How do you solve Maximum Good Subtree Score on LeetCode?

Maximum Good Subtree Score (LeetCode #3575) 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: Digit uniqueness can be efficiently tracked using bitmasks.

What is the time complexity of Maximum Good Subtree Score?

The optimal solution for Maximum Good Subtree Score runs in O(n) time and uses O(n) space. DFS visits each node once, and bit manipulation is efficient for digit checks.

What is the brute force solution for Maximum Good Subtree Score?

The brute force approach for Maximum Good Subtree Score is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all subsets of nodes in each subtree and check if they are good. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Maximum Good Subtree Score?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar tree problems.

What are common mistakes when solving Maximum Good Subtree Score?

Not handling digit overlaps correctly. Forgetting to reset bitmask when backtracking.

#3575

Maximum Good Subtree Score

Hard

ArrayDynamic ProgrammingBit ManipulationTreeDepth-First SearchBitmaskBit ManipulationDepth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Use DFS and bit manipulation to efficiently track digit usage and calculate scores.

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Algorithm

3 steps

1Step 1: Perform DFS on the tree, maintaining a bitmask for used digits and the current score.
2Step 2: For each node, attempt to include it in the current subset if it doesn't violate the digit uniqueness.
3Step 3: Update the maxScore for each node based on the best score from its children.

solution.py3 lines

1def maxGoodSubtreeScore(vals, par):
2    # Implementation here
3    ...

ℹ

Complexity note: DFS visits each node once, and bit manipulation is efficient for digit checks.

1Digit uniqueness can be efficiently tracked using bitmasks.
2DFS allows for systematic exploration of tree structures.

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