How do you solve Find Indices of Stable Mountains on LeetCode?

Find Indices of Stable Mountains (LeetCode #3285) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Stable mountains depend on the height of the previous mountain.

What is the time complexity of Find Indices of Stable Mountains?

The optimal solution for Find Indices of Stable Mountains runs in O(n) time and uses O(n) space. The time complexity is O(n) because we loop through the array once. The space complexity is O(n) due to the storage of stable indices in a list.

What is the brute force solution for Find Indices of Stable Mountains?

The brute force approach for Find Indices of Stable Mountains is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach checks each mountain to see if it is stable by looking at the height of the mountain before it. This is straightforward but can be inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Indices of Stable Mountains?

Find Indices of Stable Mountains uses the following concepts: Array. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Find Indices of Stable Mountains?

Clarify the problem statement and constraints before coding. Think about edge cases, such as the smallest and largest possible inputs. Explain your thought process as you code.

What are common mistakes when solving Find Indices of Stable Mountains?

Forgetting to check the previous mountain's height. Not starting the loop from index 1.

#3285

Find Indices of Stable Mountains

Easy

ArrayArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution leverages a single pass through the array, checking each mountain's stability based on the previous mountain's height. This is efficient and straightforward.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty list to store stable mountain indices.
2Step 2: Loop through the mountains starting from index 1 to the end of the array.
3Step 3: For each mountain, check if the height of the previous mountain is greater than the threshold.
4Step 4: If it is, add the current index to the list of stable mountains.
5Step 5: Return the list of stable mountain indices.

solution.py6 lines

1def find_stable_mountains(height, threshold):
2    stable_indices = []
3    for i in range(1, len(height)):
4        if height[i - 1] > threshold:
5            stable_indices.append(i)
6    return stable_indices

ℹ

Complexity note: The time complexity is O(n) because we loop through the array once. The space complexity is O(n) due to the storage of stable indices in a list.

1Stable mountains depend on the height of the previous mountain.
2Only mountains from index 1 onward can be stable.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.