human
I found an interesting problem on https://leetcode.com/problems/pour-water/:
We are given an elevation map, heights[i] representing the height of the terrain at that index.  The width at each index is 1.  After V units of water fall at index K, how much water is at each index?

Water first drops at index K and rests on top of the highest terrain or water at that index.  Then, it flows according to the following rules:
If the droplet would eventually fall by moving left, then move left.
Otherwise, if the droplet would eventually fall by moving right, then move right.
Otherwise, rise at it's current position.
Here, "eventually fall" means that the droplet will eventually be at a lower level if it moves in that direction.
Also, "level" means the height of the terrain plus any water in that column.

We can assume there's infinitely high terrain on the two sides out of bounds of the array.  Also, there could not be partial water being spread out evenly on more than 1 grid block - each unit of water has to be in exactly one block.

Example 1:

Input: heights = [2,1,1,2,1,2,2], V = 4, K = 3
Output: [2,2,2,3,2,2,2]
Explanation:
#       #
#       #
##  # ###
#########
 0123456    

Example 2:

Input: heights = [1,2,3,4], V = 2, K = 2
Output: [2,3,3,4]
Explanation:
The last droplet settles at index 1, since moving further left would not cause it to eventually fall to a lower height.


Example 3:

Input: heights = [3,1,3], V = 5, K = 1
Output: [4,4,4]


Note:
heights will have length in [1, 100] and contain integers in [0, 99].
V will be in range [0, 2000].
K will be in range [0, heights.length - 1].

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python code you requested.

class Solution:
     def reachNumber(self, target):
         """
         :type target: int
         :rtype: int
         """
         target=abs(target)
         
         n=math.ceil(((1+8*target)**0.5-1)/2)
         
         if target%2>0:
             if n%4==1 or n%4==2:
                 return n
             elif n%4==3:
                 return n+2
             else:
                 return n+1
         else:
             if n%4==3 or n%4==0:
                 return n
             elif n%4==1:
                 return n+2
             else:
                 return n+1