Add solution #1160

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,158 LeetCode solutions in JavaScript
+# 1,159 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -906,6 +906,7 @@
 1155|[Number of Dice Rolls With Target Sum](./solutions/1155-number-of-dice-rolls-with-target-sum.js)|Medium|
 1156|[Swap For Longest Repeated Character Substring](./solutions/1156-swap-for-longest-repeated-character-substring.js)|Medium|
 1157|[Online Majority Element In Subarray](./solutions/1157-online-majority-element-in-subarray.js)|Hard|
+1160|[Find Words That Can Be Formed by Characters](./solutions/1160-find-words-that-can-be-formed-by-characters.js)|Easy|
 1161|[Maximum Level Sum of a Binary Tree](./solutions/1161-maximum-level-sum-of-a-binary-tree.js)|Medium|
 1189|[Maximum Number of Balloons](./solutions/1189-maximum-number-of-balloons.js)|Easy|
 1200|[Minimum Absolute Difference](./solutions/1200-minimum-absolute-difference.js)|Easy|

solutions/1160-find-words-that-can-be-formed-by-characters.js

@@ -0,0 +1,42 @@
+/**
+ * 1160. Find Words That Can Be Formed by Characters
+ * https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/
+ * Difficulty: Easy
+ *
+ * You are given an array of strings words and a string chars.
+ *
+ * A string is good if it can be formed by characters from chars (each character can only be
+ * used once).
+ *
+ * Return the sum of lengths of all good strings in words.
+ */
+
+/**
+ * @param {string[]} words
+ * @param {string} chars
+ * @return {number}
+ */
+var countCharacters = function(words, availableChars) {
+  const charFrequency = new Map();
+  for (const char of availableChars) {
+    charFrequency.set(char, (charFrequency.get(char) || 0) + 1);
+  }
+
+  let result = 0;
+  for (const word of words) {
+    const tempFrequency = new Map(charFrequency);
+    let isValid = true;
+
+    for (const char of word) {
+      if (!tempFrequency.has(char) || tempFrequency.get(char) === 0) {
+        isValid = false;
+        break;
+      }
+      tempFrequency.set(char, tempFrequency.get(char) - 1);
+    }
+
+    if (isValid) result += word.length;
+  }
+
+  return result;
+};