Trapping Rain Water

/*
Basic logic:we will find water trapped onto every building by min(left_max,right_max)-arr[i];

T.C : O(n),S.C O(1),one pass
In our algo,left_max will store maximum of 0 to i-1 and right_max will store maximum of j+1 to n-1.
using two pointer(i,j),so if we think about ith index,so we have left_max for 0 to i-1 and right_max from j+1 to n-1
so,
Case 1: if left_max < right_max:
 *We will think to get water over the ith building.
 we need max of 0 to i-1 that we already have ie. left_max
 we need max of i+1 to n-1 but we have maxiumum of j+1 to n-1 only,not having maxiumum of i+1 to j
 but even if some building height is greater than right_max still we will have min(left_max,right_max) = left_max
 because left_max is smaller than even current right_max.
Case 2: if left_max>=right_max:
 *We will think to get water over jth building.
 If left_max is greater and right_max is smaller then we need only smaller one so we got the water height.
*/

class Solution:
    def maxWater(self, arr):
        i = 1
        j = len(arr) - 2
        left_max = arr[0]
        right_max = arr[-1]
        ans = 0

        while i <= j:
            if left_max > right_max:
                ans += max(0, right_max - arr[j])
                right_max = max(right_max, arr[j])
                j -= 1
            else:
                ans += max(0, left_max - arr[i])
                left_max = max(left_max, arr[i])
                i += 1

        return ans