This problem can be solved using the concept of Disjoint Set Union (DSU) or Union-Find data structure. Here are the steps to solve this problem: 1. Initialize a DSU with N nodes. In the beginning, each node is a separate component (or subgraph), so the initial number of disjoint subgraphs is N. 2. For each edge (A, B), perform the union operation on nodes A and B. This operation merges the two components containing A and B into a single component. If A and B are already in the same component, the operation has no effect. 3. After processing all edges, the number of disjoint subgraphs is equal to the number of separate components in the DSU. This can be found by iterating over all nodes and counting the number of unique parent nodes. Here is a Python code snippet that implements the above steps: ```python def find(parent, i): if parent[i] == i: return i return find(parent, parent[i]) def union(parent, rank, x, y): xroot = find(parent, x) yroot = find(parent, y) if rank[xroot] rank[yroot]: parent[yroot] = xroot else : parent[yroot] = xroot rank[xroot] += 1 def countDisjointSubgraphs(N, edges): parent = [i for i in range(N)] rank = [0 for i in range(N)] for edge in edges: union(parent, rank, edge[0], edge[1]) unique_parents = set(find(parent, i) for i in range(N)) return len(unique_parents) ``` You can use the `countDisjointSubgraphs` function to solve the problem. The function takes the number of nodes N and a list of edges as input, and returns the number of disjoint subgraphs. Each edge is represented as a pair of integers (A, B).

Question

This problem can be solved using the concept of Disjoint Set Union (DSU) or Union-Find data structure. Here are the steps to solve this problem:

1. Initialize a DSU with N nodes. In the beginning, each node is a separate component (or subgraph), so the initial number of disjoint subgraphs is N.

2. For each edge (A, B), perform the union operation on nodes A and B. This operation merges the two components containing A and B into a single component. If A and B are already in the same component, the operation has no effect.

3. After processing all edges, the number of disjoint subgraphs is equal to the number of separate components in the DSU. This can be found by iterating over all nodes and counting the number of unique parent nodes.

Here is a Python code snippet that implements the above steps:

```python
def find(parent, i):
    if parent[i] == i:
        return i
    return find(parent, parent[i])

def union(parent, rank, x, y):
    xroot = find(parent, x)
    yroot = find(parent, y)

if rank[xroot] < rank[yroot]:
        parent[xroot] = yroot
    elif rank[xroot] > rank[yroot]:
        parent[yroot] = xroot
    else :
        parent[yroot] = xroot
        rank[xroot] += 1

def countDisjointSubgraphs(N, edges):
    parent = [i for i in range(N)]
    rank = [0 for i in range(N)]

for edge in edges:
        union(parent, rank, edge[0], edge[1])

unique_parents = set(find(parent, i) for i in range(N))
    return len(unique_parents)
```

You can use the `countDisjointSubgraphs` function to solve the problem. The function takes the number of nodes N and a list of edges as input, and returns the number of disjoint subgraphs. Each edge is represented as a pair of integers (A, B).

Knowee AI · Accepted Answer