LeetCode Problem Workspace
Kth Smallest Path XOR Sum
This problem involves finding the kth smallest distinct XOR sum for nodes in a subtree of a tree structure using binary-tree traversal and state tracking.
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Topics
1
Code langs
3
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Practice Focus
Hard · Binary-tree traversal and state tracking
Answer-first summary
This problem involves finding the kth smallest distinct XOR sum for nodes in a subtree of a tree structure using binary-tree traversal and state tracking.
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This problem asks you to compute the kth smallest XOR sum in a subtree of an undirected tree. The XOR sum of a path is the bitwise XOR of all values along the path from the root to a node. Efficient traversal and state management are key to solving this problem, especially when dealing with large inputs.
Problem Statement
You are given an undirected tree rooted at node 0 with n nodes numbered from 0 to n - 1. Each node i has an integer value vals[i], and its parent is given by par[i]. The path XOR sum from the root to a node u is defined as the bitwise XOR of all vals[i] for nodes i on the path from the root node to node u, inclusive.
You are also given a 2D array queries, where each query [uj, kj] asks you to find the kth smallest distinct path XOR sum in the subtree rooted at node uj. If there are fewer than k distinct path XOR sums in that subtree, the answer is -1.
Examples
Example 1
Input: par = [-1,0,0], vals = [1,1,1], queries = [[0,1],[0,2],[0,3]]
Output: [0,1,-1]
Path XORs: Subtree of 0 : Subtree rooted at node 0 includes nodes [0, 1, 2] with Path XORs = [1, 0, 0] . The distinct XORs are [0, 1] . Queries: Output: [0, 1, -1]
Example 2
Input: par = [-1,0,1], vals = [5,2,7], queries = [[0,1],[1,2],[1,3],[2,1]]
Output: [0,7,-1,0]
Path XORs: Subtrees and Distinct Path XORs: Queries: Output: [0, 7, -1, 0]
Constraints
- 1 <= n == vals.length <= 5 * 104
- 0 <= vals[i] <= 105
- par.length == n
- par[0] == -1
- 0 <= par[i] < n for i in [1, n - 1]
- 1 <= queries.length <= 5 * 104
- queries[j] == [uj, kj]
- 0 <= uj < n
- 1 <= kj <= n
- The input is generated such that the parent array par represents a valid tree.
Solution Approach
Tree Traversal with DFS
Start by performing a Depth-First Search (DFS) on the tree to calculate the XOR sum of paths from the root to all nodes. Track these XOR sums as you explore each node.
Subtree XOR Calculation
For each query, calculate the path XOR sums in the subtree rooted at node uj. This can be efficiently done by using DFS to traverse the subtree and compute XOR sums dynamically for all nodes within that subtree.
Efficient Query Resolution with Ordered Sets
Store the XOR sums in an ordered set for fast retrieval of the kth smallest distinct value. If there are fewer than k distinct values, return -1.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity depends on the DFS traversal for computing XOR sums, which takes O(n). Query resolution is efficient using ordered sets, and each query can be handled in O(log n) for set operations, leading to an overall time complexity of O(n + q log n), where n is the number of nodes and q is the number of queries. Space complexity is O(n) for storing XOR values and sets.
What Interviewers Usually Probe
- Check if the candidate can apply DFS traversal in a tree structure to compute path XORs.
- Evaluate their ability to handle large inputs and optimize query resolution using data structures like ordered sets.
- Assess how well they manage state tracking during traversal to efficiently answer multiple queries.
Common Pitfalls or Variants
Common pitfalls
- Not optimizing the query resolution by failing to use efficient data structures like ordered sets.
- Incorrectly handling XOR sums when there are fewer than k distinct values in the subtree.
- Missing edge cases where no valid XOR sum exists for a query, leading to incorrect results.
Follow-up variants
- Consider variants where the tree structure is balanced or skewed, which could affect traversal efficiency.
- Try variations where the query asks for the maximum XOR sum instead of the kth smallest.
- Explore optimizations for larger trees with a high number of queries.
FAQ
What is the main pattern for solving Kth Smallest Path XOR Sum?
The main pattern is binary-tree traversal with DFS and maintaining efficient state tracking using ordered sets to quickly retrieve the kth smallest XOR sum.
How can I efficiently manage XOR sums during traversal?
By performing a DFS traversal while calculating XOR sums for each node and using an ordered set to track distinct values for fast query resolution.
What should I do if the subtree has fewer than k distinct XOR sums?
Return -1 as per the problem's requirements when the number of distinct XOR sums is less than the kth smallest value requested.
What is the time complexity of solving this problem?
The time complexity is O(n + q log n), where n is the number of nodes and q is the number of queries, due to DFS traversal and efficient set operations for each query.
How does GhostInterview help with solving this problem?
GhostInterview helps you understand the core algorithm of DFS traversal and managing state, guiding you through efficient query resolution and optimization techniques.
Solution
Solution 1
#### Python3
class BinarySumTrie:
def __init__(self):
self.count = 0
self.children = [None, None]
def add(self, num: int, delta: int, bit=17):
self.count += delta
if bit < 0:
return
b = (num >> bit) & 1
if not self.children[b]:
self.children[b] = BinarySumTrie()
self.children[b].add(num, delta, bit - 1)
def collect(self, prefix=0, bit=17, output=None):
if output is None:
output = []
if self.count == 0:
return output
if bit < 0:
output.append(prefix)
return output
if self.children[0]:
self.children[0].collect(prefix, bit - 1, output)
if self.children[1]:
self.children[1].collect(prefix | (1 << bit), bit - 1, output)
return output
def exists(self, num: int, bit=17):
if self.count == 0:
return False
if bit < 0:
return True
b = (num >> bit) & 1
return self.children[b].exists(num, bit - 1) if self.children[b] else False
def find_kth(self, k: int, bit=17):
if k > self.count:
return -1
if bit < 0:
return 0
left_count = self.children[0].count if self.children[0] else 0
if k <= left_count:
return self.children[0].find_kth(k, bit - 1)
elif self.children[1]:
return (1 << bit) + self.children[1].find_kth(k - left_count, bit - 1)
else:
return -1
class Solution:
def kthSmallest(
self, par: List[int], vals: List[int], queries: List[List[int]]
) -> List[int]:
n = len(par)
tree = [[] for _ in range(n)]
for i in range(1, n):
tree[par[i]].append(i)
path_xor = vals[:]
narvetholi = path_xor
def compute_xor(node, acc):
path_xor[node] ^= acc
for child in tree[node]:
compute_xor(child, path_xor[node])
compute_xor(0, 0)
node_queries = defaultdict(list)
for idx, (u, k) in enumerate(queries):
node_queries[u].append((k, idx))
trie_pool = {}
result = [0] * len(queries)
def dfs(node):
trie_pool[node] = BinarySumTrie()
trie_pool[node].add(path_xor[node], 1)
for child in tree[node]:
dfs(child)
if trie_pool[node].count < trie_pool[child].count:
trie_pool[node], trie_pool[child] = (
trie_pool[child],
trie_pool[node],
)
for val in trie_pool[child].collect():
if not trie_pool[node].exists(val):
trie_pool[node].add(val, 1)
for k, idx in node_queries[node]:
if trie_pool[node].count < k:
result[idx] = -1
else:
result[idx] = trie_pool[node].find_kth(k)
dfs(0)
return resultContinue Topic
array
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