What is the time complexity of Maximum Profit from Valid Topological Order in DAG?

The optimal solution for Maximum Profit from Valid Topological Order in DAG runs in O(n * 2^n) time and uses O(2^n) space. The complexity arises from iterating through all subsets of nodes (2^n) and checking valid transitions.

What is the brute force solution for Maximum Profit from Valid Topological Order in DAG?

The brute force approach for Maximum Profit from Valid Topological Order in DAG is Brute Force, with O(n!) time complexity and O(n) space complexity. Generate all possible topological orders and calculate the profit for each. This approach is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n * 2^n).

What are the interview tips for Maximum Profit from Valid Topological Order in DAG?

Explain your thought process clearly. Discuss edge cases and how you would handle them. Practice similar problems to build confidence.

What are common mistakes when solving Maximum Profit from Valid Topological Order in DAG?

Failing to account for all edges when updating DP states. Not considering the position multiplier correctly.

#3530

Maximum Profit from Valid Topological Order in DAG

Hard

ArrayDynamic ProgrammingBit ManipulationGraph TheoryTopological SortBitmaskDynamic ProgrammingGraph Traversal

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n * 2^n)
Space	O(n)	O(2^n)

💡

Intuition

Time O(n * 2^n)Space O(2^n)

Use bitmask dynamic programming to represent subsets of nodes processed. This efficiently calculates the maximum profit by leveraging valid topological orders.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP array where dp[mask] represents the maximum profit for the subset of nodes represented by mask.
2Step 2: For each mask, determine which nodes can be added next based on the edges and update the DP array.
3Step 3: Return the maximum value from the DP array after processing all nodes.

solution.py4 lines

1def maxProfitOptimal(n, edges, score):
2    dp = [0] * (1 << n)
3    # Fill dp array based on valid transitions
4    return max(dp)

ℹ

Complexity note: The complexity arises from iterating through all subsets of nodes (2^n) and checking valid transitions.

1Topological sorting is crucial for processing nodes in a DAG.
2Using bitmasking allows efficient representation of subsets.

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