Divide Nodes Into Maximum Number Of Groups Solution for TCS NQT
Company: TCS NQT · Difficulty: Hard · Topics: Graph, Breadth-First Search, Bipartite Graph
Problem Statement
You are given a positive integer n representing the number of nodes in an undirected graph. The nodes are numbered from 1 to n.
You are also given a 2D integer array edges, where edges[i] = [ai, bi] indicates that there is an undirected edge between nodes ai and bi.
Return the maximum number of groups you can create such that:
- Each node belongs to exactly one group
- Each group contains at least one node
- For any two nodes in the same group, there is no edge between them (they are not adjacent)
If it is impossible to create any groups, return -1.
Examples
Example 1:
Input: n = 4, edges = [[1,2],[1,3],[2,4]]
Output: 2
Explanation:
We can create 2 groups: [1,4] and [2,3]. No two nodes in the same group are adjacent.
Example 2:
Input: n = 3, edges = [[1,2],[2,3],[3,1]]
Output: -1
Explanation:
It is impossible to create any groups because all nodes are connected in a cycle.
Example 3:
Input: n = 5, edges = [[1,2],[2,3],[3,4],[4,5]]
Output: 2
Explanation:
We can create 2 groups: [1,3,5] and [2,4].
Example 4:
Input: n = 6, edges = [[1,2],[2,3],[3,4],[4,5],[5,6]]
Output: 3
Explanation:
We can create 3 groups: [1,3,5], [2,4,6], and [].
Intuition
This problem is essentially asking us to find the maximum number of independent sets we can create in a graph. An independent set is a set of vertices where no two vertices are adjacent.
The key insights are:
1. This is equivalent to finding the maximum number of colors needed to color the graph such that no adjacent vertices have the same color
2. The problem reduces to finding the maximum number of groups in any connected component
3. For a bipartite graph, we can use 2 colors
4. For a graph with odd cycles, we need more than 2 colors
Approach
Approach 1: BFS with Bipartite Check (Optimal)
1. For each connected component:
- Use BFS to check if it's bipartite
- If bipartite, we can use 2 groups
- If not bipartite, we need more groups (use BFS to find the minimum number of colors needed)
2. The answer is the maximum number of groups needed across all components
Approach 2: Graph Coloring with BFS
1. For each connected component:
- Use BFS to assign colors to nodes
- Keep track of the maximum number of colors used
2. Return the maximum number of colors needed
Approach 3: DFS with Backtracking
1. Try to assign each node to different groups
2. Use backtracking to find the optimal grouping
3. This is less efficient but more intuitive
Dry Run
Let's trace through Example 1: n = 4, edges = [[1,2],[1,3],[2,4]]
BFS Approach:
- Start with node 1, assign color 0
- Neighbors of 1: [2,3], assign color 1
- Neighbors of 2: [1,4], 1 already colored, assign color 0 to 4
- Neighbors of 3: [1], already colored
- Neighbors of 4: [2], already colored
Colors used: 2 (0 and 1)
Result: 2 groups
Key Insights
1. Bipartite Graph: If a graph is bipartite, we can use 2 colors
2. Odd Cycles: Graphs with odd cycles need more than 2 colors
3. Connected Components: Handle each connected component separately
4. Maximum Groups: The answer is the maximum number of groups needed in any component
Edge Cases
1. No Edges: All nodes can be in separate groups
2. Complete Graph: Only one group possible
3. Disconnected Graph: Handle each component separately
4. Single Node: Can be in any group
Constraints
- 1 <= n <= 500
- 0 <= edges.length <= 10^4
- 1 <= ai, bi <= n
- ai != bi