How do you solve Maximum Building Height on LeetCode?

Maximum Building Height (LeetCode #1840) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding how to manage height restrictions is key to solving this problem.

What is the time complexity of Maximum Building Height?

The optimal solution for Maximum Building Height runs in O(n log n) time and uses O(n) space. The time complexity is dominated by the sorting step, while the space complexity accounts for storing the restrictions.

What is the brute force solution for Maximum Building Height?

The brute force approach for Maximum Building Height is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible height for each building, ensuring that we respect the height differences and restrictions. This method is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Building Height?

Maximum Building Height uses the following concepts: Array, Math, Sorting. The recommended patterns to study are: Sorting, Greedy Algorithms.

What are the interview tips for Maximum Building Height?

Always clarify the problem constraints and edge cases before diving into coding. Think about how you can optimize your solution after writing the brute force version. Practice explaining your thought process clearly as you work through the problem.

What are common mistakes when solving Maximum Building Height?

Failing to consider both sides of the restrictions when calculating heights. Not including the first and last buildings in the restrictions list.

#1840

Maximum Building Height

Hard

ArrayMathSortingSortingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

The optimal solution involves processing the restrictions in two passes: one from left to right and another from right to left. This allows us to calculate the maximum possible height for each building while respecting the restrictions.

⚙️

Algorithm

4 steps

1Step 1: Add the first building with height 0 to the restrictions list and sort the restrictions based on building IDs.
2Step 2: Traverse the sorted restrictions from left to right, calculating the maximum height for each building based on the previous building's height and the current restriction.
3Step 3: Traverse the sorted restrictions from right to left, applying the same logic to ensure the heights are valid in both directions.
4Step 4: Calculate the maximum height possible between each pair of buildings based on the adjusted heights.

solution.py13 lines

1# Full working Python code
2
3def maxBuilding(n, restrictions):
4    restrictions.append([1, 0])
5    restrictions.append([n, n])
6    restrictions.sort()
7    max_height = 0
8    for i in range(1, len(restrictions)):
9        left_id, left_height = restrictions[i - 1]
10        right_id, right_height = restrictions[i]
11        height = (right_height + left_height + (right_id - left_id)) // 2
12        max_height = max(max_height, height)
13    return max_height

ℹ

Complexity note: The time complexity is dominated by the sorting step, while the space complexity accounts for storing the restrictions.

1Understanding how to manage height restrictions is key to solving this problem.
2Two passes (left to right and right to left) help ensure that all restrictions are respected.

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