42. Trapping Rain Water

Hard LeetCode

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 * 10^4
• 0 <= height[i] <= 10^5

How to solve the problem

• Two Pointers

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

Complexity

• Time Complexity: O(n) - We traverse the array once with two pointers
• Space Complexity: O(1) - Only using constant extra space for variables