1762. 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.

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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 Decresing Monotonick Stack 2021-08-18

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class Solution {
    public int[] findBuildings(int[] heights) {
        Stack<Integer> pos = new Stack<>();
        for (int i = 0; i < heights.length; i++) {
            while (pos.size() > 0 && heights[pos.peek()] <= heights[i]) {
                pos.pop();
            }
            
            pos.push(i);
        }
        int[] res = new int[pos.size()];
        
        for (int i = pos.size() - 1; i >= 0; i--) {
            res[i] = pos.pop();
        }
        return res;
    }
}

Solution Increasing Monotonick Stack 2022-01-24

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class Solution {

    public int[] findBuildings(int[] heights) {
        Stack<Integer> stack = new Stack<>();
        int n = heights.length;
        stack.push(n - 1);
        for (int i = n - 2; i >= 0; i--) {  
            if (stack.size() > 0 && heights[stack.peek()] < heights[i]) {
                stack.push(i);
            }
        }
        
        int[] res = new int[stack.size()];
        int i = 0;
        while (stack.size() > 0) {
            res[i++] = stack.pop();
        }
        return res;
    }
}