How do you solve Divide Nodes Into the Maximum Number of Groups on LeetCode?

Divide Nodes Into the Maximum Number of Groups (LeetCode #2493) 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: The problem can be reduced to checking if the graph is bipartite.

What is the brute force solution for Divide Nodes Into the Maximum Number of Groups?

The brute force approach for Divide Nodes Into the Maximum Number of Groups is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying to assign nodes to groups in all possible ways and checking if the conditions are satisfied. This method is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n + e).

What are the interview tips for Divide Nodes Into the Maximum Number of Groups?

Always clarify whether the graph is directed or undirected. Discuss edge cases, such as isolated nodes or fully connected graphs. Explain your thought process clearly while coding.

What are common mistakes when solving Divide Nodes Into the Maximum Number of Groups?

Failing to handle disconnected components. Assuming that the graph is always connected.

#2493

Divide Nodes Into the Maximum Number of Groups

Hard

Depth-First SearchBreadth-First SearchUnion-FindGraph TheoryGraph Traversal (BFS/DFS)Bipartite Graph Checking

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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 leverages the concept of bipartite graphs. We can use BFS or DFS to color the graph and determine if it can be divided into two groups. If it can, we can then calculate the maximum number of groups based on the bipartite nature.

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Algorithm

4 steps

1Step 1: Initialize a color array to store group assignments for each node.
2Step 2: For each unvisited node, perform BFS or DFS to color the graph with two colors.
3Step 3: If a conflict is found (two adjacent nodes have the same color), return -1.
4Step 4: Count the number of nodes in each color and return the total number of groups as the sum of the counts.

solution.py33 lines

1# Full working Python code
2from collections import defaultdict, deque
3
4def maxGroupsOptimal(n, edges):
5    graph = defaultdict(list)
6    for a, b in edges:
7        graph[a].append(b)
8        graph[b].append(a)
9    color = [-1] * (n + 1)
10    max_groups = 0
11
12    def bfs(start):
13        queue = deque([start])
14        color[start] = 0
15        count = [1, 0]
16        while queue:
17            node = queue.popleft()
18            for neighbor in graph[node]:
19                if color[neighbor] == -1:
20                    color[neighbor] = 1 - color[node]
21                    count[color[neighbor]] += 1
22                    queue.append(neighbor)
23                elif color[neighbor] == color[node]:
24                    return -1
25        return count[0] + count[1]
26
27    for i in range(1, n + 1):
28        if color[i] == -1:
29            result = bfs(i)
30            if result == -1:
31                return -1
32            max_groups += result
33    return max_groups

ℹ

Complexity note: The time complexity is O(n + e), where n is the number of nodes and e is the number of edges, due to the BFS traversal of the graph. Space complexity is O(n) for storing the graph and color arrays.

1The problem can be reduced to checking if the graph is bipartite.
2Each connected component can be processed independently.

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