Problem Description

You have an undirected graph with n nodes numbered from 0 to n - 1. Each node has a value given by values[i] for node i. The graph's edges are defined by a 2D array edges, where each element edges[j] = [u_j, v_j, time_j] represents an undirected edge between nodes u_j and v_j that takes time_j seconds to traverse.

Your task is to find a valid path that:

• Starts at node 0
• Ends at node 0 (returns to the starting point)
• Takes at most maxTime seconds to complete
• Can visit the same node multiple times

The quality of a path is calculated as the sum of values of all unique nodes visited during the path. This means if you visit a node multiple times, its value is only counted once in the quality calculation.

You need to return the maximum possible quality among all valid paths.

Key constraints:

• Each node has at most 4 edges connected to it
• You must start and end at node 0
• The total time of the path cannot exceed maxTime
• Node values are only counted once in the quality sum, regardless of how many times you visit them

Example: If you have a path that visits nodes 0 → 1 → 2 → 1 → 0, and the values are [5, 10, 15, ...], the quality would be 5 + 10 + 15 = 30 (node 1's value is only counted once despite being visited twice).

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem explicitly involves an undirected graph with nodes and edges. We have nodes numbered from 0 to n-1 and edges connecting them with travel times.

Since the answer is "yes" for a graph problem, the flowchart doesn't continue down the "no" path. However, let's analyze the problem characteristics to understand why backtracking is the appropriate approach:

Does the problem have small constraints?

• Yes: The problem states that each node has at most 4 edges connected to it, and with maxTime constraints, the number of edges we can traverse is limited. The reference solution notes that we can traverse at most 10 edges (maxTime/min(time_j) = 100/10 = 10), making the search space manageable.

Brute force / Backtracking?

• Yes: Given the small constraints and the need to explore all possible valid paths from node 0 back to node 0, backtracking is ideal. We need to:
  • Try different paths from the current node
  • Keep track of visited nodes (for quality calculation)
  • Backtrack when we've explored a path
  • Ensure we don't exceed the time limit

Why Backtracking fits perfectly:

1. We need to explore all possible paths that start and end at node 0
2. The search space is small enough (max 4 edges per node, limited path length)
3. We need to track state (visited nodes, current time, current quality)
4. We need to undo choices (unmark nodes as visited) when backtracking

Conclusion: The flowchart analysis confirms that backtracking is the optimal approach for this problem. The DFS with backtracking allows us to exhaustively search all valid paths while maintaining the visited state for quality calculation and ensuring we stay within the time constraints.

Intuition

The key insight comes from examining the constraints. With at most 4 edges per node and a maximum time limit, we realize the search space is actually quite small. Even in the worst case where each edge takes minimum time (10 seconds), we can only traverse about 10 edges within maxTime = 100. This means our paths won't be very long.

Since we need to find the maximum quality among all valid paths from node 0 back to node 0, and the search space is manageable, we can afford to explore every possible path. This naturally leads us to a backtracking approach.

The tricky part is handling the quality calculation - we want to count each node's value only once, even if we visit it multiple times. This suggests we need to track which nodes we've visited in our current path. When we visit a node for the first time, we add its value to our quality sum. If we visit it again, we don't add its value again.

Here's the thought process:

1. Start at node 0 with its value already counted
2. For each neighbor, check if we have enough time to visit it
3. If we visit a new node, mark it as visited and add its value
4. If we revisit a node, don't add its value again
5. When we return to node 0, we have a complete path - update our answer if this path has better quality
6. Backtrack by unmarking nodes we visited (so other paths can count them fresh)

The beauty of this approach is that the backtracking naturally handles the exploration of all paths, while the visited array ensures we correctly calculate the quality by counting each node's value at most once per path. The DFS recursion stack keeps track of our current path, and we only need to maintain the cumulative time and quality as we go.

Learn more about Graph and Backtracking patterns.

Solution Approach

The solution uses DFS with backtracking to explore all valid paths. Let's walk through the implementation step by step:

1. Graph Representation First, we build an adjacency list g to represent the graph. For each edge [u, v, t], we add (v, t) to g[u] and (u, t) to g[v] since the graph is undirected.

g = [[] for _ in range(n)]
for u, v, t in edges:
    g[u].append((v, t))
    g[v].append((u, t))

2. Visited Tracking We maintain a boolean array vis of length n to track which nodes are currently in our path. Initially, we mark node 0 as visited since we start from there.

vis = [False] * n
vis[0] = True

3. DFS Function The core logic is in the dfs(u, cost, value) function where:

• u: current node we're at
• cost: total time spent so far
• value: total quality (sum of unique node values) collected so far

4. Base Case When we reach node 0 (which means we've completed a cycle back to start), we update our answer:

if u == 0:
    ans = max(ans, value)

5. Exploring Neighbors For each neighbor v of current node u with edge time t:

• First check if we have enough time: cost + t <= maxTime
• If vis[v] is True (node already visited in current path):
  • Recursively call dfs(v, cost + t, value) without adding the node's value again
• If vis[v] is False (first time visiting this node):
  • Mark it as visited: vis[v] = True
  • Recursively call with added value: dfs(v, cost + t, value + values[v])
  • Backtrack by unmarking: vis[v] = False

6. Starting the Search We initiate the DFS from node 0 with:

• Starting cost: 0
• Starting value: values[0] (node 0's value)

The algorithm cleverly handles the requirement that each node's value should be counted at most once per path. By maintaining the vis array and only adding a node's value when we first visit it in the current path, we ensure correct quality calculation. The backtracking (setting vis[v] = False) allows other paths to count that node's value independently.

