LeetCode 2509: Cycle Length Queries in a Tree

Explanation

To solve this problem, we can simulate the process of adding an edge between two nodes in the tree, finding the cycle, and then removing the edge. We can traverse the tree to find the path from one node to another, then check if there are any common nodes in the path to determine the cycle length.

1. Build the tree structure based on the given n value.
2. For each query, add the edge between the nodes, find the path from one node to another, check for common nodes, and calculate the cycle length.
3. Remove the added edge and move to the next query.
4. Repeat the process for all queries and store the cycle lengths in the answer array.

Time complexity: O(m * n) where m is the number of queries and n is the given integer n. Space complexity: O(n) for storing the tree structure.

class Solution {
    public int[] cycleLengthQueries(int n, int[][] queries) {
        int[] answer = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int a = queries[i][0];
            int b = queries[i][1];
            answer[i] = findCycleLength(n, a, b);
        }
        return answer;
    }

    private int findCycleLength(int n, int a, int b) {
        // Implementation of finding cycle length
    }
}