How do you solve Kth Smallest Path XOR Sum on LeetCode?

Kth Smallest Path XOR Sum (LeetCode #3590) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding tree structure is crucial.

What is the time complexity of Kth Smallest Path XOR Sum?

The optimal solution for Kth Smallest Path XOR Sum runs in O(n) time and uses O(n) space. DFS visits each node once, and merging results is efficient, leading to O(n) complexity overall.

What is the brute force solution for Kth Smallest Path XOR Sum?

The brute force approach for Kth Smallest Path XOR Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. For each query, traverse the entire subtree to compute all path XOR sums. Then sort and select the k-th smallest distinct sum. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Kth Smallest Path XOR Sum?

Kth Smallest Path XOR Sum uses the following concepts: Array, Tree, Depth-First Search, Ordered Set. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Kth Smallest Path XOR Sum?

Clarify tree structure and input format. Discuss edge cases like single-node trees. Practice DFS and union-find techniques.

What are common mistakes when solving Kth Smallest Path XOR Sum?

Not handling duplicates in XOR sums. Forgetting to sort results before accessing k-th element.

#3590

Kth Smallest Path XOR Sum

Hard

ArrayTreeDepth-First SearchOrdered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Utilize DFS to compute path XORs while merging results from child nodes efficiently using a union-find approach.

⚙️

Algorithm

3 steps

1Step 1: Perform DFS from the root, maintaining a set of XOR sums for each node.
2Step 2: Use union-find to merge child XOR sets into the parent set during the DFS.
3Step 3: For each query, retrieve the k-th smallest XOR sum from the merged results.

solution.py18 lines

1def kth_smallest_path_xor(par, vals, queries):
2    from collections import defaultdict
3    def dfs(node, current_xor, results):
4        current_xor ^= vals[node]
5        results.add(current_xor)
6        for child in tree[node]:
7            dfs(child, current_xor, results)
8    tree = defaultdict(list)
9    for i in range(len(par)):
10        if par[i] != -1:
11            tree[par[i]].append(i)
12    answers = []
13    for u, k in queries:
14        results = set()
15        dfs(u, 0, results)
16        distinct_sums = sorted(results)
17        answers.append(distinct_sums[k-1] if k <= len(distinct_sums) else -1)
18    return answers

ℹ

Complexity note: DFS visits each node once, and merging results is efficient, leading to O(n) complexity overall.

1Understanding tree structure is crucial.
2XOR properties can simplify path calculations.

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