The time complexity is bounded by the small search space - with at most 4 edges per node and limited path length, the algorithm efficiently explores all possibilities despite being exponential in nature.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example Graph:

• Nodes: 4 nodes (0, 1, 2, 3)
• Values: [10, 5, 15, 20]
• Edges: [[0,1,10], [0,2,15], [1,3,20], [2,3,10]]
• maxTime: 30

Graph Visualization:

    0(10)---10---1(5)
      |           |
     15          20
      |           |
    2(15)---10---3(20)

DFS Exploration:

1. Start at node 0

   • vis = [True, False, False, False]
   • cost = 0, value = 10
   • Check neighbors: nodes 1 and 2

2. Path: 0 → 1

   • cost = 10, value = 10 + 5 = 15
   • vis = [True, True, False, False]
   • From node 1, can go to: node 0 (cost=20) or node 3 (cost=30)

3. Path: 0 → 1 → 0

   • Back at node 0, cost = 20, value = 15
   • Update answer: ans = 15
   • Continue exploring from node 1...

4. Path: 0 → 1 → 3

   • cost = 30, value = 15 + 20 = 35
   • vis = [True, True, False, True]
   • From node 3, time budget exhausted (any move would exceed 30)
   • Backtrack: vis[3] = False

5. Backtrack to node 0, explore second neighbor

   • vis = [True, False, False, False]
   • Path: 0 → 2
   • cost = 15, value = 10 + 15 = 25
   • vis = [True, False, True, False]

6. Path: 0 → 2 → 3

   • cost = 25, value = 25 + 20 = 45
   • vis = [True, False, True, True]
   • From node 3, can only afford to go back to node 2 (cost=35 > 30 for node 1)
   • Backtrack: vis[3] = False

7. Path: 0 → 2 → 0

   • Back at node 0, cost = 30, value = 25
   • Update answer: ans = max(15, 25) = 25

Key Observations:

• When we visit node 1 twice in path 0→1→0→1, we don't add value[1] again
• The vis array tracks nodes in current path, preventing double-counting
• Backtracking (vis[v] = False) allows other paths to count that node's value
• We only update the answer when we return to node 0
• The algorithm explores all feasible paths within the time limit

Final Answer: 25 (from path 0→2→0 with nodes {0, 2} visited)

Solution Implementation

Time and Space Complexity

Time Complexity: O(n + m + 4^(maxTime/min(time_j)))

The time complexity consists of:

• O(n + m) for building the adjacency list from the edges
• The DFS exploration complexity, which in the worst case can be O(4^(maxTime/min(time_j)))

The exponential part arises because:

• From each node, we can potentially explore multiple paths (up to degree of the node)
• We can revisit nodes multiple times as long as the time constraint allows
• The maximum depth of recursion is bounded by maxTime/min(time_j) where min(time_j) is the minimum edge weight
• At each level, we might branch to multiple neighbors, and since we can revisit nodes, the branching factor can be up to 4 in dense graphs (considering bidirectional edges)

Space Complexity: O(n + m + maxTime/min(time_j))

The space complexity consists of:

• O(n + m) for storing the graph adjacency list
• O(n) for the visited array
• O(maxTime/min(time_j)) for the recursion stack depth in the worst case, which represents the maximum number of edges we can traverse within the time limit

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Incorrectly Handling the Starting Node's Value

The Problem: A common mistake is forgetting to include node 0's value in the initial quality calculation or accidentally double-counting it. Some implementations might start with current_value = 0 instead of current_value = values[0], which would miss counting the starting node's value entirely.

Why It Happens: Since we mark visited[0] = True before starting the DFS, and we only add values for unvisited nodes during traversal, node 0's value would never be added if we don't initialize it properly.

The Fix: Always initialize the DFS with values[0] as the starting quality:

# Correct initialization
dfs(0, 0, values[0])  # Start with node 0's value included

# Incorrect initialization that misses node 0's value
# dfs(0, 0, 0)  # This would miss counting values[0]

Pitfall 2: Not Properly Backtracking the Visited State

The Problem: Forgetting to reset visited[neighbor] = False after exploring a path leads to incorrect results. Without proper backtracking, once a node is visited in any path, it would remain marked as visited for all subsequent path explorations, preventing other valid paths from counting that node's value.

Why It Happens: The backtracking step can be easily overlooked, especially when focusing on the forward exploration logic. The algorithm explores multiple paths that might share common nodes, and each path should independently track which nodes it has visited.

The Fix: Always reset the visited state after exploring a branch:

if not visited[neighbor]:
    visited[neighbor] = True
    dfs(neighbor, time_spent + travel_time, current_value + values[neighbor])
    visited[neighbor] = False  # Critical: Must backtrack!

Example of the Bug: Consider a graph where node 0 connects to nodes 1 and 2, and both 1 and 2 connect back to 0. Without proper backtracking:

• Path 1: 0 → 1 → 0 would mark node 1 as visited
• Path 2: 0 → 2 → 1 → 0 would not be able to count node 1's value because it's still marked as visited from Path 1

This would result in a lower quality score than the correct implementation.