Problem Description

You have an undirected tree with n nodes labeled from 0 to n - 1. The tree structure is given by an array edges where edges[i] = [ai, bi] represents an edge between nodes ai and bi.

Each node has a price given in the array price, where price[i] is the price of node i.

You need to perform multiple trips on this tree. Each trip is defined in the array trips, where trips[i] = [starti, endi] means you travel from node starti to node endi along any path in the tree.

The cost of a trip is the sum of prices of all nodes on the path from start to end (including both endpoints).

Before starting any trips, you can select some nodes and halve their prices. However, there's a restriction: the nodes you select must be non-adjacent (no two selected nodes can be directly connected by an edge).

Your goal is to minimize the total cost of all trips combined. You need to strategically choose which nodes to halve in price (while respecting the non-adjacency constraint) to achieve the minimum possible total cost.

Return the minimum total price sum to perform all the given trips.

Example:

• If you have a path from node A to node B passing through nodes with prices [10, 20, 30], the trip cost would be 60.
• If you halve the price of the middle node (assuming it's not adjacent to any other halved node), the trip cost becomes 10 + 10 + 30 = 50.

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 states we have an undirected tree with n nodes and edges connecting them. A tree is a special type of graph.

Is it a tree?

• Yes: The problem clearly mentions "There exists an undirected and unrooted tree" with n nodes and n-1 edges, which is the definition of a tree.

DFS

• Following the "yes" path from the tree node leads us directly to DFS.

Conclusion: The flowchart suggests using DFS (Depth-First Search) for this problem.

Why DFS is Appropriate

The DFS pattern fits perfectly for this problem because:

1. Path Finding in Trees: We need to find paths between pairs of nodes (start and end of each trip). DFS is ideal for traversing paths in a tree structure.

2. Counting Node Usage: We need to count how many times each node appears across all trip paths. DFS allows us to traverse each required path and accumulate the count for nodes visited.

3. Tree DP with Non-Adjacent Constraint: After counting node frequencies, we need to solve an optimization problem with a non-adjacency constraint (similar to the "House Robber on a Tree" pattern). This requires another DFS to compute two states for each node:

   • The minimum cost if we halve the current node's price
   • The minimum cost if we don't halve the current node's price

The solution uses two DFS passes:

• First DFS (dfs): Finds paths for each trip and counts node frequencies
• Second DFS (dfs2): Performs tree dynamic programming to minimize the total cost while respecting the non-adjacency constraint

Intuition

The key insight is to break this problem into two subproblems:

First, figure out which nodes matter and how much. Since we're performing multiple trips, some nodes will be visited more frequently than others. A node that appears in many trip paths contributes more to the total cost. So we need to count how many times each node is used across all trips. For each trip from start to end, we traverse the unique path in the tree and increment a counter for every node on that path.

Second, optimize which nodes to halve. Once we know the frequency of each node, the problem becomes: "Given that node i contributes frequency[i] * price[i] to the total cost, which nodes should we halve to minimize the total, with the constraint that no two adjacent nodes can both be halved?"

This second part is a classic dynamic programming pattern on trees, similar to the "House Robber III" problem. For each node, we have two choices:

• Halve this node's price: We save frequency[i] * price[i] / 2, but we cannot halve any of its neighbors
• Don't halve this node's price: We pay the full frequency[i] * price[i], but we're free to halve or not halve its neighbors

The beauty of this approach is that we can solve it recursively on the tree structure. For each node, we calculate:

• a: minimum cost for the subtree rooted at this node if we DON'T halve this node
• b: minimum cost for the subtree rooted at this node if we DO halve this node

When we don't halve the current node (a), we can choose the minimum between halving or not halving each child. When we halve the current node (b), we're forced to not halve any direct children, so we must use their a values.

The final answer is the minimum of these two options at the root node.

Learn more about Tree, Depth-First Search, Graph and Dynamic Programming patterns.

