LeetCode 2493 - Divide Nodes Into the Maximum Number of Groups
We are given an undirected graph with n nodes and a list of edges. The graph may contain multiple disconnected components. Our task is to divide all nodes into ordered groups numbered from 1 to m. The key constraint is based on graph edges.
Difficulty: 🔴 Hard
Topics: Depth-First Search, Breadth-First Search, Union-Find, Graph Theory
Solution
Problem Understanding
We are given an undirected graph with n nodes and a list of edges. The graph may contain multiple disconnected components. Our task is to divide all nodes into ordered groups numbered from 1 to m.
The key constraint is based on graph edges. If two nodes are connected by an edge, and one node belongs to group x, then the other node must belong to group x - 1 or x + 1. In other words, every edge must connect nodes whose group indices differ by exactly one.
The goal is not just to find any valid grouping, but to maximize the total number of groups.
This condition is extremely important because it effectively forces adjacent nodes to appear in consecutive layers. That immediately resembles a breadth first search layering structure.
The graph can also be disconnected, which means each connected component can be handled independently. The final answer is the sum of the maximum valid group counts for every connected component.
The constraints are relatively small:
n <= 500edges.length <= 10^4
This allows us to run BFS from every node if necessary, since 500 * (500 + 10^4) is still manageable.
There are several important edge cases:
- A graph containing an odd cycle is impossible to group correctly. Odd cycles are not bipartite, and the required layering condition cannot be satisfied.
- A disconnected graph requires independent processing of each component.
- A single isolated node contributes exactly one group.
- Multiple BFS starting points inside the same component may produce different numbers of layers, so we must find the maximum possible value.
Approaches
Brute Force Approach
A naive idea would be to try all possible assignments of nodes into groups and verify whether every edge satisfies the condition |x - y| = 1.
This approach is theoretically correct because it explores every valid grouping configuration and selects the maximum number of groups. However, the number of possible assignments grows exponentially with the number of nodes. Even for moderate graph sizes, this becomes completely infeasible.
Another brute force variation would attempt every node as a possible starting layer and recursively assign neighboring nodes to adjacent layers while checking consistency. Although slightly better, this still explodes combinatorially in dense graphs.
The core difficulty is that the constraints between neighboring nodes create a large global dependency structure.
Key Insight
The crucial observation is that the required grouping structure is equivalent to BFS levels in a bipartite graph.
Suppose we choose a starting node and assign it to group 1. Then all neighbors must belong to group 2. Their neighbors must belong to group 1 or 3, and so on.
This is exactly how BFS distances behave.
For any valid grouping:
- Adjacent nodes must differ by exactly one level.
- Therefore, nodes cannot share the same parity cycle inconsistently.
- The graph must be bipartite.
If a connected component is not bipartite, the answer is immediately -1.
For a bipartite component, the maximum number of groups equals the maximum number of BFS layers achievable from some node in that component. This corresponds to the largest shortest-path depth plus one.
Since n <= 500, we can safely run BFS from every node.
Approach Comparison
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force | Exponential | Exponential | Tries all possible group assignments |
| Optimal | O(n × (n + e)) | O(n + e) | BFS from every node, validate bipartite structure |
Algorithm Walkthrough
- Build an adjacency list for the graph.
Since the graph is undirected and sparse enough, an adjacency list is the most efficient representation. Each node stores all of its neighbors. 2. Identify connected components.
The graph may be disconnected, so we process each component independently. We can use BFS or DFS to collect all nodes belonging to the same component. 3. Check whether each component is bipartite.
During traversal, assign alternating colors to neighboring nodes. If we ever find an edge connecting two nodes with the same color, the graph contains an odd cycle, and the answer is -1.
This works because the required grouping condition inherently requires bipartite structure. 4. For every node in the component, run BFS.
We compute the shortest distance from the start node to all other reachable nodes.
During BFS:
- Store distances for each node.
- For every edge
(u, v), verify that|dist[u] - dist[v]| == 1.
If this condition fails, the graph structure is invalid. 5. Compute the number of groups produced by that BFS.
BFS levels naturally correspond to valid groups.
If the maximum distance reached is d, then the number of groups is d + 1.
6. Take the maximum BFS depth for the component.
Different starting nodes may produce different numbers of layers. We want the largest possible grouping count. 7. Add the component contribution to the global answer.
Since disconnected components do not interact, their optimal group counts can simply be summed.
Why it works
A valid grouping assigns consecutive levels to adjacent nodes. BFS layering already guarantees that neighboring nodes differ in distance by at most one. In a bipartite graph, edges never connect nodes of the same level parity, so every edge must connect adjacent levels.
Running BFS from every node guarantees that we discover the largest possible layering depth inside each component. Since components are independent, summing their maxima produces the optimal global answer.
