Introduction

In this blog post, we will explore how to solve the LeetCode problem "785" 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

There is an undirected graph with n nodes, where each node is numbered between 0 and n - 1. You are given a 2D array graph, where graph[u] is an array of nodes that node u is adjacent to. More formally, for each v in graph[u], there is an undirected edge between node u and node v. The graph has the following properties: There are no self-edges (graph[u] does not contain u). There are no parallel edges (graph[u] does not contain duplicate values). If v is in graph[u], then u is in graph[v] (the graph is undirected). The graph may not be connected, meaning there may be two nodes u and v such that there is no path between them. A graph is bipartite if the nodes can be partitioned into two independent sets A and B such that every edge in the graph connects a node in set A and a node in set B. Return true if and only if it is bipartite. Example 1: Input: graph = [[1,2,3],[0,2],[0,1,3],[0,2]] Output: false Explanation: There is no way to partition the nodes into two independent sets such that every edge connects a node in one and a node in the other. Example 2: Input: graph = [[1,3],[0,2],[1,3],[0,2]] Output: true Explanation: We can partition the nodes into two sets: {0, 2} and {1, 3}. Constraints: graph.length == n 1 <= n <= 100 0 <= graph[u].length < n 0 <= graph[u][i] <= n - 1 graph[u] does not contain u. All the values of graph[u] are unique. If graph[u] contains v, then graph[v] contains u.

Explanation

Here's a solution to determine if a graph is bipartite, along with explanations and the code:

• Coloring: The core idea is to try to color the graph using two colors (e.g., 0 and 1). We start with an arbitrary node and assign it a color. Then, we perform a Depth-First Search (DFS) or Breadth-First Search (BFS) to color its neighbors with the opposite color.
• Conflict Detection: During the coloring process, if we encounter a neighbor that has already been colored with the same color as the current node, it means the graph is not bipartite because it violates the condition that nodes connected by an edge must belong to different sets.

• Disconnected Components: The graph might be disconnected. Therefore, we need to iterate through all nodes and start the coloring process from any uncolored node.

• Runtime Complexity: O(N + E), where N is the number of nodes and E is the number of edges.

• Storage Complexity: O(N), due to the 'colors' array and the potential call stack depth during DFS/BFS.

Code

    def isBipartite(graph):
    """
    Determines if a graph is bipartite.

    Args:
        graph: A 2D array representing the adjacency list of the graph.

    Returns:
        True if the graph is bipartite, False otherwise.
    """

    n = len(graph)
    colors = [-1] * n  # -1: uncolored, 0: color 0, 1: color 1

    def dfs(node, color):
        colors[node] = color
        for neighbor in graph[node]:
            if colors[neighbor] == -1:
                if not dfs(neighbor, 1 - color):
                    return False
            elif colors[neighbor] == color:
                return False
        return True

    for i in range(n):
        if colors[i] == -1:
            if not dfs(i, 0):
                return False

    return True