Solution Approach

The implementation consists of two main DFS functions that solve our two subproblems:

Step 1: Count node frequencies using DFS path finding

The first DFS function dfs(i, fa, k) finds the path from node i to target node k, where fa is the parent (to avoid revisiting). The clever part is:

• We increment cnt[i] when we visit node i
• We recursively search all neighbors (except parent) to find the target
• If the target is found in any subtree, we return True to indicate this node is on the path
• If the target is NOT found (returns False), we backtrack by decrementing cnt[i]

This way, only nodes that are actually on the path from start to end keep their count incremented.

def dfs(i: int, fa: int, k: int) -> bool:
    cnt[i] += 1  # Assume this node is on the path
    if i == k:   # Found the target
        return True
    ok = any(j != fa and dfs(j, i, k) for j in g[i])  # Search children
    if not ok:
        cnt[i] -= 1  # Not on path, backtrack
    return ok

Step 2: Tree DP to minimize cost with non-adjacent constraint

The second DFS function dfs2(i, fa) returns a tuple (a, b) where:

• a = minimum cost if we DON'T halve node i
• b = minimum cost if we DO halve node i

The base cost for node i is cnt[i] * price[i] (frequency × price).

For each child j:

• If we don't halve current node (a): we can choose min(x, y) from child (either halve child or not)
• If we halve current node (b): we must take x from child (cannot halve adjacent nodes)

def dfs2(i: int, fa: int) -> (int, int):
    a = cnt[i] * price[i]  # Cost without halving
    b = a // 2             # Cost with halving
    for j in g[i]:
        if j != fa:
            x, y = dfs2(j, i)  # Get child's costs
            a += min(x, y)     # Can choose either for child
            b += x             # Must not halve child
    return a, b

Main Algorithm Flow:

1. Build adjacency list g from edges
2. For each trip [start, end], run dfs(start, -1, end) to accumulate node frequencies
3. Run dfs2(0, -1) from any node (we use 0) to compute the minimum cost
4. Return min(dfs2(0, -1)) which gives us the minimum between halving or not halving the root

The time complexity is O(n × t) where t is the number of trips (for counting paths) plus O(n) for the DP traversal.

Example Walkthrough

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

Example Setup:

• Tree with 4 nodes (0-3) and edges: [[0,1], [1,2], [1,3]]
• Tree structure:

    0
    |
    1
   / \
  2   3

• Prices: [2, 2, 10, 6] (node 0 costs 2, node 1 costs 2, node 2 costs 10, node 3 costs 6)
• Trips: [[0,3], [2,1], [2,3]]

Step 1: Count node frequencies

For trip [0,3] (path: 0→1→3):

• Visit node 0: cnt[0] = 1
• Visit node 1: cnt[1] = 1
• Visit node 3: cnt[3] = 1
• Nodes on path keep their counts

For trip [2,1] (path: 2→1):

• Visit node 2: cnt[2] = 1
• Visit node 1: cnt[1] = 2 (incremented again)

For trip [2,3] (path: 2→1→3):

• Visit node 2: cnt[2] = 2
• Visit node 1: cnt[1] = 3
• Visit node 3: cnt[3] = 2

Final frequencies: cnt = [1, 3, 2, 2]

Step 2: Calculate contribution of each node

• Node 0: 1 × 2 = 2
• Node 1: 3 × 2 = 6
• Node 2: 2 × 10 = 20
• Node 3: 2 × 6 = 12
• Total without halving: 40

Step 3: Tree DP to find optimal halving

Starting from node 0 and recursing:

For node 2 (leaf):

• a (don't halve): 20
• b (halve): 10
• Return (20, 10)

For node 3 (leaf):

• a (don't halve): 12
• b (halve): 6
• Return (12, 6)

For node 1:

• Base: a = 6, b = 3
• From child 2: (20, 10)
  • If don't halve 1: add min(20, 10) = 10 → a = 16
  • If halve 1: must add 20 → b = 23
• From child 3: (12, 6)
  • If don't halve 1: add min(12, 6) = 6 → a = 22
  • If halve 1: must add 12 → b = 35
• Return (22, 35)

For node 0 (root):

• Base: a = 2, b = 1
• From child 1: (22, 35)
  • If don't halve 0: add min(22, 35) = 22 → a = 24
  • If halve 0: must add 22 → b = 23
• Return (24, 23)

Final Answer: min(24, 23) = 23

The optimal strategy is to halve node 0 (saving 1) and nodes 2 and 3 (saving 10 and 6), but we can't halve both 1 and its neighbors. The algorithm finds that halving node 0 and keeping the flexibility for the subtree gives us the minimum cost of 23.

