Is Graph Bipartite

Leetcode Link

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.

Solution

A bipartite graph is a graph that can be paritiioned into two sets (A and B) such that every edge in the graph connects a node in A and a node in B. By traversing through the graph breadth first, we can partition the nodes into two sets, nodes in the even level and the odd level. We will specifically make sure that two nodes on the same level don't have a connection to guarantee that the graph is bipartite.

Implementation

Let's implement this problem by enforcing statement 3, checking whether there are links between nodes on the same level.

def isBipartite(graph: List[List[int]]) -> bool:
    visited = [False for _ in range(len(graph))] 
    def bfs(root):
        queue = deque([root])
        curlevel = set([0])
        while queue:
            length = len(queue)
            newlevel = set()
            for _ in range(length):           # for nodes in each level
                node = queue.popleft()
                for neighbor in graph[node]:
                    if neighbor in curlevel:  # edge connecting nodes on the same level
                        return False
                    if visited[neighbor]:
                        continue
                    newlevel.add(neighbor)
                    queue.append(neighbor)
                    visited[neighbor] = True
            curlevel = newlevel
        return True
    for i in range(len(graph)):
        if not visited[i] and not bfs(i):
            return False
    return True

Ready to land your dream job?

Unlock your dream job with a 2-minute evaluator for a personalized learning plan!

Start Evaluator

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

What's the output of running the following function using the following tree as input?

1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
12

1import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
17}
18

1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);
15}
16

1 2 3 6 x x

1 2 3 x x x 6 x x

1 2 6 x x x 3 x x

I don't know

Is Graph Bipartite

Solution

Implementation

Ready to land your dream job?

Unlock your dream job with a 2-minute evaluator for a personalized learning plan!

Recommended Readings