Problem Statement
In this problem, you are given a tree structure, which is a type of graph that remains connected and acyclic. This tree is described by two items: an array called parent and a string s. The parent array has n elements, indexed from 0 to n-1, where each value represents the parent of that node, and the root node (0) has a parent value of -1. The string s also has n characters, where each character at s[i] is assigned to node i in the graph.
Your challenge is to determine the length of the longest path within the tree where no two adjacent nodes on this path have the same character assigned to them. The crux of the problem lies in traversing the tree in such a way that you take into account the character constraints, all the while aiming to maximize the length of the path.
Examples
Example 1
Input:
parent = [-1,0,0,1,1,2], s = "abacbe"
Output:
3
Explanation:
The longest path where each two adjacent nodes have different characters in the tree is the path: 0 -> 1 -> 3. The length of this path is 3, so 3 is returned. It can be proven that there is no longer path that satisfies the conditions.
Example 2
Input:
parent = [-1,0,0,0], s = "aabc"
Output:
3
Explanation:
The longest path where each two adjacent nodes have different characters is the path: 2 -> 0 -> 3. The length of this path is 3, so 3 is returned.
Constraints
n == parent.length == s.length1 <= n <= 1050 <= parent[i] <= n - 1for alli >= 1parent[0] == -1parentrepresents a valid tree.sconsists of only lowercase English letters.
Approach and Intuition
The main goal of the algorithm is to find the longest path such that each consecutive node on that pathway has different characters.
Building the Tree Representation:
- First, convert the given
parentarray to a more accessible tree format, typically using an adjacency list. This allows easy traversal using techniques like Depth-First Search (DFS) or Breadth-First Search (BFS).
- First, convert the given
Traversing with Character Check:
- Implement DFS to explore the depth of different branches of the tree. Start from the root of the tree (node
0), and for each subsequent node, check if the character differs from its parent node.
- Implement DFS to explore the depth of different branches of the tree. Start from the root of the tree (node
Maintaining Maximum Path Length:
- While you perform DFS, keep track of the path length. Only increment the path length when you travel to a node with a character that differs from the character of its parent node.
Edge Cases:
- Consider scenarios where every node might have the same character or where nodes might form many shallow branches. These cases affect traversal and need careful consideration during path calculation.
The solution's efficiency will significantly depend on how it iterates over nodes and makes decisions to either continue the path or backtrack. This check is crucial for ensuring the desired property of differing adjacent characters on the path. Given the constraints, an optimal solution should ideally work within O(n) time complexity, leveraging direct adjacency list accesses and avoiding unnecessary re-computations.
Solutions
- C++
- Java
class Solution {
public:
int maxPath(vector<int>& parent, string s) {
int size = parent.size();
vector<int> childCounter(size);
for (int idx = 1; idx < size; idx++) {
childCounter[parent[idx]]++;
}
vector<vector<int>> pathLengths(size);
queue<int> leafNodes;
int maxPathLength = 1;
for (int idx = 0; idx < size; idx++) {
pathLengths[idx] = vector<int>(2);
if (childCounter[idx] == 0 && idx != 0) {
leafNodes.push(idx);
pathLengths[idx][0] = 1;
}
}
while (!leafNodes.empty()) {
int currentNode = leafNodes.front();
leafNodes.pop();
int parentNode = parent[currentNode];
int currentPathLength = pathLengths[currentNode][0];
if (s[currentNode] != s[parentNode]) {
if (currentPathLength > pathLengths[parentNode][0]) {
pathLengths[parentNode][1] = pathLengths[parentNode][0];
pathLengths[parentNode][0] = currentPathLength;
} else if (currentPathLength > pathLengths[parentNode][1]) {
pathLengths[parentNode][1] = currentPathLength;
}
}
maxPathLength = max(maxPathLength, pathLengths[parentNode][0] + pathLengths[parentNode][1] + 1);
childCounter[parentNode]--;
if (childCounter[parentNode] == 0 && parentNode != 0) {
pathLengths[parentNode][0]++;
leafNodes.push(parentNode);
}
}
return maxPathLength;
}
};
Explore the C++ solution to determine the longest path in a tree where adjacent characters (nodes) are different. The implementation accepts a parent-child relationship list and a string representing node characteristics. Here's a step-by-step overview of the approach:
Initialize child count for each node to determine tree leaves and internal nodes.
