Introduction

In this blog post, we will explore how to solve the LeetCode problem "685" 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 rooted tree is a directed graph such that, there is exactly one node (the root) for which all other nodes are descendants of this node, plus every node has exactly one parent, except for the root node which has no parents. The given input is a directed graph that started as a rooted tree with n nodes (with distinct values from 1 to n), with one additional directed edge added. The added edge has two different vertices chosen from 1 to n, and was not an edge that already existed. The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [ui, vi] that represents a directed edge connecting nodes ui and vi, where ui is a parent of child vi. Return an edge that can be removed so that the resulting graph is a rooted tree of n nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. Example 1: Input: edges = [[1,2],[1,3],[2,3]] Output: [2,3] Example 2: Input: edges = [[1,2],[2,3],[3,4],[4,1],[1,5]] Output: [4,1] Constraints: n == edges.length 3 <= n <= 1000 edges[i].length == 2 1 <= ui, vi <= n ui != vi

Explanation

Here's a breakdown of the approach, complexity, and the Python code:

• Identify potential edge(s) to remove: The added edge creates either a node with two parents or a cycle (or both). The approach identifies if there is a node with two parents. If so, among the two edges that point to that node, the later one is removed.
• Cycle Detection (if needed): If there's no node with two parents, then the added edge created a cycle. This cycle is detected using Depth First Search (DFS). The last edge that appears in the input that is a part of the cycle must be removed.

• Return the appropriate edge: After identifying the proper edge to be removed, return it.

• Time Complexity: O(N), where N is the number of edges.

• Space Complexity: O(N), for the parent array and visited set during cycle detection.

Code

    class Solution:
    def findRedundantDirectedConnection(self, edges: list[list[int]]) -> list[int]:
        """
        Finds and returns the redundant directed connection in a graph represented by a list of edges.

        Args:
            edges: A list of lists, where each inner list represents a directed edge [u, v].

        Returns:
            A list representing the redundant edge that can be removed to make the graph a rooted tree.
        """
        n = len(edges)
        parents = [0] * (n + 1)
        in_degree = [0] * (n + 1)

        # Find edges that cause a node to have two parents
        conflict_edge = None
        for i, (u, v) in enumerate(edges):
            in_degree[v] += 1
            if parents[v] == 0:
                parents[v] = (u, i)
            else:
                conflict_edge = (u, v)
                break

        # If there's a conflict (node with two parents), analyze further
        if conflict_edge:
            u, v = conflict_edge
            first_parent_u, first_parent_index = parents[v]

            # Check if removing the later edge creates a cycle
            temp_edges = edges[:parents[v][1]] + edges[parents[v][1]+1:]

            if self.detect_cycle(temp_edges, n):
                return [first_parent_u, v]
            else:
                return [u, v]

        # If no conflict, the problem is cycle detection
        else:
            # Detect the cycle and return the last edge causing it.
            for i in range(n - 1, -1, -1):
                temp_edges = edges[:i] + edges[i+1:]
                if self.detect_cycle(temp_edges, n):
                    return edges[i]
            return []

    def detect_cycle(self, edges, n):
        """
        Detects if a cycle exists in a directed graph.

        Args:
            edges: A list of edges representing the directed graph.
            n: The number of nodes in the graph.

        Returns:
            True if a cycle is detected, False otherwise.
        """

        graph = [[] for _ in range(n + 1)]
        for u, v in edges:
            graph[u].append(v)

        visited = [False] * (n + 1)
        recursion_stack = [False] * (n + 1)

        def dfs(node):
            visited[node] = True
            recursion_stack[node] = True

            for neighbor in graph[node]:
                if not visited[neighbor]:
                    if dfs(neighbor):
                        return True
                elif recursion_stack[neighbor]:
                    return True

            recursion_stack[node] = False
            return False

        for node in range(1, n + 1):
            if not visited[node]:
                if dfs(node):
                    return True

        return False