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
n == height.length 1 <= n <= 2 * 10^4 0 <= height[i] <= 10^5
class Solution:
def trap(self, height: List[int]) -> int:
left, right = 0, len(height) - 1
left_max, right_max = 0, 0
trapped_water = 0
while left < right:
if height[left] < height[right]:
if height[left] >= left_max:
left_max = height[left]
else:
trapped_water += left_max - height[left]
left += 1
else:
if height[right] >= right_max:
right_max = height[right]
else:
trapped_water += right_max - height[right]
right -= 1
return trapped_water
Strightforward no trick question, just had to cover all edge case in code. But took a lot longer than I thought it would take.
The time complexity of this solution is O(n), where n is the number of elements in the input array, because each element is visited at most once. The space complexity is O(1), as no additional data structures are used.