Solution Implementation

Time and Space Complexity

The time complexity is O(m × n + n), where m is the number of trips and n is the number of nodes in the tree.

Breaking down the analysis:

• Building the adjacency list from edges takes O(n) time since there are n-1 edges in a tree
• For each trip, the dfs function traverses a path from start to end node, which in the worst case visits all n nodes (when the path spans the entire tree)
• With m trips, the total time for all DFS traversals is O(m × n)
• The dfs2 function performs one traversal of the entire tree, visiting each node exactly once, taking O(n) time
• Overall: O(n) + O(m × n) + O(n) = O(m × n)

The space complexity is O(n).

Breaking down the analysis:

• The adjacency list g stores all edges, requiring O(n) space since there are n-1 edges in a tree
• The cnt Counter stores at most n entries (one for each node), using O(n) space
• The recursion stack depth for both DFS functions is at most O(n) in the worst case (when the tree is a linear chain)
• Overall: O(n) + O(n) + O(n) = O(n)

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

Common Pitfalls

1. Incorrect Path Counting with Multiple DFS Calls

A critical pitfall occurs when developers try to optimize the path-finding DFS by using memoization or by not properly resetting the visit counts between different trips. This can lead to incorrect frequency counts.

Wrong Approach:

def find_path(current, parent, target, path):
    path.append(current)
    if current == target:
        # Update counts for all nodes in path
        for node in path:
            visit_count[node] += 1
        return True
  
    for neighbor in graph[current]:
        if neighbor != parent:
            if find_path(neighbor, current, target, path):
                path.pop()  # Forgetting to pop before returning
                return True
  
    path.pop()
    return False

Issue: The path list manipulation can become error-prone, especially with early returns. Forgetting to clean up the path in all cases leads to incorrect counts.

Solution: Use the backtracking approach shown in the correct implementation where counts are incremented optimistically and decremented if the path doesn't lead to the target.

2. Misunderstanding the Non-Adjacent Constraint

Developers often misinterpret the constraint and think they need to find a maximum independent set first, then apply it globally.

Wrong Approach:

# Find maximum independent set first
selected_nodes = find_max_independent_set(graph)
# Then halve prices of selected nodes
for node in selected_nodes:
    price[node] //= 2
# Calculate total cost with fixed prices

Issue: The optimal set of nodes to halve depends on the visit frequencies. A node with high price but zero visits shouldn't be prioritized for halving.

Solution: Use dynamic programming that considers both the visit frequency and the non-adjacency constraint simultaneously, as shown in the calculate_min_price function.

3. Incorrect DP State Transition

A subtle but common mistake is incorrectly handling the DP transitions when a node is halved.

Wrong Approach:

def calculate_min_price(node, parent):
    price_not_halved = visit_count[node] * price[node]
    price_halved = visit_count[node] * (price[node] // 2)  # Wrong calculation
  
    for child in graph[node]:
        if child != parent:
            child_not_halved, child_halved = calculate_min_price(child, node)
            price_not_halved += min(child_not_halved, child_halved)
            price_halved += min(child_not_halved, child_halved)  # Wrong! Should only use child_not_halved
  
    return price_not_halved, price_halved

Issue: When the current node is halved, adjacent children CANNOT be halved due to the non-adjacency constraint. Using min(child_not_halved, child_halved) violates this constraint.

Solution: When the current node is halved, always use child_not_halved for children:

price_halved += child_not_halved  # Correct - enforces non-adjacency

4. Tree Root Selection Assumption

Some developers assume the tree has a specific root or that node 0 must be the root, leading to incorrect traversals.

Wrong Approach:

# Assuming node with degree 1 is always the root
root = -1
for i in range(n):
    if len(graph[i]) == 1:
        root = i
        break
return min(calculate_min_price(root, -1))

Issue: In an undirected tree, any node can serve as the root for the DP calculation. The choice doesn't affect the final result.

Solution: Simply use any node (like node 0) as the root, as the DP correctly handles the tree structure regardless of root choice.