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 :



title-img 




                        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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