How do you solve Graph Connectivity With Threshold on LeetCode?

Graph Connectivity With Threshold (LeetCode #1627) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Cities can be connected through common divisors, which can be efficiently managed using Union-Find.

What is the brute force solution for Graph Connectivity With Threshold?

The brute force approach for Graph Connectivity With Threshold is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check each pair of cities to see if they share a common divisor greater than the threshold. This involves checking all pairs and their divisors, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Graph Connectivity With Threshold?

Graph Connectivity With Threshold uses the following concepts: Array, Math, Union-Find, Number Theory. The recommended patterns to study are: Union-Find, Graph Traversal.

What are the interview tips for Graph Connectivity With Threshold?

Clearly explain your thought process and how you would approach building connections. Discuss potential optimizations early in the interview to show your understanding of efficiency. Practice implementing Union-Find as it's a common pattern in connectivity problems.

What are common mistakes when solving Graph Connectivity With Threshold?

Not considering the case where cities are the same (they are always connected). Failing to optimize the divisor checks, leading to unnecessary computations.

#1627

Graph Connectivity With Threshold

Hard

ArrayMathUnion-FindNumber TheoryUnion-FindGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

We can use a Union-Find (Disjoint Set Union) structure to efficiently group cities based on their connections. By connecting cities that share common divisors greater than the threshold, we can quickly answer the queries.

⚙️

Algorithm

3 steps

1Step 1: Initialize a Union-Find structure for n cities.
2Step 2: For each integer from threshold + 1 to n, connect all multiples of that integer.
3Step 3: For each query, check if the two cities belong to the same connected component in the Union-Find structure.

solution.py29 lines

1# Full working Python code
2class UnionFind:
3    def __init__(self, size):
4        self.parent = list(range(size))
5
6    def find(self, x):
7        if self.parent[x] != x:
8            self.parent[x] = self.find(self.parent[x])
9        return self.parent[x]
10
11    def union(self, x, y):
12        rootX = self.find(x)
13        rootY = self.find(y)
14        if rootX != rootY:
15            self.parent[rootY] = rootX
16
17
18def are_connected(n, threshold, queries):
19    uf = UnionFind(n + 1)
20    for z in range(threshold + 1, n + 1):
21        for multiple in range(2 * z, n + 1, z):
22            uf.union(z, multiple)
23    results = []
24    for a, b in queries:
25        results.append(uf.find(a) == uf.find(b))
26    return results
27
28# Example usage
29print(are_connected(6, 2, [[1,4],[2,5],[3,6]]))

ℹ

Complexity note: The complexity is due to the union-find operations and the number of multiples we process, which is efficient compared to the brute force approach.

1Cities can be connected through common divisors, which can be efficiently managed using Union-Find.
2Understanding the properties of divisors helps in optimizing the connection process.

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