## Description

https://leetcode.com/problems/buildings-with-an-ocean-view/

There are `n`

buildings in a line. You are given an integer array `heights`

of size `n`

that represents the heights of the buildings in the line.

The ocean is to the right of the buildings. A building has an ocean view if the building can see the ocean without obstructions. Formally, a building has an ocean view if all the buildings to its right have a **smaller** height.

Return a list of indices **(0-indexed)** of buildings that have an ocean view, sorted in increasing order.

**Example 1:**

Input:heights = [4,2,3,1]Output:[0,2,3]Explanation:Building 1 (0-indexed) does not have an ocean view because building 2 is taller.

**Example 2:**

Input:heights = [4,3,2,1]Output:[0,1,2,3]Explanation:All the buildings have an ocean view.

**Example 3:**

Input:heights = [1,3,2,4]Output:[3]Explanation:Only building 3 has an ocean view.

**Example 4:**

Input:heights = [2,2,2,2]Output:[3]Explanation:Buildings cannot see the ocean if there are buildings of thesameheight to its right.

**Constraints:**

`1 <= heights.length <= 10`

^{5}`1 <= heights[i] <= 10`

^{9}

## Explanation

Iterate from the right side and keep tracking of the tallest height. If a building is greater than the tallest height, append it to the ocean view buildings.

## Python Solution

```
class Solution:
def findBuildings(self, heights: List[int]) -> List[int]:
results = []
prev_max = None
for i in range(len(heights) - 1, -1, -1):
height = heights[i]
if not prev_max:
prev_max = height
results.append(i)
else:
if height > prev_max:
results.append(i)
prev_max = max(prev_max, height)
return sorted(results)
```

- Time Complexity: O(Nlog(N)).
- Space Complexity: O(N).