2467. Most Profitable Path in a Tree
Description
There is an undirected tree with n
nodes labeled from 0
to n - 1
, rooted at node 0
. You are given a 2D integer array edges
of length n - 1
where edges[i] = [ai, bi]
indicates that there is an edge between nodes ai
and bi
in the tree.
At every node i
, there is a gate. You are also given an array of even integers amount
, where amount[i]
represents:
- the price needed to open the gate at node
i
, ifamount[i]
is negative, or, - the cash reward obtained on opening the gate at node
i
, otherwise.
The game goes on as follows:
- Initially, Alice is at node
0
and Bob is at nodebob
. - At every second, Alice and Bob each move to an adjacent node. Alice moves towards some leaf node, while Bob moves towards node
0
. - For every node along their path, Alice and Bob either spend money to open the gate at that node, or accept the reward. Note that:
- If the gate is already open, no price will be required, nor will there be any cash reward.
- If Alice and Bob reach the node simultaneously, they share the price/reward for opening the gate there. In other words, if the price to open the gate is
c
, then both Alice and Bob payc / 2
each. Similarly, if the reward at the gate isc
, both of them receivec / 2
each.
- If Alice reaches a leaf node, she stops moving. Similarly, if Bob reaches node
0
, he stops moving. Note that these events are independent of each other.
Return the maximum net income Alice can have if she travels towards the optimal leaf node.
Example 1:
![](https://spcdn.pages.dev/leetcode/problems/2467.Most%20Profitable%20Path%20in%20a%20Tree/images/eg1.png)
Input: edges = [[0,1],[1,2],[1,3],[3,4]], bob = 3, amount = [-2,4,2,-4,6] Output: 6 Explanation: The above diagram represents the given tree. The game goes as follows: - Alice is initially on node 0, Bob on node 3. They open the gates of their respective nodes. Alice's net income is now -2. - Both Alice and Bob move to node 1. Since they reach here simultaneously, they open the gate together and share the reward. Alice's net income becomes -2 + (4 / 2) = 0. - Alice moves on to node 3. Since Bob already opened its gate, Alice's income remains unchanged. Bob moves on to node 0, and stops moving. - Alice moves on to node 4 and opens the gate there. Her net income becomes 0 + 6 = 6. Now, neither Alice nor Bob can make any further moves, and the game ends. It is not possible for Alice to get a higher net income.
Example 2:
![](https://spcdn.pages.dev/leetcode/problems/2467.Most%20Profitable%20Path%20in%20a%20Tree/images/eg2.png)
Input: edges = [[0,1]], bob = 1, amount = [-7280,2350] Output: -7280 Explanation: Alice follows the path 0->1 whereas Bob follows the path 1->0. Thus, Alice opens the gate at node 0 only. Hence, her net income is -7280.
Constraints:
2 <= n <= 105
edges.length == n - 1
edges[i].length == 2
0 <= ai, bi < n
ai != bi
edges
represents a valid tree.1 <= bob < n
amount.length == n
amount[i]
is an even integer in the range[-104, 104]
.
Solutions
Solution 1: Two DFS Traversals
According to the problem, we know that Bob’s moving path is fixed, that is, starting from node $bob$ and finally reaching node $0$. Therefore, we can first run a DFS to find out the time it takes for Bob to reach each node, which we record in the array $ts$.
Then we run another DFS to find the maximum score for each of Alice’s moving paths. We denote the time for Alice to reach node $i$ as $t$, and the current cumulative score as $v$. After Alice passes node $i$, the cumulative score has three cases:
- The time $t$ for Alice to reach node $i$ is the same as the time $ts[i]$ for Bob to reach node $i$. In this case, Alice and Bob open the door at node $i$ at the same time, and the score Alice gets is $v + \frac{amount[i]}{2}$.
- The time $t$ for Alice to reach node $i$ is less than the time $ts[i]$ for Bob to reach node $i$. In this case, Alice opens the door at node $i$, and the score Alice gets is $v + amount[i]$.
- The time $t$ for Alice to reach node $i$ is greater than the time $ts[i]$ for Bob to reach node $i$. In this case, Alice does not open the door at node $i$, and the score Alice gets is $v$, which remains unchanged.
When Alice reaches a leaf node, update the maximum score.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes.
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