Add solution #2054

JoshCrozier · JoshCrozier · commit 33c42ad472bd · 2025-07-25T00:24:50.000-05:00
diff --git a/README.md b/README.md
@@ -1862,6 +1862,7 @@
 2050|[Parallel Courses III](./solutions/2050-parallel-courses-iii.js)|Hard|
 2052|[Minimum Cost to Separate Sentence Into Rows](./solutions/2052-minimum-cost-to-separate-sentence-into-rows.js)|Medium|
 2053|[Kth Distinct String in an Array](./solutions/2053-kth-distinct-string-in-an-array.js)|Medium|
+2054|[Two Best Non-Overlapping Events](./solutions/2054-two-best-non-overlapping-events.js)|Medium|
 2055|[Plates Between Candles](./solutions/2055-plates-between-candles.js)|Medium|
 2056|[Number of Valid Move Combinations On Chessboard](./solutions/2056-number-of-valid-move-combinations-on-chessboard.js)|Hard|
 2057|[Smallest Index With Equal Value](./solutions/2057-smallest-index-with-equal-value.js)|Easy|
diff --git a/solutions/2054-two-best-non-overlapping-events.js b/solutions/2054-two-best-non-overlapping-events.js
@@ -0,0 +1,58 @@
+/**
+ * 2054. Two Best Non-Overlapping Events
+ * https://leetcode.com/problems/two-best-non-overlapping-events/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed 2D integer array of events where
+ * events[i] = [startTimei, endTimei, valuei]. The ith event starts at
+ * startTimei and ends at endTimei, and if you attend this event, you will
+ * receive a value of valuei. You can choose at most two non-overlapping
+ * events to attend such that the sum of their values is maximized.
+ *
+ * Return this maximum sum.
+ *
+ * Note that the start time and end time is inclusive: that is, you cannot
+ * attend two events where one of them starts and the other ends at the
+ * same time. More specifically, if you attend an event with end time t,
+ * the next event must start at or after t + 1.
+ */
+
+/**
+ * @param {number[][]} events
+ * @return {number}
+ */
+var maxTwoEvents = function(events) {
+  events.sort((a, b) => a[1] - b[1]);
+
+  const maxValueUpTo = new Array(events.length);
+  maxValueUpTo[0] = events[0][2];
+
+  for (let i = 1; i < events.length; i++) {
+    maxValueUpTo[i] = Math.max(maxValueUpTo[i - 1], events[i][2]);
+  }
+
+  let result = 0;
+  for (let i = 0; i < events.length; i++) {
+    const [start, end, value] = events[i];
+    result = Math.max(result, value);
+
+    let left = 0;
+    let right = i - 1;
+    let bestIndex = -1;
+    while (left <= right) {
+      const mid = Math.floor((left + right) / 2);
+      if (events[mid][1] < start) {
+        bestIndex = mid;
+        left = mid + 1;
+      } else {
+        right = mid - 1;
+      }
+    }
+
+    if (bestIndex !== -1) {
+      result = Math.max(result, value + maxValueUpTo[bestIndex]);
+    }
+  }
+
+  return result;
+};
