Optimal Grid Path Protocol 4 — Problem Statement & Solution Guide

TreesMediumBinary Lifting LCA

TimeO(N log N + Q log N)

SpaceO(N log N)

Problem Description

You are tasked with optimizing the routing protocol for a hierarchical network of servers. The network is structured as a rooted binary tree where each node represents a server. To ensure efficient data retrieval, the system requires calculating the Lowest Common Ancestor (LCA) for multiple pairs of nodes. Given the structure of the tree and a list of query pairs, determine the LCA for each pair. The LCA of two nodes u and v is defined as the deepest node that is an ancestor of both u and v. If one node is an ancestor of the other, the LCA is the ancestor node itself. You must process all queries efficiently to minimize latency in the network protocol.

DSA Pattern Breakdown

"Optimal Grid Path Protocol 4"

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WHY DOES IT MATTER?

LCA is a cornerstone of many hierarchical problems—permission checks, network routing, version control merges, and taxonomy queries—all of which require fast ancestor resolution. Mastering LCA patterns demonstrates a candidate’s ability to convert a seemingly O(N) per‑query problem into a scalable solution, a skill prized in large‑scale systems.

OPTIMIZATION CHALLENGE

The breakthrough is recognizing that ancestor relationships are repeatable across queries; by precomputing a sparse set of jump pointers (binary lifting) or a depth‑ordered traversal (Euler tour), you replace repeated upward walks with constant‑time table lookups or range‑minimum queries, collapsing O(N·Q) work into O(N log N + Q).

REAL-WORLD CONNECTION

Think of a corporate org chart where you need to find the nearest common manager of two employees. In distributed systems, this mirrors locating the nearest common router or service node that can coordinate two requests, minimizing latency and hop count.

During an interview, first sketch the DFS that records depth and parent, then immediately mention the 2^k table—this signals you know the optimal path. If the interviewer pushes for O(1) queries, transition to the Euler‑tour‑RMQ idea and discuss the ±1 property for a linear‑time sparse table.

COMPLEXITY AT A GLANCE

⏱ Time:O(N log N + Q log N)

💾 Space:O(N log N)

Core Theory — Why This Approach?

The Lowest Common Ancestor (LCA) problem asks for the deepest node that is an ancestor of two given nodes in a rooted tree. A naive solution walks up from each node to the root, marking visited ancestors, which leads to O(N) time per query and quickly becomes a bottleneck when the number of queries Q approaches 10^5 or the tree size N is large. The optimal paradigm leverages preprocessing to answer each query in logarithmic or constant time, turning the overall complexity into O(N log N + Q log N) or even O(N + Q) depending on the chosen technique. Two dominant strategies are binary lifting (also called jump pointers) and Euler tour + Range Minimum Query (RMQ). Binary lifting builds a 2^k ancestor table for every node, enabling the algorithm to lift the deeper node up to the same depth and then simultaneously jump both nodes upward in O(log N) steps until they converge. The Euler tour method flattens the tree into a depth‑first traversal array, records the first occurrence of each node, and reduces the LCA query to a minimum‑depth query over a sub‑array, which can be answered in O(1) after an O(N log N) RMQ preprocessing (or O(N) with a sparse table and the ±1 property). Both approaches dramatically reduce per‑query cost compared to the linear walk, making them suitable for massive query workloads typical in networking, compiler design, and hierarchical data analysis.

Interview Questions on This Problem

Q1How would you compute LCA for 10^5 queries on a static binary tree with 10^5 nodes in under 1 second?

Preprocess the tree using binary lifting: run a DFS to compute depth and a 2^k ancestor table for each node (k up to ⌊log2 N⌋). Each query then lifts the deeper node to the same depth and climbs both nodes together in O(log N) time, yielding overall O(N log N + Q log N) which fits the time limit.

Q2Explain the difference between binary lifting and Euler‑tour‑RMQ for LCA and when you would prefer one over the other.

Binary lifting stores O(N log N) jump pointers and answers each query in O(log N) without extra data structures, making it simple to implement and memory‑friendly for moderate N. Euler‑tour‑RMQ flattens the tree into an array of size 2N‑1 and reduces LCA to a range‑minimum query; with a sparse table it answers queries in O(1) after O(N log N) preprocessing, which is faster per query but uses more memory. Choose binary lifting when memory is constrained or when log‑factor per query is acceptable; choose Euler‑tour‑RMQ when you need constant‑time queries and can afford the extra array.

Q3A company wants to support dynamic edge insertions (adding new leaf nodes) while still answering LCA queries efficiently. Which LCA technique adapts best and why?

