Add solution #1052

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,121 LeetCode solutions in JavaScript
+# 1,122 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -857,6 +857,7 @@
 1048|[Longest String Chain](./solutions/1048-longest-string-chain.js)|Medium|
 1049|[Last Stone Weight II](./solutions/1049-last-stone-weight-ii.js)|Medium|
 1051|[Height Checker](./solutions/1051-height-checker.js)|Easy|
+1052|[Grumpy Bookstore Owner](./solutions/1052-grumpy-bookstore-owner.js)|Medium|
 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|
 1073|[Adding Two Negabinary Numbers](./solutions/1073-adding-two-negabinary-numbers.js)|Medium|

solutions/1052-grumpy-bookstore-owner.js

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+/**
+ * 1052. Grumpy Bookstore Owner
+ * https://leetcode.com/problems/grumpy-bookstore-owner/
+ * Difficulty: Medium
+ *
+ * There is a bookstore owner that has a store open for n minutes. You are given an integer array
+ * customers of length n where customers[i] is the number of the customers that enter the store
+ * at the start of the ith minute and all those customers leave after the end of that minute.
+ *
+ * During certain minutes, the bookstore owner is grumpy. You are given a binary array grumpy
+ * where grumpy[i] is 1 if the bookstore owner is grumpy during the ith minute, and is 0 otherwise.
+ *
+ * When the bookstore owner is grumpy, the customers entering during that minute are not satisfied.
+ * Otherwise, they are satisfied.
+ *
+ * The bookstore owner knows a secret technique to remain not grumpy for minutes consecutive
+ * minutes, but this technique can only be used once.
+ *
+ * Return the maximum number of customers that can be satisfied throughout the day.
+ */
+
+/**
+ * @param {number[]} customers
+ * @param {number[]} grumpy
+ * @param {number} minutes
+ * @return {number}
+ */
+var maxSatisfied = function(customers, grumpy, minutes) {
+  let baseSatisfied = 0;
+  let windowGain = 0;
+  let maxGain = 0;
+
+  for (let i = 0; i < customers.length; i++) {
+    if (grumpy[i] === 0) {
+      baseSatisfied += customers[i];
+    }
+    if (i < minutes) {
+      windowGain += grumpy[i] * customers[i];
+    } else {
+      windowGain += grumpy[i] * customers[i] - grumpy[i - minutes] * customers[i - minutes];
+    }
+    maxGain = Math.max(maxGain, windowGain);
+  }
+
+  return baseSatisfied + maxGain;
+};