Python Solution
from collections import deque
from typing import List
class Solution:
def magnificentSets(self, n: int, edges: List[List[int]]) -> int:
graph = [[] for _ in range(n + 1)]
for u, v in edges:
graph[u].append(v)
graph[v].append(u)
color = [-1] * (n + 1)
def get_component(start: int) -> List[int]:
queue = deque([start])
component = []
color[start] = 0
while queue:
node = queue.popleft()
component.append(node)
for neighbor in graph[node]:
if color[neighbor] == -1:
color[neighbor] = color[node] ^ 1
queue.append(neighbor)
elif color[neighbor] == color[node]:
return []
return component
def bfs_depth(start: int) -> int:
queue = deque([start])
distance = [-1] * (n + 1)
distance[start] = 0
max_distance = 0
while queue:
node = queue.popleft()
for neighbor in graph[node]:
if distance[neighbor] == -1:
distance[neighbor] = distance[node] + 1
max_distance = max(
max_distance,
distance[neighbor]
)
queue.append(neighbor)
elif abs(distance[neighbor] - distance[node]) != 1:
return -1
return max_distance + 1
answer = 0
for node in range(1, n + 1):
if color[node] != -1:
continue
component = get_component(node)
if not component:
return -1
best = 0
for start in component:
depth = bfs_depth(start)
if depth == -1:
return -1
best = max(best, depth)
answer += best
return answer
The implementation begins by constructing the adjacency list representation of the graph.
The get_component function performs two jobs simultaneously. It collects all nodes belonging to the same connected component and checks whether the component is bipartite. The color array stores alternating colors for neighboring nodes. If two adjacent nodes receive the same color, the graph contains an odd cycle, so the function returns an empty list to signal failure.
The bfs_depth function computes the number of BFS layers starting from a chosen node. The distance array stores shortest path distances from the start node.
As BFS expands outward:
- Unvisited neighbors receive distance
distance[node] + 1. - The maximum distance encountered determines the number of groups.
- Every edge is checked to ensure adjacent nodes differ by exactly one BFS level.
Finally, the outer loop processes each connected component independently and sums the maximum layer counts.
Go Solution
package main
import "container/list"
func magnificentSets(n int, edges [][]int) int {
graph := make([][]int, n+1)
for _, edge := range edges {
u, v := edge[0], edge[1]
graph[u] = append(graph[u], v)
graph[v] = append(graph[v], u)
}
color := make([]int, n+1)
for i := 0; i <= n; i++ {
color[i] = -1
}
getComponent := func(start int) []int {
queue := list.New()
queue.PushBack(start)
component := []int{}
color[start] = 0
for queue.Len() > 0 {
front := queue.Front()
queue.Remove(front)
node := front.Value.(int)
component = append(component, node)
for _, neighbor := range graph[node] {
if color[neighbor] == -1 {
color[neighbor] = color[node] ^ 1
queue.PushBack(neighbor)
} else if color[neighbor] == color[node] {
return []int{}
}
}
}
return component
}
bfsDepth := func(start int) int {
queue := list.New()
queue.PushBack(start)
distance := make([]int, n+1)
for i := 0; i <= n; i++ {
distance[i] = -1
}
distance[start] = 0
maxDistance := 0
for queue.Len() > 0 {
front := queue.Front()
queue.Remove(front)
node := front.Value.(int)
for _, neighbor := range graph[node] {
if distance[neighbor] == -1 {
distance[neighbor] = distance[node] + 1
if distance[neighbor] > maxDistance {
maxDistance = distance[neighbor]
}
queue.PushBack(neighbor)
} else {
diff := distance[neighbor] - distance[node]
if diff < 0 {
diff = -diff
}
if diff != 1 {
return -1
}
}
}
}
return maxDistance + 1
}
answer := 0
for node := 1; node <= n; node++ {
if color[node] != -1 {
continue
}
component := getComponent(node)
if len(component) == 0 {
return -1
}
best := 0
for _, start := range component {
depth := bfsDepth(start)
if depth == -1 {
return -1
}
if depth > best {
best = depth
}
}
answer += best
}
return answer
}
The Go implementation closely mirrors the Python solution.
The main difference is queue handling. Go does not have a built-in deque structure like Python's collections.deque, so we use container/list.
Distance and color arrays are initialized manually because Go slices default to zero values instead of -1.
Integer overflow is not a concern because the maximum graph size is only 500.
Worked Examples
Example 1
Input:
n = 6
edges = [[1,2],[1,4],[1,5],[2,6],[2,3],[4,6]]
Step 1: Build Graph
| Node | Neighbors |
|---|---|
| 1 | 2, 4, 5 |
| 2 | 1, 6, 3 |
| 3 | 2 |
| 4 | 1, 6 |
| 5 | 1 |
| 6 | 2, 4 |
Step 2: Check Bipartite Structure
Start BFS coloring from node 1.
| Node | Color |
|---|---|
| 1 | 0 |
| 2 | 1 |
| 4 | 1 |
| 5 | 1 |
| 6 | 0 |
| 3 | 0 |
No conflicts occur, so the graph is bipartite.
Step 3: Run BFS from Every Node
BFS from Node 5
| Node | Distance |
|---|---|
| 5 | 0 |
| 1 | 1 |
| 2 | 2 |
| 4 | 2 |
| 3 | 3 |
| 6 | 3 |
Maximum distance is 3.
Therefore, this BFS produces 4 groups.
