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3285. Find Indices of Stable Mountains

Description

There are n mountains in a row, and each mountain has a height. You are given an integer array height where height[i] represents the height of mountain i, and an integer threshold.

A mountain is called stable if the mountain just before it (if it exists) has a height strictly greater than threshold. Note that mountain 0 is not stable.

Return an array containing the indices of all stable mountains in any order.

 

Example 1:

Input: height = [1,2,3,4,5], threshold = 2

Output: [3,4]

Explanation:

  • Mountain 3 is stable because height[2] == 3 is greater than threshold == 2.
  • Mountain 4 is stable because height[3] == 4 is greater than threshold == 2.

Example 2:

Input: height = [10,1,10,1,10], threshold = 3

Output: [1,3]

Example 3:

Input: height = [10,1,10,1,10], threshold = 10

Output: []

 

Constraints:

  • 2 <= n == height.length <= 100
  • 1 <= height[i] <= 100
  • 1 <= threshold <= 100

Solutions

Solution 1: Traversal

We directly traverse the mountains starting from index $1$. If the height of the mountain to its left is greater than $threshold$, we add its index to the result array.

After the traversal, we return the result array.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{height}$. Ignoring the space consumption of the result array, the space complexity is $O(1)$.

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class Solution:
    def stableMountains(self, height: List[int], threshold: int) -> List[int]:
        return [i for i in range(1, len(height)) if height[i - 1] > threshold]

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class Solution {
    public List<Integer> stableMountains(int[] height, int threshold) {
        List<Integer> ans = new ArrayList<>();
        for (int i = 1; i < height.length; ++i) {
            if (height[i - 1] > threshold) {
                ans.add(i);
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> stableMountains(vector<int>& height, int threshold) {
        vector<int> ans;
        for (int i = 1; i < height.size(); ++i) {
            if (height[i - 1] > threshold) {
                ans.push_back(i);
            }
        }
        return ans;
    }
};

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func stableMountains(height []int, threshold int) (ans []int) {
    for i := 1; i < len(height); i++ {
        if height[i-1] > threshold {
            ans = append(ans, i)
        }
    }
    return
}

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function stableMountains(height: number[], threshold: number): number[] {
    const ans: number[] = [];
    for (let i = 1; i < height.length; ++i) {
        if (height[i - 1] > threshold) {
            ans.push(i);
        }
    }
    return ans;
}

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