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 the same height to its right.
Constraints:
1 <= heights.length <= 1051 <= heights[i] <= 109
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).
in place of returning sorted list fucking reverse the list mate
Hey nice post! I think you could get your time complexity down to O(n) if you simply reverse your list instead of calling sort(), since you’re appending from the right 🙂
In reality, the time complexity is probably going to be preeeety similar, but just thought you’d like to know.