## Description

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. The above elevation map 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. Thanks Marcos for contributing this image!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6


## Solutions

### 1. Brute Force

# Time: O(n^2)
# Space: O(n)
class Solution:
def trap(self, height: List[int]) -> int:
res = 0
n = len(height)
for i in range(n):
if not height[:i] or not height[i+1:]:
continue
lower = min(max(height[:i]), max(height[i:]))
if lower < height[i]:
continue
res += lower - height[i]
return res
# 315/315 cases passed (2636 ms)
# Your runtime beats 5.01 % of python3 submissions
# Your memory usage beats 88.37 % of python3 submissions (13.4 MB)


## 2. DP

# Time: O(n)
# Space: O(n)
class Solution:
def trap(self, height: List[int]) -> int:
n = len(height)
l =  * n
r =  * n
for i in range(n):
l[i] = height[i] if i == 0 else max(l[i - 1], height[i])
for i in range(n-1, -1, -1):
r[i] = height[i] if i == n - 1 else max(r[i + 1], height[i])
res = 0
for i in range(n):
res += min(l[i], r[i]) - height[i]
return res

# 315/315 cases passed (48 ms)
# Your runtime beats 85.43 % of python3 submissions
# Your memory usage beats 90.7 % of python3 submissions (13.4 MB)


### 3. Two Pointers

# Time: O(n)
# Space: O(1)
class Solution:
def trap(self, height: List[int]) -> int:
n = len(height)
if n == 0:
return 0
l, r = 0, n - 1
max_l, max_r = height, height[r]
res = 0
while l < r:
if max_l < max_r:
res += max_l - height[l]
l += 1
max_l = max(max_l,height[l])
else:
res += max_r - height[r]
r -= 1
max_r = max(max_r, height[r])
return res

# 315/315 cases passed (52 ms)
# Your runtime beats 66.24 % of python3 submissions
# Your memory usage beats 97.67 % of python3 submissions (13.4 MB)