns For Maximum Number of Equal Rows](./solutions/1072-flip-columns-for-maximum-number-of-equal-rows.js)|Medium|

solutions/1048-longest-string-chain.js

@@ -0,0 +1,44 @@
+/**
+ * 1048. Longest String Chain
+ * https://leetcode.com/problems/longest-string-chain/
+ * Difficulty: Medium
+ *
+ * You are given an array of words where each word consists of lowercase English letters.
+ *
+ * wordA is a predecessor of wordB if and only if we can insert exactly one letter anywhere in
+ * wordA without changing the order of the other characters to make it equal to wordB.
+ *
+ * For example, "abc" is a predecessor of "abac", while "cba" is not a predecessor of "bcad".
+ *
+ * A word chain is a sequence of words [word1, word2, ..., wordk] with k >= 1, where word1 is
+ * a predecessor of word2, word2 is a predecessor of word3, and so on. A single word is
+ * trivially a word chain with k == 1.
+ *
+ * Return the length of the longest possible word chain with words chosen from the given list
+ * of words.
+ */
+
+/**
+ * @param {string[]} words
+ * @return {number}
+ */
+var longestStrChain = function(words) {
+  const map = new Map();
+  let result = 1;
+
+  words.sort((a, b) => a.length - b.length);
+
+  for (const word of words) {
+    let currentLength = 1;
+    for (let i = 0; i < word.length; i++) {
+      const prev = word.slice(0, i) + word.slice(i + 1);
+      if (map.has(prev)) {
+        currentLength = Math.max(currentLength, map.get(prev) + 1);
+      }
+    }
+    map.set(word, currentLength);
+    result = Math.max(result, currentLength);
+  }
+
+  return result;
+};