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42. Trapping Rain Water

Array Two Pointers Dynamic Programming Stack Monotonic Stack

Problem - Trapping Rain Water

Hard

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

Solutions

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class Solution:
    def trap(self, height: List[int]) -> int:
        stack = []
        water = 0

        for i, h in enumerate(height):
            while stack and h >= height[stack[-1]]:
                bottom = stack.pop()
                if not stack:
                    break
                left = stack[-1]
                width = i - left - 1

                height_check = min(height[left], h) - height[bottom]
                water += width * height_check
            stack.append(i)

        return water

Submission Stats:

  • Runtime: 11 ms (71.17%)
  • Memory: 19.7 MB (7.78%)