LeetCode #3530 — HARD

Maximum Profit from Valid Topological Order in DAG

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given a Directed Acyclic Graph (DAG) with n nodes labeled from 0 to n - 1, represented by a 2D array edges, where edges[i] = [ui, vi] indicates a directed edge from node ui to vi. Each node has an associated score given in an array score, where score[i] represents the score of node i.

You must process the nodes in a valid topological order. Each node is assigned a 1-based position in the processing order.

The profit is calculated by summing up the product of each node's score and its position in the ordering.

Return the maximum possible profit achievable with an optimal topological order.

A topological order of a DAG is a linear ordering of its nodes such that for every directed edge u → v, node u comes before v in the ordering.

Example 1:

Input: n = 2, edges = [[0,1]], score = [2,3]

Output: 8

Explanation:

Node 1 depends on node 0, so a valid order is [0, 1].

Node Processing Order Score Multiplier Profit Calculation
0 1st 2 1 2 × 1 = 2
1 2nd 3 2 3 × 2 = 6

The maximum total profit achievable over all valid topological orders is 2 + 6 = 8.

Example 2:

Input: n = 3, edges = [[0,1],[0,2]], score = [1,6,3]

Output: 25

Explanation:

Nodes 1 and 2 depend on node 0, so the most optimal valid order is [0, 2, 1].

Node Processing Order Score Multiplier Profit Calculation
0 1st 1 1 1 × 1 = 1
2 2nd 3 2 3 × 2 = 6
1 3rd 6 3 6 × 3 = 18

The maximum total profit achievable over all valid topological orders is 1 + 6 + 18 = 25.

Constraints:

  • 1 <= n == score.length <= 22
  • 1 <= score[i] <= 105
  • 0 <= edges.length <= n * (n - 1) / 2
  • edges[i] == [ui, vi] denotes a directed edge from ui to vi.
  • 0 <= ui, vi < n
  • ui != vi
  • The input graph is guaranteed to be a DAG.
  • There are no duplicate edges.
Patterns Used
📊Dynamic Programming→⚡Bit Manipulation→📋Topological Sort→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a Directed Acyclic Graph (DAG) with n nodes labeled from 0 to n - 1, represented by a 2D array edges, where edges[i] = [ui, vi] indicates a directed edge from node ui to vi. Each node has an associated score given in an array score, where score[i] represents the score of node i. You must process the nodes in a valid topological order. Each node is assigned a 1-based position in the processing order. The profit is calculated by summing up the product of each node's score and its position in the ordering. Return the maximum possible profit achievable with an optimal topological order. A topological order of a DAG is a linear ordering of its nodes such that for every directed edge u → v, node u comes before v in the ordering.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Bit Manipulation · Topological Sort

Example 1

2
[[0,1]]
[2,3]

Example 2

3
[[0,1],[0,2]]
[1,6,3]
Step 02

Core Insight

What unlocks the optimal approach

  • Use bitmask dynamic programming.
  • States are <code>mask</code> = (bits such that if a bit is set, it means the corresponding node is removed).
  • Try maintaining the <code>degrees</code> across function calls.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3530: Maximum Profit from Valid Topological Order in DAG
class Solution {
  public int maxProfit(int n, int[][] edges, int[] score) {
    final int maxMask = 1 << n;
    // need[i] := the bitmask representing all nodes that must be placed before node i
    int[] need = new int[n];
    // dp[mask] := the maximum profit achievable by placing the set of nodes represented by `mask`
    int[] dp = new int[maxMask];
    Arrays.fill(dp, -1);
    dp[0] = 0;

    for (int[] edge : edges) {
      final int u = edge[0];
      final int v = edge[1];
      need[v] |= 1 << u;
    }

    // Iterate over all subsets of nodes (represented by bitmask `mask`)
    for (int mask = 0; mask < maxMask; ++mask) {
      if (dp[mask] == -1)
        continue;
      // Determine the position of the next node to be placed (1-based).
      int pos = Integer.bitCount(mask) + 1;
      // Try to place each node `i` that hasn't been placed yet.
      for (int i = 0; i < n; ++i) {
        if ((mask >> i & 1) == 1)
          continue;
        // Check if all dependencies of node `i` are already placed.
        if ((mask & need[i]) == need[i]) {
          final int newMask = mask | 1 << i; // Mark node `i` as placed.
          dp[newMask] = Math.max(dp[newMask], dp[mask] + score[i] * pos);
        }
      }
    }

    return dp[maxMask - 1];
  }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × m)
Space
O(n × m)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

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