Sync LeetCode submission Runtime - 18 ms (44.05%), Memory - 19.2 MB (56.11%)

Author: jimit105

0042-trapping-rain-water/README.md

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+<p>Given <code>n</code> non-negative integers representing an elevation map where the width of each bar is <code>1</code>, compute how much water it can trap after raining.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img src="https://assets.leetcode.com/uploads/2018/10/22/rainwatertrap.png" style="width: 412px; height: 161px;" />
+<pre>
+<strong>Input:</strong> height = [0,1,0,2,1,0,1,3,2,1,2,1]
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> The above elevation map (black section) is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> height = [4,2,0,3,2,5]
+<strong>Output:</strong> 9
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == height.length</code></li>
+	<li><code>1 &lt;= n &lt;= 2 * 10<sup>4</sup></code></li>
+	<li><code>0 &lt;= height[i] &lt;= 10<sup>5</sup></code></li>
+</ul>

0042-trapping-rain-water/solution.py

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+# Approach 4: Using 2 pointers
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def trap(self, height: List[int]) -> int:
+        left, right = 0, len(height) - 1
+        ans = 0
+        left_max, right_max = 0, 0
+
+        while left < right:
+            if height[left] < height[right]:
+                left_max = max(left_max, height[left])
+                ans += left_max - height[left]
+                left += 1
+            else:
+                right_max = max(right_max, height[right])
+                ans += right_max - height[right]
+                right -= 1
+
+        return ans