0042.py

# Source: https://leetcode.com/problems/trapping-rain-water
# Title: Trapping Rain Water
# Difficulty: Hard
# Author: Mu Yang <http://muyang.pro>

################################################################################################################################
# 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:
#
#   https://assets.leetcode.com/uploads/2018/10/22/rainwatertrap.png
#   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 * 10^4
#   0 <= height[i] <= 10^5
#
################################################################################################################################

class Solution:
    def trap(self, height: List[int]) -> int:

        i = 0
        j = len(height)-1

        if i >= j:
            return 0

        left_val = height[i]
        right_val = height[j]
        ans = 0

        while i < j:
            if height[i] < height[j]:
                i += 1
                if height[i] > left_val:
                    left_val = height[i]
                else:
                    ans += left_val - height[i]
            else:
                j -= 1
                if height[j] > right_val:
                    right_val = height[j]
                else:
                    ans += right_val - height[j]

        return ans