Add solution #948

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,032 LeetCode solutions in JavaScript
+# 1,033 LeetCode solutions in JavaScript
 
 [https://leetcode.com/](https://leetcode.com/)
 
@@ -757,6 +757,7 @@
 945|[Minimum Increment to Make Array Unique](./solutions/0945-minimum-increment-to-make-array-unique.js)|Medium|
 946|[Validate Stack Sequences](./solutions/0946-validate-stack-sequences.js)|Medium|
 947|[Most Stones Removed with Same Row or Column](./solutions/0947-most-stones-removed-with-same-row-or-column.js)|Medium|
+948|[Bag of Tokens](./solutions/0948-bag-of-tokens.js)|Medium|
 966|[Vowel Spellchecker](./solutions/0966-vowel-spellchecker.js)|Medium|
 970|[Powerful Integers](./solutions/0970-powerful-integers.js)|Easy|
 976|[Largest Perimeter Triangle](./solutions/0976-largest-perimeter-triangle.js)|Easy|

solutions/0948-bag-of-tokens.js

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+/**
+ * 948. Bag of Tokens
+ * https://leetcode.com/problems/bag-of-tokens/
+ * Difficulty: Medium
+ *
+ * You start with an initial power of power, an initial score of 0, and a bag of tokens given as
+ * an integer array tokens, where each tokens[i] denotes the value of tokeni.
+ *
+ * Your goal is to maximize the total score by strategically playing these tokens. In one move,
+ * you can play an unplayed token in one of the two ways (but not both for the same token):
+ * - Face-up: If your current power is at least tokens[i], you may play tokeni, losing tokens[i]
+ *   power and gaining 1 score.
+ * - Face-down: If your current score is at least 1, you may play tokeni, gaining tokens[i] power
+ *   and losing 1 score.
+ *
+ * Return the maximum possible score you can achieve after playing any number of tokens.
+ */
+
+/**
+ * @param {number[]} tokens
+ * @param {number} power
+ * @return {number}
+ */
+var bagOfTokensScore = function(tokens, power) {
+  let score = 0;
+  let result = 0;
+  let left = 0;
+  let right = tokens.length - 1;
+
+  tokens.sort((a, b) => a - b);
+
+  while (left <= right && (power >= tokens[left] || score > 0)) {
+    while (left <= right && power >= tokens[left]) {
+      power -= tokens[left];
+      score++;
+      left++;
+    }
+    result = Math.max(result, score);
+
+    if (left <= right && score > 0) {
+      power += tokens[right];
+      score--;
+      right--;
+    }
+  }
+
+  return result;
+};