Is Graph Bipartite
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 containu
). - There are no parallel edges (
graph[u]
does not contain duplicate values). - If
v
is ingraph[u]
, thenu
is ingraph[v]
(the graph is undirected). - The graph may not be connected, meaning there may be two nodes
u
andv
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 containu
.- All the values of
graph[u]
are unique. - If
graph[u]
containsv
, thengraph[v]
containsu
.
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.
1def isBipartite(graph: List[List[int]]) -> bool:
2 visited = [False for _ in range(len(graph))]
3 def bfs(root):
4 queue = deque([root])
5 curlevel = set([0])
6 while queue:
7 length = len(queue)
8 newlevel = set()
9 for _ in range(length): # for nodes in each level
10 node = queue.popleft()
11 for neighbor in graph[node]:
12 if neighbor in curlevel: # edge connecting nodes on the same level
13 return False
14 if visited[neighbor]:
15 continue
16 newlevel.add(neighbor)
17 queue.append(neighbor)
18 visited[neighbor] = True
19 curlevel = newlevel
20 return True
21 for i in range(len(graph)):
22 if not visited[i] and not bfs(i):
23 return False
24 return True
How many times is a tree node visited in a depth first search?
How many ways can you arrange the three letters A, B and C?
Which of the following problems can be solved with backtracking (select multiple)
Which of the two traversal algorithms (BFS and DFS) can be used to find whether two nodes are connected?
Recommended Readings
Top Patterns to Conquer the Technical Coding Interview Should the written word bore you fear not A delightful video alternative awaits iframe width 560 height 315 src https www youtube com embed LW8Io6IPYHw title YouTube video player frameborder 0 allow accelerometer autoplay clipboard write encrypted media gyroscope picture in picture
Recursion Recursion is one of the most important concepts in computer science Simply speaking recursion is the process of a function calling itself Using a real life analogy imagine a scenario where you invite your friends to lunch https algomonster s3 us east 2 amazonaws com recursion jpg You first
Runtime Overview When learning about algorithms and data structures you'll frequently encounter the term time complexity This concept is fundamental in computer science and offers insights into how long an algorithm takes to complete given a certain input size What is Time Complexity Time complexity represents the amount of time
Got a question? Ask the Teaching Assistant anything you don't understand.
Still not clear? Ask in the Forum, Discord or Submit the part you don't understand to our editors.