Create TrappingRainWater.py

LeetCode/TrappingRainWater.py

@@ -0,0 +1,65 @@
+"""
+Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap 
+after raining.
+The above elevation map 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. Thanks Marcos for contributing this image!
+Example:
+Input: [0,1,0,2,1,0,1,3,2,1,2,1]
+Output: 6
+"""
+
+#not working yet
+class Solution:
+    def trap(self, height: List[int]) -> int:
+
+        water = 0
+        maxnum = 0
+        maxlist = []
+
+        for num in height:
+            if num > maxnum:
+                maxlist.append(num)
+                maxnum = num
+
+        maxnum = 0
+        maxlistb = []
+        for num in reversed(height):
+            if num > maxnum:
+                maxlistb.append(num)
+                maxnum = num
+
+        for num in reversed(maxlistb[:len(maxlistb)-1]):
+            maxlist.append(num)
+
+        #maxlist.pop(0) #first value
+        lp = 0 #left pointer
+        rp = 0 #right pointer
+        if maxlist == [] or height == []:
+            return water
+
+        while lp < len(height)-1:
+            print("lp",lp,"rp",rp,"maxnums",maxlist,"water",water)
+            if rp < len(height)-1:
+                rp += 1
+            if maxlist == []:
+                break
+            if height[rp] >= maxlist[0]:
+                ht = min(height[lp],height[rp])
+                for x in range(lp+1,rp):
+                    water += ht - height[x]
+                lp = rp
+                maxlist.pop(0)
+                pass
+
+        return water
+
+
+"""
+My Solution
+
+"""
+
+"""
+Fastest Solution
+
+"""