Introduction

In this blog post, we will explore how to solve the LeetCode problem "684" 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

In this problem, a tree is an undirected graph that is connected and has no cycles. You are given a graph that started as a tree with n nodes labeled from 1 to n, with one additional edge added. The added edge has two different vertices chosen from 1 to n, and was not an edge that already existed. The graph is represented as an array edges of length n where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the graph. Return an edge that can be removed so that the resulting graph is a tree of n nodes. If there are multiple answers, return the answer that occurs last in the input. Example 1: Input: edges = [[1,2],[1,3],[2,3]] Output: [2,3] Example 2: Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]] Output: [1,4] Constraints: n == edges.length 3 <= n <= 1000 edges[i].length == 2 1 <= ai < bi <= edges.length ai != bi There are no repeated edges. The given graph is connected.

Explanation

Here's a breakdown of the solution:

• Cycle Detection: The core idea is to find the edge that creates a cycle in the graph. Since we know there's exactly one redundant edge that causes a cycle, removing any edge from that cycle will restore the tree property.
• Union-Find (Disjoint Set Union): An efficient way to detect cycles in a graph is by using the Union-Find data structure. We iterate through the edges. For each edge, if the two nodes are already in the same set (i.e., connected), then adding this edge creates a cycle.

• Last Occurrence: Because the prompt asks for the edge that appears last in the input, we simply keep track of the first edge that creates a cycle, and then continue iterating. The final edge that closes the cycle is the one we return.

• Time Complexity: O(n), Space Complexity: O(n) - where n is the number of edges (and nodes). The Union-Find operations (find and union) take nearly constant time on average due to path compression.

Code

    def findRedundantConnection(edges):
    """
    Finds and returns the redundant edge in a graph represented as an array of edges.

    Args:
        edges: A list of lists, where each inner list represents an edge [ai, bi].

    Returns:
        A list representing the redundant edge that can be removed to form a tree.
    """

    n = len(edges)
    parent = list(range(n + 1))  # Initialize parent array for Union-Find

    def find(i):
        """Finds the representative of the set that element i belongs to (with path compression)."""
        if parent[i] == i:
            return i
        parent[i] = find(parent[i])  # Path compression
        return parent[i]

    def union(i, j):
        """Unions the sets containing elements i and j."""
        root_i = find(i)
        root_j = find(j)
        if root_i != root_j:
            parent[root_i] = root_j
            return False  # No cycle formed
        return True  # Cycle formed

    redundant_edge = None
    for u, v in edges:
        if union(u, v):
            redundant_edge = [u, v]

    return redundant_edge