2581. Count Number of Possible Root Nodes

Problem Description

Alice has an undirected tree consisting of n nodes, labeled from 0 to n - 1. This tree is represented by a 2D integer array named edges, where each element edges[i] consists of two nodes a[i] and b[i] that are connected by an edge. Bob is tasked with finding the root of the tree but only knows the edges.

To find the root, Bob can make several guesses. Each guess is an assumption about who the parent of a particular node is. These guesses are represented by a 2D integer array guesses, with each guesses[j] containing two elements: u[j] and v[j], where Bob guesses that u[j] is the parent of v[j]. Alice, instead of confirming each guess, only tells Bob that at least k of his guesses are true.

The goal is to find the number of possible nodes that can be the root of the tree based on Bob’s guesses and Alice's confirmation that at least k of these guesses are true. If no tree can satisfy the condition given the guesses and value of k, the answer should be 0.

Intuition

To solve this problem, we need to consider each node as a potential root and verify if at least k guesses are true when that node is the root. The core idea is to use Depth-First Search (DFS) to explore each node starting from an arbitrary node (in this case, node 0) and count the number of correct guesses associated with the path from the current node to all its descendants. This number is stored in a variable cnt.

We start the process by considering node 0 as the root and use DFS (dfs1) to calculate the count cnt of correct guesses reachable from node 0. After we have the initial count of correct guesses for the subtree with the root 0, we need to examine whether other nodes can also serve as the root while maintaining at least k correct guesses (dfs2).

To achieve this, as we traverse each node, we adjust the count cnt of correct guesses by:

• Subtracting 1 from cnt if the current edge is in Bob's guesses but is not in the correct direction for the new potential root.
• Adding 1 to cnt if the edge connecting the current node to its parent is a correct guess in the reversed direction for the current subtree's root.

After the adjustment, if cnt is greater than or equal to k, it means the current node could be a root, so we count that as a possibility. We perform this process for every node by changing the root of the subtree during the DFS traversal, allowing us to identify all possible root nodes that satisfy Alice's condition of at least k correct guesses.

The provided solution uses a hash map gs to represent Bob's guesses for quick lookup during the count adjustment, and the final answer ans is incremented each time a valid root is encountered. DFS ensures that all nodes are visited, and potential roots are evaluated in a single pass through the tree.

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

Solution Approach

The solution relies on a tree Depth-First Search (DFS) algorithm to walk through the tree and count the number of guesses that are correct. There are two DFS methods implemented here, dfs1 and dfs2, both working together to figure out all possible nodes that can be the root.

• Initially, we convert the edges list to an adjacency list g which maps each node to its connected nodes (children and parents), enabling easy traversal. This is a common pattern when working with graph-based structures such as trees.

• "Tree DP" mentioned in the reference solution approach refers to Tree Dynamic Programming, which is used to cleverly cache results that can be reused to avoid redundant calculations.

• A hash map gs (Counter object in Python) is created from the guesses to quickly check if a guess (a directed edge) exists and to maintain counts efficiently. The key in gs is a tuple (u, v), representing a guess where u is assumed to be the parent of v.

• dfs1 is used initially to set the baseline count cnt of correct guesses assuming the root is node 0. As dfs1 traverses the tree, it increments cnt if it encounters an edge in gs.

• After establishing the baseline cnt, dfs2 traverses the tree again. As it moves from a node i to an adjacent node j, it adjusts cnt to reflect the number of correct guesses if node j were the root. Essentially, it changes the perspective of the root for each subtree, recalculating the guesses accordingly. The rule is:

  • cnt -= gs[(i, j)]: If the "parent-to-child" guess (i, j) was previously counted as correct for node i, it must be subtracted when considering node j as the root since (i, j) would no longer represent a "parent-to-child" relation.
  • cnt += gs[(j, i)]: Conversely, if the "child-to-parent" guess (j, i) exists, it is now a correct "parent-to-child" relation in the context of node j being the root, hence cnt is incremented.

