Skip to main content
Talentd

Distinct Gate Paths to LCA - Solution & Explanation

HardPremiumFree on FleetCode6 min read
Practice this problem

Problem Statement

You are given an undirected tree rooted at node 0 with n nodes numbered from 0 to n - 1, represented by an array parent where parent[i] is the parent of node i.

Each node i has three types of gates, given in a 2D array gates where gates[i] = [redi, bluei, whitei] which represents the number of red, blue, and white gates at node i.

  • Red gate: usable only with a red card.
  • Blue gate: usable only with a blue card.
  • White gate: usable with either card, but flips the card color when used.

Alice and Bob start at given nodes with either a red or blue card (1 = red, 0 = blue). They must independently move upward to their lowest common ancestor (LCA).

At each node, a person may move to their parent only if they can use at least one gate at that node with their current card. White gates may be used any number of times to flip the card color.

Movement rules (one move = from u to parent[u]):

  • Movement is only upward toward the root.
  • At node u, pick exactly one specific gate instance. Identical gates are treated as separate and counted individually.
  • If holding a red card: use a red gate to remain red, or a white gate to change to blue.
  • If holding a blue card: use a blue gate to remain blue, or a white gate to change to red.
  • If no usable gate exists at u, the sequence ends.

You are also given a 2D array queries where queries[i] = [aNodei, aCardi, bNodei, bCardi]:

  • aNodei, aCardi: Alice's starting node and card.
  • bNodei, bCardi: Bob's starting node and card.

For each query, count the number of distinct valid ways modulo 109 + 7 for both to reach their LCA.

After computing the result for all queries, return the bitwise XOR of those values.

Note:

  • Two ways are distinct if the set of gates used differs for either Alice or Bob.
  • If any person is already at the LCA, then the number of ways for them is 1.
  • The lowest common ancestor (LCA) is defined between two nodes a and b as the lowest node in a tree that has both a and b as descendants (where a node is allowed to be a descendant of itself).

 

Example 1:

Input: n = 3, parent = [-1,0,0], gates = [[1,0,1],[0,1,1],[1,1,0]], queries = [[1,0,2,0],[1,1,2,0],[1,0,2,1]]

Output: 1

Explanation:

i Alice
[Node, Card]		 Bob
[Node, Card]		 LCA Alice
Path		 Bob
Path		 Alice Ways Bob Ways Total Ways
0 [1, 0]: Blue [2, 0]: Blue 0 1 → 0 2 → 0 2 (1 Blue + 1 White at node 1) 1 (1 Blue at node 2) 2 × 1 = 2
1 [1, 1]: Red [2, 0]: Blue 0 1 → 0 2 → 0 1 (1 White at node 1) 1 (1 Blue at node 2) 1 × 1 = 1
2 [1, 0]: Blue [2, 1]: Red 0 1 → 0 2 → 0 2 (1 Blue + 1 White at node 1) 1 (1 Red at node 2) 2 × 1 = 2

Thus, the XOR of all values: 2 XOR 1 XOR 2 = 1.

Example 2:

Input: n = 3, parent = [-1,0,1], gates = [[0,1,2],[1,0,1],[0,0,3]], queries = [[2,0,1,0],[2,1,0,0],[1,1,2,1]]

Output: 3

Explanation:

i Alice
[Node, Card]		 Bob
[Node, Card]		 LCA Alice Path Bob Path Alice Ways Bob Ways Total Ways
0 [2, 0]: Blue [1, 0]: Blue 1 2 → 1 1 3 (3 White at node 2) 1 (no move) 3 × 1 = 3
1 [2, 1]: Red [0, 0]: Blue 0 2 → 1 → 0 0 3 (3 White at node 2) × 1 (1 White at node 1) = 3 1 (no move) 3 × 1 = 3
2 [1, 1]: Red [2, 1]: Red 1 1 2 → 1 1 (no move) 3 (3 White at node 2) 1 × 3 = 3

Thus, the XOR of all values: 3 XOR 3 XOR 3 = 3.

 

Constraints:​​​​​​​

  • 2 <= n <= 2 * 104
  • n == parent.length == gates.length
  • parent[0] == -1
  • 0 <= parent[i] < n for i in [1, n - 1]
  • gates[i] == [redi, bluei, whitei]
  • 0 <= redi, bluei, whitei <= 10
  • 1 <= queries.length <= 2 * 104
  • queries[i] = [aNodei, aCardi, bNodei, bCardi]
  • 0 <= aNodei, bNodei <= n - 1
  • 0 <= aCardi, bCardi <= 1
  • The input is generated such that the array parent represents a valid tree.

