How do you solve Is Graph Bipartite? on LeetCode?

Is Graph Bipartite? (LeetCode #785) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + e) time complexity and O(n) space complexity. Key insight: Bipartite graphs can be colored with two colors without adjacent nodes sharing the same color.

What is the brute force solution for Is Graph Bipartite??

The brute force approach for Is Graph Bipartite? is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible way to color the graph's nodes with two colors. If we can color the graph without any adjacent nodes sharing the same color, then it is bipartite. The optimal Optimal Solution approach improves this to O(n + e).

What algorithm or data structure is used to solve Is Graph Bipartite??

Is Graph Bipartite? uses the following concepts: Depth-First Search, Breadth-First Search, Union-Find, Graph Theory. The recommended patterns to study are: Graph Traversal (BFS, DFS), Coloring Problems, Union-Find.

What are the interview tips for Is Graph Bipartite??

Always consider edge cases, such as disconnected graphs. Explain your thought process clearly while coding. Practice both BFS and DFS approaches, as they can be applied to many graph problems.

What are common mistakes when solving Is Graph Bipartite??

Failing to handle disconnected components of the graph. Not checking for adjacent nodes having the same color.

#785

Is Graph Bipartite?

Medium

Depth-First SearchBreadth-First SearchUnion-FindGraph TheoryGraph Traversal (BFS, DFS)Coloring ProblemsUnion-Find

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + e)
Space	O(1)	O(n)

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Intuition

Time O(n + e)Space O(n)

The optimal solution uses BFS or DFS to color the graph in one pass, ensuring that no two adjacent nodes have the same color. This is efficient and leverages the properties of bipartite graphs.

⚙️

Algorithm

3 steps

1Step 1: Initialize a color array to track the color of each node.
2Step 2: For each uncolored node, start a BFS or DFS to color the graph.
3Step 3: If you find two adjacent nodes with the same color, return false. If all nodes are colored successfully, return true.

solution.py18 lines

1# Full working Python code
2from collections import deque
3
4def isBipartite(graph):
5    color = [-1] * len(graph)
6    for i in range(len(graph)):
7        if color[i] == -1:
8            queue = deque([i])
9            color[i] = 0
10            while queue:
11                node = queue.popleft()
12                for neighbor in graph[node]:
13                    if color[neighbor] == -1:
14                        color[neighbor] = 1 - color[node]
15                        queue.append(neighbor)
16                    elif color[neighbor] == color[node]:
17                        return False
18    return True

ℹ

Complexity note: The time complexity is O(n + e) where n is the number of nodes and e is the number of edges, as we traverse each node and edge once. The space complexity is O(n) due to the color array.

1Bipartite graphs can be colored with two colors without adjacent nodes sharing the same color.
2Using BFS or DFS efficiently checks for bipartiteness in a single pass.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.