42. Trapping Rain Water

Problem Description

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

This problem has 3 different approach, during the interview, don’t directly give the optimal solution except this is Facebook or Google. Optimize your algorithm along with your thoughts, talk to your interviewer.

• Brute Force O(n^2), O(1)
• Forward & Backward traversal O(n), O(n)
• Two Pointers O(n), O(1)

Brute Force

class Solution:
    def trap(self, heights: List[int]) -> int:
        answer = 0
        for i, val in enumerate(height):
            l = 0 if i == 0 else max(height[:i])
            r = 0 if i == len(height) - 1 else max(height[i+1:])
            answer += max(0, min(l, r) - val)
        
        return answer

Forward & Backward traversal

class Solution:
    def trap(self, heights: List[int]) -> int:
        f = [0]
        b = [0]
        res = 0
        n = len(heights)

        for i in range(1, n):
            curr_max = max(f[-1], heights[i - 1])
            f.append(curr_max)

        for j in range(n - 1 - 1, -1, -1):
            curr_max = max(b[-1], heights[j + 1])
            b.append(curr_max)

        for i in range(n):
            curr_water = min(f[i], b[n - i - 1]) - heights[i]
            curr_water = curr_water if curr_water > 0 else 0
            res += curr_water

        return res

Two Pointers

class Solution:
    def trap(self, heights: List[int]) -> int:
        if not heights or len(heights) == 0:
            return 0

        left, right = 0, len(heights) - 1
        left_max, right_max = heights[0], heights[-1]
        water = 0
        while left <= right:
            if left_max < right_max:
                left_max = max(left_max, heights[left])
                water += (left_max - heights[left])
                left += 1
            else:
                right_max = max(right_max, heights[right])
                water += (right_max - heights[right])
                right -= 1

        return water

Complexity Analysis

• Time Complexity
  • O(N)
• Space Complexity
  • O(1)