The O(n) Club: Is Your Graph Bipartite, or Just Awkward?

The O(n) Club: Is Your Graph Bipartite, or Just Awkward?

⚡ TL;DR

Want to know if your graph can join the exclusive Bipartite Club? Try coloring each node with one of two colors as you traverse. If you ever have to give two frenemies the same trendy shade, your graph gets bounced at the door. Here’s a Java snippet that does this with BFS—no artistic talent required:

// Java BFS coloring for bipartite
boolean isBipartite(int[][] graph) {
    int[] color = new int[graph.length];
    Arrays.fill(color, -1);
    for (int i = 0; i < graph.length; i++) {
        if (color[i] == -1 && !bfs(graph, color, i)) return false;
    }
    return true;
}
// see full bfs() below

🧠 How Devs Usually Mess This Up

You might think a disconnected graph is instantly sketchy. Wrong! Each fragment can live its own peaceful, bipartite life. Equally tragic: thinking both groups must have the same size—nope. The worst? Missing that a single triangle (odd cycle) blows up the whole operation. Don’t let evil little 3-cycles sneak past your radar!

🔍 What’s Actually Going On

Picture a reality show dinner party. Your job: seat squabbling guests so no two enemies (edges) sit together (same group). Try two-coloring the guest list—blues on one table, reds on the other. If things get awkward and you’re forced to put rivals on the same team, the party’s over. And if there’s a table of three (odd cycle), you’ll be reaching for the antacids.

🛠️ PseudoCode

• Initialize color array: Everyone starts uncolored, aka “I don’t know them.”

  int[] color = new int[n];
  Arrays.fill(color, -1);

• Loop through all nodes: The social scene might have isolated cliques—visit every lonely developer!

  for (each node not colored):
      if (fail at coloring from here) return false
  return true

• BFS or DFS traverse: For each friend, pick the opposite color. If you run into a coloring clash, give up and flip the nearest office table.

  queue = new Queue()
  color[start] = 0
  while (queue not empty):
      node = queue.poll()
      for neighbor in graph[node]:
          if (color[neighbor] == -1):
              color[neighbor] = 1 - color[node]
              queue.add(neighbor)
          else if (color[neighbor] == color[node]):
              return false

💻 The Code

import java.util.*;
 public class BipartiteChecker {
    public boolean isBipartite(int[][] graph) {
        int n = graph.length;
        int[] color = new int[n];
        Arrays.fill(color, -1);
        for (int i = 0; i < n; i++) {
            if (color[i] == -1) {
                if (!bfs(graph, color, i)) {
                    return false;
                }
            }
        }
        return true;
    }
     private boolean bfs(int[][] graph, int[] color, int start) {
        Queue
<integer> queue = new LinkedList<>();
        queue.offer(start);
        color[start] = 0;
        while (!queue.isEmpty()) {
            int node = queue.poll();
            for (int neighbor : graph[node]) {
                if (color[neighbor] == -1) {
                    color[neighbor] = 1 - color[node];
                    queue.offer(neighbor);
                } else if (color[neighbor] == color[node]) {
                    return false;
                }
            }
        }
        return true;
    }
}</integer>

⚠️ Pitfalls, Traps & Runtime Smacks

• Disconnected graphs: Remember, one non-bipartite patch ruins the whole thing. Always scan all connected components!
• Self-loops: If a node is its own BFF (has an edge to itself), this is an automatic fail. Sorry, narcissists.
• Efficiency: We’re talking O(N + E) time, so unless your graph is bigger than your caffeine bill, it’ll finish before you can say “LeetCode.”

🧠 Interviewer Brain-Tattoo

“A graph without odd cycles? It’s just two awkward parties that never mix.”