# Straightforward O(n) solution python

• A straightforward solution is to move from left to right and calculate water collected if the left elevation is the smaller one. Do the same thing from the right assuming right to be the smaller one, taking care of equal elevations

``````    def trap(self, height):
"""
:type height: List[int]
:rtype: int
"""
if height == None or len(height) == 0 : return 0
#move front and collect all rainwater
water,puddle = 0,0
prev = height[0]
for each in range(1,len(height)):
if height[each] >= prev: #collected only when 2 elevations are high enough
#print water,s
water+=puddle
puddle = 0
prev = height[each]
else:
puddle+= prev-height[each] #possible rainwater collection

#move from the other direction and collect the water
puddle = 0
prev = height[-1]
for each in range(len(height)-2,-1,-1):

if height[each] > prev: #same as before but dont double count '=' heights
print water,puddle
water+=puddle
puddle=0
prev = height[each]
else:
puddle+= prev-height[each]
return water``````

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