Formatted question description: https://leetcode.ca/all/1762.html

# 1762. Buildings With an Ocean View

Medium

## Description

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: 

Explanation: Only building 3 has an ocean view.

Example 4:

Input: heights = [2,2,2,2]

Output: 

Explanation: Buildings cannot see the ocean if there are buildings of the same height to its right.

Constraints:

• 1 <= heights.length <= 10^5
• 1 <= heights[i] <= 10^9

## Solution

For index i where 0 <= i < n, the building at index i has an ocean view if and only if for all i < j < n, there is heights[i] > heights[j]. Obviously, the building at index n - 1 has an ocean view.

Loop over heights backwards and maintain the maximum height. For the building at index i, if heights[i] is greater than the maximum height, then the building has an ocean view. Update the maximum height using heights[i] after determining whether the building at index i has an ocean view.

Finally, return the indices of buildings that have an ocean view.

class Solution {
public int[] findBuildings(int[] heights) {
int length = heights.length;
boolean[] oceanView = new boolean[length];
int count = 0;
int maxHeight = 0;
for (int i = length - 1; i >= 0; i--) {
if (heights[i] > maxHeight) {
oceanView[i] = true;
count++;
}
maxHeight = Math.max(maxHeight, heights[i]);
}
int[] buildings = new int[count];
int index = 0;
for (int i = 0; i < length; i++) {
if (oceanView[i])
buildings[index++] = i;
}
return buildings;
}
}