https://leetcode.com/problems/trapping-rain-water/
brute-force
water at any x = min( max_left_of_x, max_right_of_x) - height_of_x
two-pointer solution
when keeping
1)
- p_start, at 0
- p_end at 11 [length-1] of array
- check,
- p_start height is less. => 0
- p_end height is less => 1
- answer: p_start => 0 is lesser
- height at 0 is less
- p_start becomes our p_current pointer.
- with respect to p_current
- we check
- max_left
- max_right
- whenever we check which is lesser p_start, p_end, we are actually evaluating min(p_start, p_end)
- by checking min, of two, we get the level, which water can be filled.
formula
Level => check min of p_start, p_end
current_level = min(p_start, p_end)
if p_start is less, we start from left, and
- make max_left = height_at_p_start
- make max_right = height_at_p_end
- we consider level = max_left
-
***incomplete post
