**Unobstructed Buildings - Facebook Top Interview Questions**

### Problem Statement :

You are given a list of integers heights representing building heights. A building heights[i] can see the ocean if every building on its right has shorter height. Return the building indices where you can see the ocean, in ascending order. Constraints 0 ≤ n ≤ 100,000 where n is the length of heights Example 1 Input heights = [1, 5, 5, 2, 3] Output [2, 4] Explanation We can see the ocean in building heights[2] and heights[4]. Example 2 Input heights = [5, 4, 3, 2, 1] Output [0, 1, 2, 3, 4] Explanation We can see the ocean in every building since each building is taller than every other on its right. Example 3 Input heights = [1, 1, 1, 1, 1] Output [4] Explanation We can't see the ocean in any building other than the last one.

### Solution :

` ````
Solution in C++ :
vector<int> solve(vector<int>& heights) {
vector<int> ret;
int maxi = INT_MIN;
for (int i = heights.size() - 1; i >= 0; i--) {
if (maxi < heights[i]) {
maxi = heights[i];
ret.push_back(i);
}
}
reverse(ret.begin(), ret.end());
return ret;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int[] solve(int[] heights) {
int N = heights.length;
int[] max = new int[N + 1];
for (int i = N - 1; i >= 0; i--) max[i] = Math.max(max[i + 1], heights[i]);
ArrayList<Integer> ans = new ArrayList<Integer>();
for (int i = 0; i < N; i++) {
if (heights[i] > max[i + 1])
ans.add(i);
}
return convert(ans);
}
public int[] convert(ArrayList<Integer> arr) {
int[] ans = new int[arr.size()];
for (int i = 0; i < arr.size(); i++) ans[i] = arr.get(i);
return ans;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, heights):
ans = []
highest = -1
for i in range(len(heights) - 1, -1, -1):
if heights[i] > highest:
ans.append(i)
highest = heights[i]
return ans[::-1]
```

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