feat: add No.1642

BaffinLee · BaffinLee · commit ef5667c58e68 · 2024-02-17T21:56:23.000+08:00
diff --git a/1601-1700/1642. Furthest Building You Can Reach.md b/1601-1700/1642. Furthest Building You Can Reach.md
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+# 1642. Furthest Building You Can Reach
+
+- Difficulty: Medium.
+- Related Topics: Array, Greedy, Heap (Priority Queue).
+- Similar Questions: Make the Prefix Sum Non-negative, Find Building Where Alice and Bob Can Meet.
+
+## Problem
+
+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.**
+
+ 
+Example 1:
+
+![](https://assets.leetcode.com/uploads/2020/10/27/q4.gif)
+
+```
+Input: heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1
+Output: 4
+Explanation: Starting at building 0, you can follow these steps:
+- Go to building 1 without using ladders nor bricks since 4 >= 2.
+- Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 < 7.
+- Go to building 3 without using ladders nor bricks since 7 >= 6.
+- Go to building 4 using your only ladder. You must use either bricks or ladders because 6 < 9.
+It is impossible to go beyond building 4 because you do not have any more bricks or ladders.
+```
+
+Example 2:
+
+```
+Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2
+Output: 7
+```
+
+Example 3:
+
+```
+Input: heights = [14,3,19,3], bricks = 17, ladders = 0
+Output: 3
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= heights.length <= 105`
+	
+- `1 <= heights[i] <= 106`
+	
+- `0 <= bricks <= 109`
+	
+- `0 <= ladders <= heights.length`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} heights
+ * @param {number} bricks
+ * @param {number} ladders
+ * @return {number}
+ */
+var furthestBuilding = function(heights, bricks, ladders) {
+    var queue = new MinPriorityQueue();
+    for (var i = 0; i < heights.length - 1; i++) {
+        if (heights[i + 1] <= heights[i]) continue;
+        queue.enqueue(heights[i + 1] - heights[i], heights[i + 1] - heights[i]);
+        if (queue.size() > ladders) {
+            var height = queue.dequeue().element;
+            if (bricks < height) return i;
+            bricks -= height;
+        }
+    }
+    return heights.length - 1;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n * log(n)).
+* Space complexity : O(n).
