trapping-rain-water py3

0011_container-with-most-water/NOTES.md

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+## Approaches:
+
+1. Brute force $O(N^2)$
+2. Two-pointer $O(N)$
+
+## Follow-up questions
+
+1. What if you could slant the container?

0042_trapping-rain-water/README.md

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+Given `n` non-negative integers representing an elevation map where the width of each bar is `1`, compute how much water it can trap after raining.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2018/10/22/rainwatertrap.png)
+
+    Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
+    Output: 6
+    Explanation: 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.
+    
+
+**Example 2:**
+
+    Input: height = [4,2,0,3,2,5]
+    Output: 9
+    
+
+**Constraints:**
+
+*   `n == height.length`
+*   `1 <= n <= 2 * 104`
+*   `0 <= height[i] <= 105`

0042_trapping-rain-water/python3/imaginary_grid.py

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+# Time: O(N · H) where H is the maximum height
+# Space: O(1)
+
+class Solution:
+    def trap(self, height: List[int]) -> int:
+        maxh = 0
+        
+        for h in height:
+            maxh = max(h, maxh)
+        
+        land_width = len(height)
+        output = 0
+        
+        for i in range(maxh):
+            row_output = 0
+            trough_output = 0
+            trough_start = -1
+            
+            for j in range(land_width):
+                # A block is present
+                if height[j] >= i + 1:
+                    row_output += trough_output
+                    trough_start = j
+                    trough_output = 0
+                elif trough_start >= 0:
+                    trough_output += 1
+        
+            output += row_output
+        
+        return output
+            
+                
+