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.
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.
Input: heights = [4,3,2,1] Output: [0,1,2,3] Explanation: All the buildings have an ocean view.
Input: heights = [1,3,2,4] Output:  Explanation: Only building 3 has an ocean view.
Input: heights = [2,2,2,2] Output:  Explanation: Buildings cannot see the ocean if there are buildings of the same height to its right.
1 <= heights.length <= 105
1 <= heights[i] <= 109
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.
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).
2 Thoughts to “LeetCode 1762. Buildings With an Ocean View”
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.