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

Path Existence Queries in a Graph II (LeetCode #3534) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + m) time complexity and O(n) space complexity. Key insight: Sorting helps group nodes efficiently.

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

The optimal solution for Path Existence Queries in a Graph II runs in O(n log n + m) time and uses O(n) space. Sorting takes O(n log n), and processing each query is O(1) after that.

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

The brute force approach for Path Existence Queries in a Graph II is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check all possible paths between nodes using BFS or DFS. This method is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n log n + m).

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

Explain your thought process clearly. Use examples to illustrate your approach. Practice similar graph problems.

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

Not considering disconnected components. Overlooking edge cases with maxDiff.

#3534

Path Existence Queries in a Graph II

Hard

ArrayTwo PointersBinary SearchDynamic ProgrammingGreedyBit ManipulationGraph TheorySortingGraph TraversalUnion-Find

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n + m)Space O(n)

Sort nodes based on their values and use a two-pointer technique to efficiently find connected components. This reduces unnecessary checks.

⚙️

Algorithm

3 steps

1Step 1: Create a list of nodes sorted by their values in nums.
2Step 2: Use two pointers to group nodes into connected components based on maxDiff.
3Step 3: For each query, check if both nodes belong to the same component and return the distance.

solution.py12 lines

1def min_distance_optimal(n, nums, maxDiff, queries):
2    sorted_nodes = sorted(range(n), key=lambda x: nums[x])
3    components = [-1] * n
4    component_id = 0
5    for i in range(n):
6        if components[sorted_nodes[i]] == -1:
7            j = i
8            while j < n and nums[sorted_nodes[j]] - nums[sorted_nodes[i]] <= maxDiff:
9                components[sorted_nodes[j]] = component_id
10                j += 1
11            component_id += 1
12    return [1 if components[u] == components[v] else -1 for u, v in queries]

ℹ

Complexity note: Sorting takes O(n log n), and processing each query is O(1) after that.

1Sorting helps group nodes efficiently.
2Two-pointer technique reduces unnecessary checks.

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