LeetCode 3486: Longest Special Path II Solution

Master LeetCode problem 3486 (Longest Special Path II), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3486. Longest Special Path II

Hard

Array Hash Table Tree Depth-First Search Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can perform a depth-first search (DFS) on the tree starting from node 0. During the DFS, we keep track of the longest special path found so far and the minimum number of nodes in all such paths. We need to handle the special case where a node value may appear twice in the path.

Algorithm

Build an adjacency list representation of the tree using the given edges array.
Perform a DFS starting from node 0.
During the DFS, keep track of the following for each node:
- the length of the longest special path ending at that node.
- the number of nodes in the longest special path ending at that node.
- the number of distinct values in the path ending at that node.
Update the global result based on the above information at each node.

Time Complexity

The time complexity of this solution is O(n), where n is the number of nodes in the tree.

Space Complexity

The space complexity of this solution is O(n) for storing the adjacency list and the visited array.

Solution Code

class Solution {
    int maxLength = 0;
    int minNodes = Integer.MAX_VALUE;
    List<Integer>[] adjList;
    int[] nums;

    public int[] countLongestSpecialPath(int[][] edges, int[] nums) {
        int n = nums.length;
        adjList = new List[n];
        this.nums = nums;

        for (int i = 0; i < n; i++) {
            adjList[i] = new ArrayList<>();
        }

        for (int[] edge : edges) {
            adjList[edge[0]].add(edge[1]);
            adjList[edge[1]].add(edge[0]);
        }

        dfs(0, -1, new HashSet<>(), new HashSet<>(), 0, 0);

        return new int[]{maxLength, minNodes};
    }

    private void dfs(int node, int parent, Set<Integer> visited, Set<Integer> values, int length, int nodes) {
        visited.add(node);
        if (nums[node] == 0 || values.contains(nums[node])) {
            values.clear();
            length = 0;
            nodes = 0;
        }
        values.add(nums[node]);
        length++;
        nodes++;
        maxLength = Math.max(maxLength, length);
        if (length == maxLength) {
            minNodes = Math.min(minNodes, nodes);
        }

        for (int neighbor : adjList[node]) {
            if (neighbor != parent && !visited.contains(neighbor)) {
                dfs(neighbor, node, new HashSet<>(visited), new HashSet<>(values), length, nodes);
            }
        }
    }
}

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