What is the brute force solution for Count Unreachable Pairs of Nodes in an Undirected Graph?

The brute force approach for Count Unreachable Pairs of Nodes in an Undirected Graph is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each pair of nodes to see if they are connected. If they are not connected, we count them as unreachable. This is straightforward but inefficient for large graphs. The optimal Optimal Solution approach improves this to O(n + e).

What are the interview tips for Count Unreachable Pairs of Nodes in an Undirected Graph?

Always clarify the problem and constraints before diving into coding. Think about edge cases, such as fully connected or completely disconnected graphs. Explain your thought process clearly; it helps the interviewer follow your logic.

What are common mistakes when solving Count Unreachable Pairs of Nodes in an Undirected Graph?

Not considering disconnected components, leading to incorrect pair counts. Failing to optimize the search for connections, leading to timeouts on large inputs.

#2316

Count Unreachable Pairs of Nodes in an Undirected Graph

Medium

Depth-First SearchBreadth-First SearchUnion-FindGraph TheoryDepth-First Search (DFS)Breadth-First Search (BFS)Graph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal approach uses the concept of connected components in the graph. By finding the size of each connected component, we can calculate the number of unreachable pairs efficiently without checking every pair individually.

⚙️

Algorithm

3 steps

1Step 1: Create an adjacency list from the edges.
2Step 2: Use DFS or BFS to find all connected components and their sizes.
3Step 3: Calculate the total number of pairs using the sizes of the components: for each component size, calculate pairs with nodes in other components.

solution.py36 lines

1# Full working Python code
2from collections import defaultdict
3
4def count_unreachable_pairs(n, edges):
5    graph = defaultdict(list)
6    for a, b in edges:
7        graph[a].append(b)
8        graph[b].append(a)
9
10    visited = [False] * n
11    component_sizes = []
12
13    def dfs(node):
14        stack = [node]
15        size = 0
16        while stack:
17            curr = stack.pop()
18            if not visited[curr]:
19                visited[curr] = True
20                size += 1
21                stack.extend(graph[curr])
22        return size
23
24    for i in range(n):
25        if not visited[i]:
26            size = dfs(i)
27            component_sizes.append(size)
28
29    total_pairs = 0
30    total_nodes = 0
31    for size in component_sizes:
32        total_pairs += size * total_nodes
33        total_nodes += size
34
35    return total_pairs
36

ℹ

Complexity note: The time complexity is O(n + e) where e is the number of edges, as we traverse each node and edge once. The space complexity is O(n) for the graph representation and visited array.

1Understanding connected components is crucial for solving graph-related problems.
2Using efficient traversal methods (DFS/BFS) can significantly reduce time complexity.

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