Introduction

In this blog post, we will explore how to solve the LeetCode problem "2509" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer n. There is a complete binary tree with 2n - 1 nodes. The root of that tree is the node with the value 1, and every node with a value val in the range [1, 2n - 1 - 1] has two children where: The left node has the value 2 val, and The right node has the value 2 val + 1. You are also given a 2D integer array queries of length m, where queries[i] = [ai, bi]. For each query, solve the following problem: Add an edge between the nodes with values ai and bi. Find the length of the cycle in the graph. Remove the added edge between nodes with values ai and bi. Note that: A cycle is a path that starts and ends at the same node, and each edge in the path is visited only once. The length of a cycle is the number of edges visited in the cycle. There could be multiple edges between two nodes in the tree after adding the edge of the query. Return an array answer of length m where answer[i] is the answer to the ith query. Example 1: Input: n = 3, queries = [[5,3],[4,7],[2,3]] Output: [4,5,3] Explanation: The diagrams above show the tree of 23 - 1 nodes. Nodes colored in red describe the nodes in the cycle after adding the edge. - After adding the edge between nodes 3 and 5, the graph contains a cycle of nodes [5,2,1,3]. Thus answer to the first query is 4. We delete the added edge and process the next query. - After adding the edge between nodes 4 and 7, the graph contains a cycle of nodes [4,2,1,3,7]. Thus answer to the second query is 5. We delete the added edge and process the next query. - After adding the edge between nodes 2 and 3, the graph contains a cycle of nodes [2,1,3]. Thus answer to the third query is 3. We delete the added edge. Example 2: Input: n = 2, queries = [[1,2]] Output: [2] Explanation: The diagram above shows the tree of 22 - 1 nodes. Nodes colored in red describe the nodes in the cycle after adding the edge. - After adding the edge between nodes 1 and 2, the graph contains a cycle of nodes [2,1]. Thus answer for the first query is 2. We delete the added edge. Constraints: 2 <= n <= 30 m == queries.length 1 <= m <= 105 queries[i].length == 2 1 <= ai, bi <= 2n - 1 ai != bi

Explanation

Here's the approach to solve this problem:

• Find the Lowest Common Ancestor (LCA): For each query (a, b), the cycle formed involves the path from 'a' to the root, the path from 'b' to the root, and the edge (a, b). The length of the cycle is the sum of the lengths of these paths. The key is to efficiently find the LCA of 'a' and 'b'. The paths from a and b to root will overlap from the LCA node.
• Calculate Path Lengths: Determine the number of edges from 'a' to the LCA and from 'b' to the LCA. This can be done by counting the number of steps it takes to reach the LCA from each node.

• Cycle Length: The length of the cycle is then the path length from a to LCA + path length from b to LCA + 1 (the edge between a and b).

• Time Complexity: O(m log(2ⁿ)), where m is the number of queries and n constrains the max node number. The log factor comes from finding the path to the root in the binary tree for each query and finding the LCA. *Space Complexity: O(1)

Code

    def solve():
    n = int(input())
    queries = []
    num_queries = int(input())  # Read the number of queries
    for _ in range(num_queries):
        queries.append(list(map(int, input().split())))

    def get_parent(node):
        if node == 1:
            return 0  # No parent
        return node // 2

    def get_path_to_root(node):
        path = []
        curr = node
        while curr != 0:
            path.append(curr)
            curr = get_parent(curr)
        return path[::-1]

    def find_lca(a, b):
        path_a = get_path_to_root(a)
        path_b = get_path_to_root(b)

        lca = 1  # Initialize LCA to root
        i = 0
        while i < min(len(path_a), len(path_b)) and path_a[i] == path_b[i]:
            lca = path_a[i]
            i += 1
        return lca

    def path_length(start, end):  # nodes not indices
        length = 0
        curr = start
        path = get_path_to_root(start)
        lca = end

        path_to_lca = get_path_to_root(lca)

        while curr != lca:
           curr = get_parent(curr)
           length +=1

        return length

    results = []
    for a, b in queries:
        lca = find_lca(a, b)
        len_a_to_lca = path_length(a, lca)
        len_b_to_lca = path_length(b, lca)
        cycle_length = len_a_to_lca + len_b_to_lca + 1
        results.append(cycle_length)

    print(results)

# Example Usage (for testing purposes)
# Input format:
# n
# num_queries
# a1 b1
# a2 b2
# ...
# an bn

# Example 1 Input:
# 3
# 3
# 5 3
# 4 7
# 2 3
# Output: [4, 5, 3]

# Example 2 Input:
# 2
# 1
# 1 2
# Output: [2]

# Read input and execute if not running in an online environment
if __name__ == "__main__":
    solve()