Add solution #1124

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,141 LeetCode solutions in JavaScript
+# 1,142 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -887,6 +887,7 @@
 1111|[Maximum Nesting Depth of Two Valid Parentheses Strings](./solutions/1111-maximum-nesting-depth-of-two-valid-parentheses-strings.js)|Medium|
 1122|[Relative Sort Array](./solutions/1122-relative-sort-array.js)|Easy|
 1123|[Lowest Common Ancestor of Deepest Leaves](./solutions/1123-lowest-common-ancestor-of-deepest-leaves.js)|Medium|
+1124|[Longest Well-Performing Interval](./solutions/1124-longest-well-performing-interval.js)|Medium|
 1137|[N-th Tribonacci Number](./solutions/1137-n-th-tribonacci-number.js)|Easy|
 1143|[Longest Common Subsequence](./solutions/1143-longest-common-subsequence.js)|Medium|
 1161|[Maximum Level Sum of a Binary Tree](./solutions/1161-maximum-level-sum-of-a-binary-tree.js)|Medium|

solutions/1124-longest-well-performing-interval.js

@@ -0,0 +1,42 @@
+/**
+ * 1124. Longest Well-Performing Interval
+ * https://leetcode.com/problems/longest-well-performing-interval/
+ * Difficulty: Medium
+ *
+ * We are given hours, a list of the number of hours worked per day for a given employee.
+ *
+ * A day is considered to be a tiring day if and only if the number of hours worked is
+ * (strictly) greater than 8.
+ *
+ * A well-performing interval is an interval of days for which the number of tiring days
+ * is strictly larger than the number of non-tiring days.
+ *
+ * Return the length of the longest well-performing interval.
+ */
+
+/**
+ * @param {number[]} hours
+ * @return {number}
+ */
+var longestWPI = function(hours) {
+  let result = 0;
+  let score = 0;
+  const seen = new Map();
+
+  for (let i = 0; i < hours.length; i++) {
+    score += hours[i] > 8 ? 1 : -1;
+
+    if (score > 0) {
+      result = i + 1;
+    } else {
+      if (!seen.has(score)) {
+        seen.set(score, i);
+      }
+      if (seen.has(score - 1)) {
+        result = Math.max(result, i - seen.get(score - 1));
+      }
+    }
+  }
+
+  return result;
+};