Add solution #1642

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,443 LeetCode solutions in JavaScript
+# 1,444 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1266,6 +1266,7 @@
 1639|[Number of Ways to Form a Target String Given a Dictionary](./solutions/1639-number-of-ways-to-form-a-target-string-given-a-dictionary.js)|Hard|
 1640|[Check Array Formation Through Concatenation](./solutions/1640-check-array-formation-through-concatenation.js)|Easy|
 1641|[Count Sorted Vowel Strings](./solutions/1641-count-sorted-vowel-strings.js)|Medium|
+1642|[Furthest Building You Can Reach](./solutions/1642-furthest-building-you-can-reach.js)|Medium|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|
 1668|[Maximum Repeating Substring](./solutions/1668-maximum-repeating-substring.js)|Easy|
 1669|[Merge In Between Linked Lists](./solutions/1669-merge-in-between-linked-lists.js)|Medium|

solutions/1642-furthest-building-you-can-reach.js

@@ -0,0 +1,46 @@
+/**
+ * 1642. Furthest Building You Can Reach
+ * https://leetcode.com/problems/furthest-building-you-can-reach/
+ * Difficulty: Medium
+ *
+ * You are given an integer array heights representing the heights of buildings, some bricks,
+ * and some ladders.
+ *
+ * You start your journey from building 0 and move to the next building by possibly using
+ * bricks or ladders.
+ *
+ * While moving from building i to building i+1 (0-indexed):
+ * - If the current building's height is greater than or equal to the next building's height,
+ *   you do not need a ladder or bricks.
+ * - If the current building's height is less than the next building's height, you can either
+ *   use one ladder or (h[i+1] - h[i]) bricks.
+ *
+ * Return the furthest building index (0-indexed) you can reach if you use the given ladders
+ * and bricks optimally.
+ */
+
+/**
+ * @param {number[]} heights
+ * @param {number} bricks
+ * @param {number} ladders
+ * @return {number}
+ */
+var furthestBuilding = function(heights, bricks, ladders) {
+  const minHeap = new MinPriorityQueue();
+
+  for (let i = 0; i < heights.length - 1; i++) {
+    const climb = heights[i + 1] - heights[i];
+    if (climb > 0) {
+      minHeap.push(climb);
+      if (minHeap.size() > ladders) {
+        bricks -= minHeap.pop();
+        if (bricks < 0) {
+          return i;
+        }
+      }
+    }
+  }
+
+  return heights.length - 1;
+};
+