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1762. Buildings With an Ocean View

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.

 

Constraints:

  • 1 <= heights.length <= 105
  • 1 <= heights[i] <= 109

Solutions

  • class Solution {
    public int[] findBuildings(int[] heights) {
        int n = heights.length;
        List<Integer> ans = new ArrayList<>();
        int mx = 0;
        for (int i = heights.length - 1; i >= 0; --i) {
            if (heights[i] > mx) {
                ans.add(i);
                mx = heights[i];
            }
        }
        Collections.reverse(ans);
        return ans.stream().mapToInt(Integer::intValue).toArray();
    }
}

  • class Solution {
public:
    vector<int> findBuildings(vector<int>& heights) {
        vector<int> ans;
        int mx = 0;
        for (int i = heights.size() - 1; ~i; --i) {
            if (heights[i] > mx) {
                ans.push_back(i);
                mx = heights[i];
            }
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};

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


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

  • 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();
}


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

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