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Author: NinePiece2

Solutions/2096-find-the-longest-valid-obstacle-course-at-each-position/README.md

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+<p>You want to build some obstacle courses. You are given a <strong>0-indexed</strong> integer array <code>obstacles</code> of length <code>n</code>, where <code>obstacles[i]</code> describes the height of the <code>i<sup>th</sup></code> obstacle.</p>
+
+<p>For every index <code>i</code> between <code>0</code> and <code>n - 1</code> (<strong>inclusive</strong>), find the length of the <strong>longest obstacle course</strong> in <code>obstacles</code> such that:</p>
+
+<ul>
+	<li>You choose any number of obstacles between <code>0</code> and <code>i</code> <strong>inclusive</strong>.</li>
+	<li>You must include the <code>i<sup>th</sup></code> obstacle in the course.</li>
+	<li>You must put the chosen obstacles in the <strong>same order</strong> as they appear in <code>obstacles</code>.</li>
+	<li>Every obstacle (except the first) is <strong>taller</strong> than or the <strong>same height</strong> as the obstacle immediately before it.</li>
+</ul>
+
+<p>Return <em>an array</em> <code>ans</code> <em>of length</em> <code>n</code>, <em>where</em> <code>ans[i]</code> <em>is the length of the <strong>longest obstacle course</strong> for index</em> <code>i</code><em> as described above</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> obstacles = [1,2,3,2]
+<strong>Output:</strong> [1,2,3,3]
+<strong>Explanation:</strong> The longest valid obstacle course at each position is:
+- i = 0: [<u>1</u>], [1] has length 1.
+- i = 1: [<u>1</u>,<u>2</u>], [1,2] has length 2.
+- i = 2: [<u>1</u>,<u>2</u>,<u>3</u>], [1,2,3] has length 3.
+- i = 3: [<u>1</u>,<u>2</u>,3,<u>2</u>], [1,2,2] has length 3.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> obstacles = [2,2,1]
+<strong>Output:</strong> [1,2,1]
+<strong>Explanation: </strong>The longest valid obstacle course at each position is:
+- i = 0: [<u>2</u>], [2] has length 1.
+- i = 1: [<u>2</u>,<u>2</u>], [2,2] has length 2.
+- i = 2: [2,2,<u>1</u>], [1] has length 1.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> obstacles = [3,1,5,6,4,2]
+<strong>Output:</strong> [1,1,2,3,2,2]
+<strong>Explanation:</strong> The longest valid obstacle course at each position is:
+- i = 0: [<u>3</u>], [3] has length 1.
+- i = 1: [3,<u>1</u>], [1] has length 1.
+- i = 2: [<u>3</u>,1,<u>5</u>], [3,5] has length 2. [1,5] is also valid.
+- i = 3: [<u>3</u>,1,<u>5</u>,<u>6</u>], [3,5,6] has length 3. [1,5,6] is also valid.
+- i = 4: [<u>3</u>,1,5,6,<u>4</u>], [3,4] has length 2. [1,4] is also valid.
+- i = 5: [3,<u>1</u>,5,6,4,<u>2</u>], [1,2] has length 2.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == obstacles.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= obstacles[i] &lt;= 10<sup>7</sup></code></li>
+</ul>

Solutions/2096-find-the-longest-valid-obstacle-course-at-each-position/solution.py

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+# class BinaryIndexedTree:
+#     def  __init__ (self, n: int):
+#         self.n = n
+#         self.tree = [0] * (n + 1)
+    
+#     def update(self, index: int, val: int):
+#         while index <= self.n:
+#             self.tree[index] = max(self.tree[index], val)
+#             index += index & -index
+    
+#     def query(self, index: int):
+#         result = 0
+#         while index:
+#             result = max(self.tree[index], result)
+#             index -= index & -index
+#         return result
+
+# class Solution:
+#     def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
+#         nums = sorted(set(obstacles))
+#         n = len(nums)
+
+#         tree = BinaryIndexedTree(n)
+#         result = []
+#         for val in obstacles:
+#             idx = bisect_left(nums, val) + 1
+#             result.append(tree.query(idx) + 1)
+#             tree.update(idx, result[-1])
+#         return result
+
+class Solution:
+    def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
+        count = 0
+        dp, result = [], []
+
+        for ob in obstacles:
+            if not dp or ob >= dp[-1]:
+                dp.append(ob)
+                count += 1
+                result.append(count)
+            else:
+                idx = bisect_right(dp, ob)
+                dp[idx] = ob
+                result.append(idx + 1)
+        return result