Approach Overview

Problem Overview: You are given a tree where certain nodes represent gates. For pairs or groups of nodes, determine how many distinct paths reach their lowest common ancestor (LCA). The challenge is identifying unique gate-to-LCA routes efficiently without recomputing paths for every query.

Approach 1: Brute Force DFS per Query (O(n) per query time, O(n) space)

For every query, run a DFS from each gate node up toward the root and explicitly track the path until the LCA is found. Store visited nodes in a set to avoid double counting. This works because trees guarantee a single path between nodes, but repeatedly traversing the tree makes it expensive when queries are large. This approach is mainly useful for validating correctness or when the number of queries is very small. Traversal can be implemented using standard DFS on the tree.

Approach 2: Path Reconstruction Using Parent Pointers (O(h) per query time, O(n) space)

Precompute each node's parent and depth using a single DFS. For a query, walk both nodes upward until they meet at the LCA. While climbing, track which gates appear on the path using a hash set or frequency map. Because each step moves toward the root, the traversal cost depends on the tree height h. This reduces repeated full traversals but still becomes slow on skewed trees where h ≈ n. It demonstrates how path intersection naturally reveals the LCA.

Approach 3: Binary Lifting LCA with Prefix Path State (O((n + q) log n) time, O(n log n) space)

Preprocess the tree using binary lifting to answer LCA queries in O(log n). During the initial DFS, maintain prefix information along the root-to-node path—such as counts, bitmasks, or hash signatures representing which gates appear. For a query, compute the LCA using the binary lifting table. The distinct path contribution from each node can be derived using prefix values from the two nodes and subtracting the prefix of the LCA's parent. This transforms repeated path exploration into constant-time prefix arithmetic after the LCA lookup.

Recommended for interviews: The binary lifting + prefix state approach is what most interviewers expect for a hard tree problem. The brute-force DFS shows you understand path structure in trees, but the optimized solution proves you can combine preprocessing, prefix aggregation, and LCA queries to handle large inputs efficiently.

Solutions for this problem are being prepared.

Try solving it yourself

Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Brute Force DFS per QueryO(n) per queryO(n)Small trees or very few queries where simplicity matters
Parent Pointer Path ReconstructionO(h) per queryO(n)Moderate constraints when tree height is small
Binary Lifting + Prefix Path TrackingO((n + q) log n)O(n log n)Large inputs with many queries; standard optimal solution

Frequently Asked Questions

Is Distinct Gate Paths to LCA easy or hard?
Distinct Gate Paths to LCA is classified as a hard problem because it combines multiple advanced concepts: tree traversal, LCA preprocessing, and prefix aggregation along paths. Efficient handling of multiple queries requires careful preprocessing and logarithmic-time queries.
Distinct Gate Paths to LCA Python/Java solution
Most implementations build an adjacency list, run DFS to compute depth and parent relationships, and construct a binary lifting table. Python typically uses lists for ancestor tables, while Java uses 2D arrays. Both follow the same logic: preprocess once, answer each query using LCA and prefix differences.
How to solve Distinct Gate Paths to LCA in O(n log n)?
Run a DFS to compute node depths, parent pointers, and prefix path information from the root. Build a binary lifting table so the LCA between any two nodes can be found in O(log n). The number of distinct gate paths is derived using prefix values of the two nodes and subtracting the prefix contribution of their LCA.
What is the best approach for Distinct Gate Paths to LCA?
Binary lifting combined with prefix path tracking is the most efficient approach. Precompute ancestor tables and maintain prefix information along root-to-node paths during DFS. Each query finds the LCA in O(log n) time and derives the number of distinct gate paths using prefix differences.
Is Distinct Gate Paths to LCA asked at Google/Amazon/Meta?
Tree problems involving LCA, path counting, and prefix aggregation are common in interviews at companies like Google, Amazon, and Meta. Variants of LCA with path queries frequently appear in senior-level or hard algorithm rounds.
What data structure is used in Distinct Gate Paths to LCA?
The core structure is a tree represented with an adjacency list. The optimized solution also uses a binary lifting table for LCA queries and prefix arrays, bitmasks, or hash maps to track path information along DFS traversal.
What is the time complexity of Distinct Gate Paths to LCA?
The optimal solution runs in O((n + q) log n) time where n is the number of nodes and q is the number of queries. Preprocessing for binary lifting takes O(n log n), and each LCA query requires O(log n). Space complexity is also O(n log n) for the ancestor table.

Ready to solve this problem?

Practice Distinct Gate Paths to LCA with our built-in code editor and test cases.

Practice on FleetCode