2246. Longest Path With Different Adjacent Characters

Problem Description

In this LeetCode problem, we are given a special type of graph called a tree. This tree has n nodes, each numbered from 0 to n - 1, and node 0 serves as the root node. The relationships between nodes and their parents are represented by an array parent where parent[i] indicates the parent node of node i. By definition, since node 0 is the root, it has no parent, which is denoted by parent[0] == -1.

Along with the tree structure, each node i is assigned a character given by the string s[i]. The goal is to identify the longest path in the tree where adjacent nodes on the path have different characters. The length of the path is defined as the number of nodes in that path.

The problem is ultimately asking to find the maximum length path in the tree that meets the criteria of having no two consecutive nodes with the same character assigned to them.

Intuition

To solve the problem, one can take a recursive approach, which is commonly used to traverse trees. The intuition here is to use Depth-First Search (DFS) to explore the tree from the root node, tracking the longest path that meets the condition along the way.

The recursive DFS function explores each child of the current node and determines the maximum path length within the subtrees of those children. While traversing, it keeps track of the length of the longest path ending at the current node (mx) and updates the global answer (ans) if a longer path is found.

The critical insight is that, if the current node and its child have different characters, the path can be extended by one. We compare the length of the longest path through each child node and pick the two longest paths to possibly update our answer. The trick is to add the path lengths of two longest non-similar child paths to the global answer. After completing the DFS traversal, the final answer is adjusted by adding one to account for the length of a single node path.

The algorithm essentially constructs a graph from the parent array to track the children of each node and performs DFS starting from the root. During DFS, it checks the characters assigned to the nodes and determines the longest path where adjacent nodes have different characters.

Learn more about Tree, Depth-First Search, Graph and Topological Sort patterns.

Solution Approach

The solution approach involves depth-first search (DFS), a common technique for exploring all the nodes in a tree from a given starting point. The implementation uses a helper function dfs(i) that is designed to recursively travel through the tree starting from node i.

Here's a step-by-step explanation of how the code achieves this:

1. A dictionary type defaultdict(list) called g is created to store the graph representation of the tree, with each key corresponding to a parent node and its value being the list of child nodes.

2. The graph g is populated by iterating over the indices of the parent list, starting from index 1 (since index 0 is the root and has no parent), and appending each index i to the list g[parent[i]]. This effectively builds the adjacency list for each node.

3. A dfs(i) function is defined to use recursive DFS traversal through the tree starting from node i. The function returns the maximum length of the path ending at node i that meets the non-adjacent-character condition.

4. Within dfs(i), we loop through each child j of the current node i by accessing g[i], and recursively call dfs(j) to continue the exploration further down the tree. The returned value from the recursive call represents the length of the maximum path from child j to a leaf that satisfies the condition, plus one (accounting for the edge to node i).

5. The function checks if the characters at the current node and its child node are different using s[i] != s[j]. If so, the ans variable (which is tracking the global maximum length) is updated with the sum of mx (the current maximum path length ending at node i) and x (the path length from the child node j), since this forms a valid path with distinct adjacent characters.

6. The path length x is compared with mx and updates mx if it's longer, ensuring mx always contains the length of the longest path that can be extended from the current node i.

After defining the dfs(i) function and initializing the adjacency list, the DFS traversal is kicked off at the root, dfs(0). Since dfs only counts the length of the path without including the starting node, ans + 1 is returned to account for the root node itself, giving the final answer.

Key Points:

• Recursion: The DFS algorithm is implemented using recursion, a natural fit for exploring trees.
• Graph Representation: Even though the tree is initially represented as a parent array, it's converted into a graph using adjacency lists for easier traversal.
• Non-local Variable: The nonlocal keyword is used for variable ans to allow its modification within the nested dfs function.
• Max Tracking: Two local maximum path lengths (mx and x) are used to keep track of the paths and to update the global maximum length ans.

Using DFS and careful updates to the maximum path lengths allow for an efficient search through the tree, yielding the longest path with the required property.

Example Walkthrough

Let's take a small example to illustrate the solution approach:

Suppose we have a tree with n = 5 nodes and the following parent relationship array: parent = [-1, 0, 0, 1, 1]. This means that node 0 is the root, node 1 and node 2 are children of node 0, and nodes 3 and 4 are children of node 1. The characters assigned to each node are represented by the string s = "ababa".

Now we will walk through the steps of the DFS solution to find the longest path with different adjacent characters:

1. First, we'll create the graph g from the parent array, which will look like this:

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

2. We define the dfs(i) function to start the DFS traversal. We will also initialize ans = 0 to store the length of the longest path.

3. We start the DFS from the root node dfs(0):

   • Since node 0 has two children (1 and 2), we call dfs(1) and dfs(2).

4. In the dfs(1) call:

   • Node 1 has children 3 and 4. We call dfs(3) and dfs(4) respectively.
   • The character at node 1 is b, and at node 3 it's a. Since they are different, ans can be updated to 2 if dfs(3) returns 1.
   • Similarly, dfs(4) returns 1 because node 4's character is similar to 1, making the path length 1. Now ans would be updated to 3 because mx + x = 2 + 1 (path through node 1 to node 3 and then from node 1 to node 4).

5. For dfs(2), since node 2 has no child and its character is a which is different from the root's character a, dfs(2) will simply return 1.

6. As we return back to dfs(0), we check the character at node 0, which is a, and compare it with its children's characters. Node 2 has the same character, so we cannot form a longer path through node 2. The longest path at this point is from node 0 to node 1 to node 3, and node 1 to node 4.

7. After traversing all nodes, ans + 1 will give us the final answer. We add one because ans is tracking the number of edges in the longest path, and we want to count the number of nodes, which is one more than the number of edges.

In this particular example, the longest path with different adjacent characters has a length of 4: through the nodes 0 -> 1 -> 3 and 1 -> 4. Our ans was updated to 3 at most during the DFS traversal, and thus the final answer will be ans + 1 = 4.

Key Points of Clarification:

• While calculating the path lengths, we consider the length as the number of edges between nodes on the path.
• We need to return ans + 1 at the end of the traversal since the counting starts at 0 and the problem asks for the number of nodes in the path, not the number of edges.
• This example demonstrates how dfs helps in efficiently finding and updating the longest path in the tree with the desired property.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code is O(N), where N is the number of nodes in the input list parent. This complexity arises because the code visits every node exactly once through depth-first search (DFS). Each node processing (not counting the DFS recursion) takes constant time, leading to a linear time complexity relative to the number of nodes.

Space Complexity

The space complexity of the code is also O(N). This space is required for the adjacency list g and the call stack during the recursive DFS calls. Each node can contribute at most one frame to the call stack (in the case of a linear tree), and the adjacency list can store up to 2(N - 1) edges (considering an undirected representation of the tree for understanding, although the actual directed edges are less and do not contribute to space more than O(N)). As a result, the overall space complexity remains linear with respect to N.

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