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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: [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 <= 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;
}
}

############

class Solution {
public int[] findBuildings(int[] heights) {
int mx = 0;
for (int i = heights.length - 1; i >= 0; --i) {
int v = heights[i];
if (mx < v) {
mx = v;
}
}
return ans.stream().mapToInt(i -> i).toArray();
}
}

• // OJ: https://leetcode.com/problems/buildings-with-an-ocean-view/
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> findBuildings(vector<int>& A) {
vector<int> ans{(int)A.size() - 1};
for (int i = A.size() - 2; i >= 0; --i) {
if (A[i] > A[ans.back()]) ans.push_back(i);
}
reverse(begin(ans), end(ans));
return ans;
}
};

• class Solution:
def findBuildings(self, heights: List[int]) -> List[int]:
mx = 0
ans = []
for i in range(len(heights) - 1, -1, -1):
v = heights[i]
if mx < v:
ans.append(i)
mx = v
return ans[::-1]


• func findBuildings(heights []int) []int {
mx := 0
ans := []int{}
for i := len(heights) - 1; i >= 0; i-- {
v := heights[i]
if mx < v {
ans = append(ans, i)
mx = v
}
}
for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
ans[i], ans[j] = ans[j], ans[i]
}
return ans
}

• /**
* @param {number[]} heights
* @return {number[]}
*/
var findBuildings = function (heights) {
let mx = 0;
let ans = [];
for (let i = heights.length - 1; i >= 0; --i) {
const v = heights[i];
if (mx < v) {
ans.push(i);
mx = v;
}
}
return ans.reverse();
};


• function findBuildings(heights: number[]): number[] {
const ans: number[] = [];
let mx = 0;
for (let i = heights.length - 1; ~i; --i) {
if (heights[i] > mx) {
ans.push(i);
mx = heights[i];
}
}
return ans.reverse();
}