• Following the adjustment, if cnt is greater than or equal to k, it indicates that node j can be a possible root. We use the ans variable to keep track of the number of nodes that meet this condition.

• This process is repeated for each node by recursively calling dfs2 as the traversal moves through the tree.

Using DFS and these adjustments, we can cover all nodes and their subtrees, ensuring that the potential roots are counted without redundant recalculations. The hash map gs ensures quick adjustments to the cnt as the DFS progresses, avoiding the need to recount guesses from scratch. The final value of ans will be the total number of possible roots found, which is returned as the solution.

Example Walkthrough

Let's take a small example to illustrate the solution approach. Suppose we have a tree with n = 4 nodes and the tree structure is defined by the following edges:

1edges = [[0, 1], [0, 2], [2, 3]]

This means node 0 is connected to nodes 1 and 2, and node 2 is connected to node 3. Now, let's imagine Bob makes the following two guesses:

1guesses = [[0, 2], [2, 3]]

Here, Bob guesses that node 0 is the parent of node 2, and node 2 is the parent of node 3. Alice tells him that at least k=1 of his guesses are true.

According to the solution approach, we would first build an adjacency list from the edges:

1g = {
2  0: [1, 2],
3  1: [0],
4  2: [0, 3],
5  3: [2]
6}

Then, we create a hash map gs from Bob's guesses:

1gs = {
2  (0, 2): 1,
3  (2, 3): 1
4}

We begin DFS with dfs1 starting from node 0, which we arbitrarily assume to be the root:

• From 0, we explore its children 1 and 2. The guess (0, 2) exists, so we increment cnt to 1.
• There are no guesses involving node 1, so visiting that node does not change cnt.
• From node 2, we visit 3, and since the guess (2, 3) is present, we increment cnt to 2.

Now we know that there are 2 correct guesses in the subtree with 0 as the root, which is more than k=1. Therefore, node 0 is a potential root.

Next, we execute dfs2 to check other nodes:

• While exploring from node 2, we consider it as a potential root. We adjust cnt:
  • We subtract 1 for the edge (0, 2) because if 2 were the root, then (0, 2) would be wrong, so cnt becomes 1.
  • We add 1 for the edge (2, 3) which was already counted before. Therefore, cnt stays at 1.

Since cnt is still 1, node 2 could also be a root.

• For nodes 1 and 3, there are no incoming guesses, so current cnt will not satisfy k=1 when considering them as roots.

Therefore, the possible node roots based on the minimum k=1 correct guess from Bob would be [0, 2]. The final answer ans would be 2 as there are two possible nodes that could be roots.

Solution Implementation

Time and Space Complexity

The time complexity of the algorithm primarily consists of two depth-first search (DFS) operations, dfs1 and dfs2. The dfs1 function is called once for each vertex to calculate the initial count of the guesses in cnt. The dfs2 function is called recursively to traverse all vertices in the graph, incrementing or decrementing cnt as necessary and updating ans.

• Both dfs1 and dfs2 visit each vertex once and each edge twice (once from each vertex it connects). Since there are n nodes and each node is connected by one edge, in other words there are n - 1 edges (for a tree). Therefore, each DFS will take O(n) time, totaling O(2n) which is simplified to O(n) for the whole process of DFS.
• The construction of graph g and guesses gs will be O(n + m), where n is the length of edges and m is the length of guesses. Each edge and guess is processed only once.

As a result, the overall time complexity is O(n + m), since we have to consider time for both DFS operations and the initial construction of graph structures.

For the space complexity, the algorithm uses additional data structures that store graph information and the counts of guesses:

• Graph g stores adjacency lists, which in total will have 2 * (n - 1) entries (each edge is stored twice). This equates to O(n).
• Guesses gs are stored as a Counter (a type of dictionary in Python), which in the worst case, will have m unique entries corresponding to each unique guess. This equates to O(m).

Therefore, together with the recursion call stack (which in the worst case, can go as deep as n for a skewed tree), the overall space complexity would be O(n + m), accounting for storing the graph g, the Counter gs, and the depth of the recursion stack.

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