How do you solve Properties Graph on LeetCode?

Properties Graph (LeetCode #3493) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Using sets for intersections is efficient.

What is the time complexity of Properties Graph?

The optimal solution for Properties Graph runs in O(n²) time and uses O(n) space. The intersection check is O(n) for each pair, leading to O(n²) in total. Space is used for the graph and visited array.

What is the brute force solution for Properties Graph?

The brute force approach for Properties Graph is Brute Force, with O(n²) time complexity and O(n) space complexity. We can compare every pair of property arrays to check for common elements. This straightforward method is easy to understand but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n²).

#3493

Properties Graph

Medium

ArrayHash TableDepth-First SearchBreadth-First SearchUnion-FindGraph Theory

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n²)Space O(n)

Using sets allows us to efficiently compute intersections. We can leverage DFS to find connected components in the graph formed by valid edges.

⚙️

Algorithm

3 steps

1Step 1: Create a graph using adjacency lists based on intersections.
2Step 2: Use DFS to explore each component, marking nodes as visited.
3Step 3: Count the number of times a new DFS starts, which indicates a new component.

solution.py21 lines

1def intersect(a, b): return len(set(a) & set(b))
2
3def countComponents(properties, k):
4    n = len(properties)
5    graph = [[] for _ in range(n)]
6    for i in range(n):
7        for j in range(i + 1, n):
8            if intersect(properties[i], properties[j]) >= k:
9                graph[i].append(j)
10                graph[j].append(i)
11    visited = [False] * n
12    def dfs(node):
13        visited[node] = True
14        for neighbor in graph[node]:
15            if not visited[neighbor]: dfs(neighbor)
16    components = 0
17    for i in range(n):
18        if not visited[i]:
19            dfs(i)
20            components += 1
21    return components

ℹ

Complexity note: The intersection check is O(n) for each pair, leading to O(n²) in total. Space is used for the graph and visited array.

1Using sets for intersections is efficient.

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