LeetCode 1762. Buildings With an Ocean View

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 <= 105
  • 1 <= 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).

2 Thoughts to “LeetCode 1762. Buildings With an Ocean View”

  1. 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.

Leave a Reply

Your email address will not be published. Required fields are marked *