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Author: Chris Wu

problems/candy.py

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+class Solution(object):
+    def candy(self, ratings):
+        N = len(ratings)
+        
+        l2r = [1]*N
+        r2l = [1]*N
+        
+        for i in xrange(1, N):
+            if ratings[i]>ratings[i-1]:
+                l2r[i] = l2r[i-1]+1
+        
+        for i in xrange(N-2, -1, -1):
+            if ratings[i]>ratings[i+1]:
+                r2l[i] = r2l[i+1]+1
+        
+        ans = 0
+        for i in xrange(N):
+            ans += max(l2r[i], r2l[i])
+        
+        return ans
+        

problems/trapping-rain-water.py

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+"""
+Let's calculate the answer considering only the left boandary.
+The water each i can contain is wall[i-1]-height[i]
+if no water at i-1 the wall[i-1] will be height[i-1], else it should be height[i-1]+water[i-1]
+=> water[i] = (height[i-1] + water[i-1]) - height[i] (l2r below)
+
+Do the same from right to left, this time considering only the right boandary.
+
+The amount of water at i considering both left and right wall will be min(l2r[i], r2l[i])
+"""
+class Solution(object):
+    def trap(self, height):
+        N = len(height)
+        l2r = [0]*N #water
+        r2l = [0]*N #water
+        
+        for i in xrange(1, N):
+            l2r[i] = max(l2r[i-1]+height[i-1]-height[i], 0)
+        
+        for i in xrange(N-2, -1, -1):
+            r2l[i] = max(r2l[i+1]+height[i+1]-height[i], 0)
+        
+        ans = 0
+        for i in xrange(N):
+            ans += min(l2r[i], r2l[i])
+        return ans