42-Trapping-Rain-Water.md

42. Trapping Rain Water

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
• 0 <= n <= 3 * 104
• 0 <= height[i] <= 105

Solution

class Solution:
    def trap(self, height: List[int]) -> int:
        ans = 0
        left, right = 0, len(height) - 1
        left_max, right_max = 0, 0
        while left < right:
            left_max, right_max = max(left_max, height[left]), max(right_max, height[right])
            if left_max < right_max:
                ans += max(0, left_max - height[left])
                left += 1
            else:
                ans += max(0, right_max - height[right])
                right -= 1
        return ans

Use two pointers and record max heights from left and right.