Trapping Rain Water in Python

Suppose we have an array of n non-negative integers. These are representing an elevation map where the width of each bar is 1, we have to compute how much water it is able to trap after raining. So the map will be like −

Here we can see there are 6 blue boxes, so the output will be 6.

To solve this, we will follow these steps −

• Define a stack st, water := 0 and i := 0

• while i < size of height

  • if is stack is empty or height[stack top] >= height[i], then push i into stack, increase i by 1

  • otherwise

    • x := stack top element, delete top from stack

    • if stack is not empty, then

      • temp := min of height[stack top element] and height[i]

      • dest := i – stack top element – 1

      • water := water + dist * (temp – height[x])

• return water

Example

Let us see the following implementation to get a better understanding −

 Live Demo

class Solution(object):
   def trap(self, height):
      stack = []
      water = 0
      i=0
      while i<len(height):
         if len(stack) == 0 or height[stack[-1]]>=height[i]:
            stack.append(i)
            i+=1
         else:
            x = stack[-1]
            stack.pop()
            if len(stack) != 0:
               temp = min(height[stack[-1]],height[i])
               dist = i - stack[-1]-1
               water += dist*(temp - height[x])
      return water
ob = Solution()
print(ob.trap([0,1,0,2,1,0,1,3,2,1,2,1]))

Input

[0,1,0,2,1,0,1,3,2,1,2,1]

Output

6

Arnab Chakraborty

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Updated on: 26-May-2020

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