Trapping rain water

class Solution:
    def trap(self, height: List[int]) -> int:
        res = 0
        lmax = [0] * len(height)
        rmax = [0] * len(height)
        curr = 0
        for i in range(len(height)):
            curr = max(curr, height[i])
            lmax[i] = curr
        curr = 0
        for i in range(len(height) - 1, -1, -1):
            curr = max(curr, height[i])
            rmax[i] = curr

        for i in range(1, len(height) - 1):
            a = lmax[i]
            b = rmax[i]
            res += min(a, b) - height[i]
        return res

Trapping Rain Water

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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:

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