How do you solve Beautiful Towers II on LeetCode?

Beautiful Towers II (LeetCode #2866) 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: The peak can be any index, and the heights must respect the mountain condition.

What is the brute force solution for Beautiful Towers II?

The brute force approach for Beautiful Towers II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves iterating through each possible peak index and calculating the maximum possible heights on both sides of the peak. This is straightforward but inefficient as it checks every combination. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Beautiful Towers II?

Beautiful Towers II uses the following concepts: Array, Stack, Monotonic Stack. The recommended patterns to study are: Dynamic Programming, Two-Pointer Technique.

What are the interview tips for Beautiful Towers II?

Clearly explain your thought process and how you are approaching the problem. Consider edge cases, such as arrays with only one element or all elements being the same. Practice breaking down the problem into smaller parts before jumping into coding.

What are common mistakes when solving Beautiful Towers II?

Not considering the bounds of the mountain when filling heights. Overlooking the need to respect maxHeights constraints.

#2866

Beautiful Towers II

Medium

ArrayStackMonotonic StackDynamic ProgrammingTwo-Pointer Technique

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a two-pass approach to calculate the maximum possible heights for each peak. This reduces the complexity by avoiding redundant calculations and ensures we respect the mountain condition efficiently.

⚙️

Algorithm

4 steps

1Step 1: Create an array left that stores the maximum heights from the left up to each index.
2Step 2: Create an array right that stores the maximum heights from the right up to each index.
3Step 3: For each index i, calculate the maximum height as the minimum of left[i] and right[i].
4Step 4: Sum the heights to get the maximum possible sum.

solution.py20 lines

1# Full working Python code
2
3def maxBeautifulSum(maxHeights):
4    n = len(maxHeights)
5    left = [0] * n
6    right = [0] * n
7    # Fill left array
8    left[0] = maxHeights[0]
9    for i in range(1, n):
10        left[i] = min(left[i - 1], maxHeights[i])
11    # Fill right array
12    right[n - 1] = maxHeights[n - 1]
13    for i in range(n - 2, -1, -1):
14        right[i] = min(right[i + 1], maxHeights[i])
15    # Calculate maximum sum
16    max_sum = 0
17    for i in range(n):
18        max_sum += min(left[i], right[i])
19    return max_sum
20

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a constant number of times. The space complexity is O(n) due to the additional arrays used to store left and right heights.

1The peak can be any index, and the heights must respect the mountain condition.
2Using two arrays to track maximum heights from both sides allows efficient calculation.

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