Binary lifting adapts more naturally to dynamic insertions because you can compute the new node’s ancestors on the fly using its parent’s jump table, updating O(log N) entries. Euler‑tour‑RMQ requires rebuilding the tour or updating the RMQ structure, which is costly for frequent insertions.

Examples

Example 1

Input

tree = [3, 5, 1, 6, 2, 0, 8, null, null, 7, 4], queries = [[6, 2], [5, 1], [7, 4]]

Output

[5, 3, 2]

Explanation: For query [6, 2]: The path from root 3 to 6 is 3->5->6. The path from root 3 to 2 is 3->5->2. The deepest common node is 5. For query [5, 1]: Node 5 is a child of 3, and node 1 is a child of 3. The common ancestor is 3. For query [7, 4]: Node 7 is a child of 2, and node 4 is a child of 2. The common ancestor is 2.

Example 2

Input

tree = [1, 2, 3, 4, 5, 6, 7], queries = [[4, 5], [6, 7], [1, 7]]

Output

[2, 3, 1]

Explanation: For query [4, 5]: Both 4 and 5 are children of 2. The LCA is 2. For query [6, 7]: Both 6 and 7 are children of 3. The LCA is 3. For query [1, 7]: Node 1 is the root and an ancestor of all nodes. The LCA is 1.

Example 3

Input

tree = [10, 20, 30, 40, 50, 60, 70, null, null, null, null, 80, 90, null, null, null, null], queries = [[80, 90], [40, 60], [50, 70]]

Output

[60, 20, 30]

Explanation: For query [80, 90]: 80 and 90 are children of 60. LCA is 60. For query [40, 60]: 40 is in the left subtree of 20, and 60 is in the right subtree of 30 (which is right of 10). The path to 40 is 10->20->40. The path to 60 is 10->30->60. The common ancestor is 10? Wait, 20 is left child of 10, 30 is right child of 10. So LCA is 10. Correction: Let's re-evaluate. 40 is child of 20. 60 is child of 30. 20 and 30 are children of 10. So LCA is 10. Let's adjust the example to be clearer. Let's use [40, 50]. 40 is child of 20. 50 is child of 20. LCA is 20. Let's use [40, 60]. LCA is 10. Let's use [80, 90]. LCA is 60. Let's use [40, 50]. LCA is 20. Let's use [60, 70]. LCA is 30. So queries = [[80, 90], [40, 50], [60, 70]]. Output = [60, 20, 30].

Constraints

1 <= number of nodes <= 10^5
1 <= number of queries <= 10^5
Node values are unique integers in the range [1, 10^9]
The tree is a valid binary tree
All query nodes exist in the tree

Optimal Approach & Strategy

Preprocess the tree with binary lifting (or Euler‑tour‑RMQ) to enable ancestor jumps in O(log N) (or O(1)) time, reducing total runtime to O(N log N + Q log N) (or O(N + Q)).

Brute Force Approach

For each query, repeatedly move the deeper node up one parent at a time until both nodes are at the same depth, then ascend both together until they match, costing O(N) per query.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N log N + Q log N)

function solution(nums) {
   let maxSum = nums[0];
   let currentSum = nums[0];
   for (let i = 1; i < nums.length; i++) {
       currentSum = Math.max(nums[i], currentSum + nums[i]);
       maxSum = Math.max(maxSum, currentSum);
   }
   return maxSum;
}

class Solution {
   public:
       int solution(vector<int>& nums) {
           int maxSum = nums[0];
           int currentSum = nums[0];
           for (int i = 1; i < nums.size(); i++) {
               currentSum = max(nums[i], currentSum + nums[i]);
               maxSum = max(maxSum, currentSum);
           }
           return maxSum;
       }
   };

class Solution {
   public int solution(int[] nums) {
       int maxSum = nums[0];
       int currentSum = nums[0];
       for (int i = 1; i < nums.length; i++) {
           currentSum = Math.max(nums[i], currentSum + nums[i]);
           maxSum = Math.max(maxSum, currentSum);
       }
       return maxSum;
   }
}

def solution(nums):
   max_sum = nums[0]
   current_sum = nums[0]
   for i in range(1, len(nums)):
       current_sum = max(nums[i], current_sum + nums[i])
       max_sum = max(max_sum, current_sum)
   return max_sum

function solution(nums) {
   let maxSum = nums[0];
   let currentSum = nums[0];
   for (let i = 1; i < nums.length; i++) {
       currentSum = Math.max(nums[i], currentSum + nums[i]);
       maxSum = Math.max(maxSum, currentSum);
   }
   return maxSum;
}

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Optimal Grid Path Protocol 4 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Trees

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Optimal Grid Path Protocol 4 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor