How do you solve Path Existence Queries in a Graph I on LeetCode?

Path Existence Queries in a Graph I (LeetCode #3532) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + q * α(n)) time complexity and O(n) space complexity. Key insight: Connected components can be efficiently managed with Union-Find.

What is the time complexity of Path Existence Queries in a Graph I?

The optimal solution for Path Existence Queries in a Graph I runs in O(n + q * α(n)) time and uses O(n) space. Union-Find operations are nearly constant time, making it efficient for multiple queries.

What is the brute force solution for Path Existence Queries in a Graph I?

The brute force approach for Path Existence Queries in a Graph I is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every pair of nodes to see if they can be connected directly based on the maxDiff condition. If they can, use a DFS or BFS to explore connectivity. The optimal Optimal Solution approach improves this to O(n + q * α(n)).

What algorithm or data structure is used to solve Path Existence Queries in a Graph I?

Path Existence Queries in a Graph I uses the following concepts: Array, Hash Table, Binary Search, Union-Find, Graph Theory. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Path Existence Queries in a Graph I?

Explain your thought process clearly. Discuss edge cases and how you would handle them. Practice coding the Union-Find structure.

What are common mistakes when solving Path Existence Queries in a Graph I?

Not considering all possible connections. Confusing direct and indirect connections.

#3532

Path Existence Queries in a Graph I

Medium

ArrayHash TableBinary SearchUnion-FindGraph TheoryHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + q * α(n))
Space	O(1)	O(n)

💡

Intuition

Time O(n + q * α(n))Space O(n)

Utilize Union-Find to group connected nodes based on the maxDiff condition. Preprocess the nodes into connected components for efficient query resolution.

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Algorithm

3 steps

1Step 1: Initialize a Union-Find structure to group nodes based on the maxDiff condition.
2Step 2: Iterate through the sorted nums array, connecting nodes that satisfy the maxDiff condition.
3Step 3: For each query, check if the two nodes belong to the same connected component.

solution.py16 lines

1class UnionFind:
2    def __init__(self, size):
3        self.parent = list(range(size))
4    def find(self, x):
5        if self.parent[x] != x:
6            self.parent[x] = self.find(self.parent[x])
7        return self.parent[x]
8    def union(self, x, y):
9        self.parent[self.find(x)] = self.find(y)
10
11def pathExist(n, nums, maxDiff, queries):
12    uf = UnionFind(n)
13    for i in range(n - 1):
14        if nums[i + 1] - nums[i] <= maxDiff:
15            uf.union(i, i + 1)
16    return [uf.find(u) == uf.find(v) for u, v in queries]

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Complexity note: Union-Find operations are nearly constant time, making it efficient for multiple queries.

1Connected components can be efficiently managed with Union-Find.
2Sorting helps in grouping nodes with similar values.

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