Introduction

In this blog post, we will explore how to solve the LeetCode problem "2493" 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 a positive integer n representing the number of nodes in an undirected graph. The nodes are labeled from 1 to n. You are also given a 2D integer array edges, where edges[i] = [ai, bi] indicates that there is a bidirectional edge between nodes ai and bi. Notice that the given graph may be disconnected. Divide the nodes of the graph into m groups (1-indexed) such that: Each node in the graph belongs to exactly one group. For every pair of nodes in the graph that are connected by an edge [ai, bi], if ai belongs to the group with index x, and bi belongs to the group with index y, then |y - x| = 1. Return the maximum number of groups (i.e., maximum m) into which you can divide the nodes. Return -1 if it is impossible to group the nodes with the given conditions. Example 1: Input: n = 6, edges = [[1,2],[1,4],[1,5],[2,6],[2,3],[4,6]] Output: 4 Explanation: As shown in the image we: - Add node 5 to the first group. - Add node 1 to the second group. - Add nodes 2 and 4 to the third group. - Add nodes 3 and 6 to the fourth group. We can see that every edge is satisfied. It can be shown that that if we create a fifth group and move any node from the third or fourth group to it, at least on of the edges will not be satisfied. Example 2: Input: n = 3, edges = [[1,2],[2,3],[3,1]] Output: -1 Explanation: If we add node 1 to the first group, node 2 to the second group, and node 3 to the third group to satisfy the first two edges, we can see that the third edge will not be satisfied. It can be shown that no grouping is possible. Constraints: 1 <= n <= 500 1 <= edges.length <= 104 edges[i].length == 2 1 <= ai, bi <= n ai != bi There is at most one edge between any pair of vertices.

Explanation

Here's the breakdown of the solution:

• Connected Components and Bipartiteness Check: The problem involves checking if each connected component of the graph can be colored using two colors such that no two adjacent nodes have the same color (bipartiteness). If any component is not bipartite, it's impossible to satisfy the condition |y - x| = 1, thus we return -1.

• Finding Maximum Length of Each Component: For each bipartite connected component, perform a Breadth-First Search (BFS) starting from any node. During the BFS, assign group numbers (which can be thought of as distances from the starting node). Track the minimum and maximum group numbers reached in each component. The length of the component (in terms of group differences) is the maximum group number minus the minimum group number plus 1.

• Summing Component Lengths: The maximum number of groups is the sum of the lengths of all the bipartite connected components.

• Time Complexity: O(N + E) where N is the number of nodes and E is the number of edges. This is due to graph traversal (BFS). Storage Complexity: O(N).

Code

    from collections import defaultdict, deque

def maximumGroups(n, edges):
    """
    Divides the nodes of a graph into groups such that for every edge (a, b),
    if a belongs to group x and b belongs to group y, then |y - x| = 1.

    Args:
        n: The number of nodes in the graph.
        edges: A list of edges, where each edge is a tuple (a, b).

    Returns:
        The maximum number of groups, or -1 if it's impossible to group the nodes.
    """

    graph = defaultdict(list)
    for a, b in edges:
        graph[a].append(b)
        graph[b].append(a)

    visited = set()
    max_groups = 0

    for i in range(1, n + 1):
        if i not in visited:
            min_group = float('inf')
            max_group = float('-inf')
            is_bipartite = True
            q = deque([(i, 1)])  # (node, group)
            component_visited = {i: 1}
            visited.add(i)

            while q:
                node, group = q.popleft()
                min_group = min(min_group, group)
                max_group = max(max_group, group)

                for neighbor in graph[node]:
                    if neighbor not in component_visited:
                        component_visited[neighbor] = group + (1 if len(graph[node]) > 0 else 0) * (-1 if group % 2 == 0 else 1)
                        q.append((neighbor, component_visited[neighbor]))
                        visited.add(neighbor)
                    elif abs(component_visited[neighbor] - group) != 1:
                        is_bipartite = False
                        break
                if not is_bipartite:
                    break

            if not is_bipartite:
                return -1

            max_groups += (max_group - min_group + 1)

    return max_groups