BFS from Node 1
| Node | Distance |
|---|---|
| 1 | 0 |
| 2 | 1 |
| 4 | 1 |
| 5 | 1 |
| 3 | 2 |
| 6 | 2 |
Maximum distance is 2.
This produces only 3 groups.
The best value over all starting nodes is 4.
Final answer:
4
Example 2
Input:
n = 3
edges = [[1,2],[2,3],[3,1]]
Step 1: Build Graph
| Node | Neighbors |
|---|---|
| 1 | 2, 3 |
| 2 | 1, 3 |
| 3 | 1, 2 |
Step 2: Bipartite Check
Start coloring from node 1.
| Node | Color |
|---|---|
| 1 | 0 |
| 2 | 1 |
| 3 | 1 |
Now examine edge (2,3).
Both nodes have color 1, which violates bipartite structure.
Therefore, the graph contains an odd cycle.
Final answer:
-1
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n × (n + e)) | BFS is executed from every node |
| Space | O(n + e) | Adjacency list, queues, and helper arrays |
The graph has at most 500 nodes, so performing BFS from every node is acceptable. Each BFS traverses all vertices and edges in the connected component. The adjacency list dominates memory usage.
Test Cases
from typing import List
class Solution:
def magnificentSets(self, n: int, edges: List[List[int]]) -> int:
from collections import deque
graph = [[] for _ in range(n + 1)]
for u, v in edges:
graph[u].append(v)
graph[v].append(u)
color = [-1] * (n + 1)
def get_component(start):
queue = deque([start])
component = []
color[start] = 0
while queue:
node = queue.popleft()
component.append(node)
for neighbor in graph[node]:
if color[neighbor] == -1:
color[neighbor] = color[node] ^ 1
queue.append(neighbor)
elif color[neighbor] == color[node]:
return []
return component
def bfs_depth(start):
queue = deque([start])
dist = [-1] * (n + 1)
dist[start] = 0
best = 0
while queue:
node = queue.popleft()
for neighbor in graph[node]:
if dist[neighbor] == -1:
dist[neighbor] = dist[node] + 1
best = max(best, dist[neighbor])
queue.append(neighbor)
elif abs(dist[neighbor] - dist[node]) != 1:
return -1
return best + 1
answer = 0
for node in range(1, n + 1):
if color[node] != -1:
continue
component = get_component(node)
if not component:
return -1
best = 0
for start in component:
depth = bfs_depth(start)
if depth == -1:
return -1
best = max(best, depth)
answer += best
return answer
sol = Solution()
assert sol.magnificentSets(
6,
[[1,2],[1,4],[1,5],[2,6],[2,3],[4,6]]
) == 4 # official example
assert sol.magnificentSets(
3,
[[1,2],[2,3],[3,1]]
) == -1 # odd cycle
assert sol.magnificentSets(
1,
[]
) == 1 # single isolated node
assert sol.magnificentSets(
4,
[[1,2],[3,4]]
) == 4 # two disconnected edges
assert sol.magnificentSets(
5,
[[1,2],[2,3],[3,4],[4,5]]
) == 5 # simple path graph
assert sol.magnificentSets(
4,
[[1,2],[2,3],[3,4],[4,1]]
) == 3 # even cycle
assert sol.magnificentSets(
7,
[[1,2],[2,3],[4,5]]
) == 7 # disconnected with isolated nodes
assert sol.magnificentSets(
6,
[[1,2],[2,3],[3,1],[4,5]]
) == -1 # one invalid component invalidates all
assert sol.magnificentSets(
8,
[[1,2],[2,3],[3,4],[4,5],[5,6],[6,7],[7,8]]
) == 8 # long chain
assert sol.magnificentSets(
6,
[[1,2],[1,3],[1,4],[1,5],[1,6]]
) == 3 # star graph
Test Summary
| Test | Why |
|---|---|
| Official example 1 | Valid bipartite graph with optimal layering |
| Official example 2 | Odd cycle detection |
| Single node | Minimum input size |
| Two disconnected edges | Independent component handling |
| Path graph | Maximum layering depth |
| Even cycle | Bipartite cyclic graph |
| Disconnected with isolated nodes | Mixed component structures |
| Invalid component mixed with valid component | Ensures immediate failure |
| Long chain | Stress BFS depth |
| Star graph | Central hub structure |
Edge Cases
One important edge case is an odd cycle, such as a triangle. These graphs are not bipartite, which means it is impossible to assign consecutive group indices consistently. A naive implementation that only checks BFS depths might incorrectly accept such graphs. The bipartite coloring step prevents this issue by detecting same-color adjacent nodes.
Another important case is disconnected graphs. Since components do not interact, their optimal group counts should be added independently. Forgetting this can lead to undercounting or incorrect traversal logic. The implementation explicitly extracts connected components before processing them.
Isolated nodes are also easy to mishandle. A node with no edges still forms a valid component and contributes one group. The BFS logic correctly returns a depth of 1 because the maximum distance from the node to itself is 0.
A subtle edge case involves graphs where different BFS starting nodes produce different numbers of layers. For example, in a path graph, starting from the middle gives fewer layers than starting from an endpoint. The implementation handles this correctly by running BFS from every node in the component and selecting the maximum result.