# Time: O(N)
# Space: O(N)

class Solution:
    def trap(self, heights: List[int]) -> int:
        # Two pointer approach
        left, right = 0, len(heights) - 1
        
        # Since while scanning the array, at a given index i, we
        # only need to consider the minimum of max_left_heights and
        # max_right_heights, we just need to keep moving from both
        # ends and keep track of the maximum height seen so far. We
        # can then compare these two heights and decide from which
        # side to move next.
        max_left = max_right = 0
        
        output = 0
        while left <= right:
            # By the formula from the DP approach:
            #
            # min(max_left, max_right) - heights[i]
            #
            # If max_left <= max_right, formula becomes:
            #
            # max_left - heights[i]
            #
            if max_left <= max_right:
                # Shorter form to avoid -ve values
                output += max(max_left - heights[left], 0)
                
                max_left = max(max_left, heights[left])
                left += 1
            # By the formula from the DP approach:
            #
            # min(max_left, max_right) - heights[i]
            #
            # If max_left > max_right, formula becomes:
            #
            # max_right - heights[i]
            #
            else:
                output += max(max_right - heights[right], 0)
                
                max_right = max(max_right, heights[right])
                right -= 1
            
        
        return output