Add solution #1255

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,199 LeetCode solutions in JavaScript
+# 1,200 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -957,6 +957,7 @@
 1252|[Cells with Odd Values in a Matrix](./solutions/1252-cells-with-odd-values-in-a-matrix.js)|Easy|
 1253|[Reconstruct a 2-Row Binary Matrix](./solutions/1253-reconstruct-a-2-row-binary-matrix.js)|Medium|
 1254|[Number of Closed Islands](./solutions/1254-number-of-closed-islands.js)|Medium|
+1255|[Maximum Score Words Formed by Letters](./solutions/1255-maximum-score-words-formed-by-letters.js)|Hard|
 1261|[Find Elements in a Contaminated Binary Tree](./solutions/1261-find-elements-in-a-contaminated-binary-tree.js)|Medium|
 1267|[Count Servers that Communicate](./solutions/1267-count-servers-that-communicate.js)|Medium|
 1268|[Search Suggestions System](./solutions/1268-search-suggestions-system.js)|Medium|

solutions/1255-maximum-score-words-formed-by-letters.js

@@ -0,0 +1,51 @@
+/**
+ * 1255. Maximum Score Words Formed by Letters
+ * https://leetcode.com/problems/maximum-score-words-formed-by-letters/
+ * Difficulty: Hard
+ *
+ * Given a list of words, list of  single letters (might be repeating) and score of every character.
+ *
+ * Return the maximum score of any valid set of words formed by using the given letters (words[i]
+ * cannot be used two or more times).
+ *
+ * It is not necessary to use all characters in letters and each letter can only be used once.
+ * Score of letters 'a', 'b', 'c', ... ,'z' is given by score[0], score[1], ... , score[25]
+ * respectively.
+ */
+
+/**
+ * @param {string[]} words
+ * @param {character[]} letters
+ * @param {number[]} score
+ * @return {number}
+ */
+var maxScoreWords = function(words, letters, score) {
+  const letterCount = new Array(26).fill(0);
+  for (const letter of letters) {
+    letterCount[letter.charCodeAt(0) - 97]++;
+  }
+  return calculateMaxScore(0, letterCount);
+
+  function calculateMaxScore(index, available) {
+    if (index === words.length) return 0;
+
+    const skipScore = calculateMaxScore(index + 1, available);
+    const word = words[index];
+    const tempCount = [...available];
+    let canUse = true;
+    let wordScore = 0;
+
+    for (const char of word) {
+      const charIndex = char.charCodeAt(0) - 97;
+      if (tempCount[charIndex] === 0) {
+        canUse = false;
+        break;
+      }
+      tempCount[charIndex]--;
+      wordScore += score[charIndex];
+    }
+
+    const useScore = canUse ? wordScore + calculateMaxScore(index + 1, tempCount) : 0;
+    return Math.max(skipScore, useScore);
+  }
+};