Maintain a queue for leaf nodes and discover paths from the leaves to the root. This ensures that the path calculation proceeds only when all children of a node are processed.
Store two longest distinct paths for each node using a 2D vector— this allows tracking both the longest and second longest paths that can be used for path extension.
For each leaf node:
- Process the node until the queue is empty.
- For each node, compare and dynamically store the longest path lengths based on current node path lengths and the parent's.
- If the current node and its parent have different characters, potentially update the parent's longest path lengths.
- Continuously adjust the recorded maximum path length.
Compute and adjust path lengths when moving from leaf to the root, ensuring the longest potential paths accounting for character differences are recorded.
The core of the solution involves handling and updating states from the bottom-up (from leaves towards the root) in the tree, leveraging queue data structure to process each node level-by-level. This methodology ensures the highest efficiency by only finalizing a node's path contribution once all its child contributions are known. The emphasis on distinct adjacent characters drives the conditional checks and updates during the path length calculations.
class Solution {
public int maxPathLength(int[] parent, String chars) {
int size = parent.length;
int[] childCounter = new int[size];
// Children calculation starts from the first index, ignoring the root.
for (int i = 1; i < size; i++) {
childCounter[parent[i]]++;
}
Queue<Integer> queue = new LinkedList<>();
int maxPath = 1;
int[][] maxChains = new int[size][2];
for (int i = 1; i < size; i++) {
// Initialize queue with leaf nodes
if (childCounter[i] == 0) {
maxChains[i][0] = 1;
queue.offer(i);
}
}
while (!queue.isEmpty()) {
int current = queue.poll();
int parentNode = parent[current];
int currentChainLength = maxChains[current][0];
if (chars.charAt(current) != chars.charAt(parentNode)) {
// Update the max chains for the parent node
if (currentChainLength > maxChains[parentNode][0]) {
maxChains[parentNode][1] = maxChains[parentNode][0];
maxChains[parentNode][0] = currentChainLength;
} else if (currentChainLength > maxChains[parentNode][1]) {
maxChains[parentNode][1] = currentChainLength;
}
}
// Calculate potential max path reaching up to this point
maxPath = Math.max(maxPath, maxChains[parentNode][0] + maxChains[parentNode][1] + 1);
childCounter[parentNode]--;
// If no more childs, add parent to queue
if (childCounter[parentNode] == 0 && parentNode != 0) {
maxChains[parentNode][0]++;
queue.offer(parentNode);
}
}
return maxPath;
}
}
The Java program defines a Solution class with the function maxPathLength(int[] parent, String chars). This function calculates the length of the longest path in a tree such that no two adjacent characters are the same. Here's a breakdown of the process:
- Initializes necessary data structures like arrays for child node count (
childCounter), a queue (queue) for processing nodes level by level, and a 2D array (maxChains) to track maximum chain lengths. - Iterates over the given
parentarray to count the number of children for each node starting from the root's children. - Populates the queue with leaf nodes (nodes having zero children) and initializes their chain lengths in
maxChains. - Processes each node from the queue:
- Retrieves and updates chain information of the node's parent based on the character comparison between the node and its parent.
- Updates the global
maxPathvariable after adding the two largest chain lengths from the current node's parent. - Decreases the child count for the parent node and adds the parent to the queue if all its children have been processed and it is not the root.
- Continues until all nodes are processed and returns the computed maximum path length.
This approach ensures an efficient calculation of the longest path using BFS (Breadth-First Search) by systematically processing levels of the tree and selectively updating potential paths. Adjustments are made based on character comparisons, factoring in only nodes that contribute to forming paths with distinct adjacent characters.
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