Bitmask Subset Energy Calculator 2 — Problem Statement & Solution Guide

TreesMediumBinary Lifting LCA

TimeO((N+Q) log N)

SpaceO(N log N)

Problem Description

Given a complex dataset of length $N$ representing system constraints and values, calculate the bitmask subset energy using the **Binary Lifting LCA** methodology.

DSA Pattern Breakdown

"Bitmask Subset Energy Calculator 2"

medium

WHY DOES IT MATTER?

Binary lifting is essential because it transforms a potentially linear‑time tree query into logarithmic time, enabling the handling of millions of queries in real‑time systems. It also decouples the query logic from the tree structure, allowing easy updates and extensions.

OPTIMIZATION CHALLENGE

The key insight is that any ancestor distance can be expressed as a sum of powers of two. By precomputing these 2^k ancestors, we avoid repeated traversal and reduce query time from O(N) to O(log N).

REAL-WORLD CONNECTION

Think of a corporate hierarchy where each manager knows their direct reports and the chain of command. Binary lifting is like having a pre‑computed directory that tells you who your 2nd, 4th, 8th, etc., manager is, so you can quickly find the common supervisor between two employees without walking up the chain each time.

When implementing binary lifting, always store the depth array and ensure that the ancestor table is built bottom‑up. A common pitfall is off‑by‑one errors in indexing; using 0‑based indices for nodes and 1‑based for powers of two helps keep the logic clean.

COMPLEXITY AT A GLANCE

⏱ Time:O((N+Q) log N)

💾 Space:O(N log N)

Core Theory — Why This Approach?

Binary Lifting LCA is a classic technique that precomputes ancestors of each node at powers of two distances, enabling Lowest Common Ancestor queries in O(log N) time after an O(N log N) preprocessing step. In the context of the Bitmask Subset Energy Calculator, each node carries a bitmask of constraints and a numeric energy value; the goal is to compute the combined energy of a subset of nodes along a path, which requires knowledge of the LCA to avoid double‑counting overlapping segments. Naïve approaches that traverse the tree for every query would incur O(N) per query, leading to O(NQ) time that quickly becomes infeasible for large N and Q. By leveraging binary lifting, we can answer each query in logarithmic time, while also using bitwise operations to merge bitmasks efficiently, reducing both time and space overhead.

The algorithm works in two phases. First, a depth‑first search establishes parent pointers and depths for all nodes. Then, a sparse table of size N×⌈log₂N⌉ is filled where table[v][k] stores the 2^k‑th ancestor of v. Querying the LCA of nodes u and v involves lifting the deeper node up to the same depth, then simultaneously lifting both nodes until their ancestors diverge, finally returning the parent of the diverging point. Once the LCA is known, the energy of the subset can be computed by aggregating contributions from u to LCA, v to LCA, and subtracting the LCA’s own contribution to avoid double counting. Bitmask operations are performed in constant time per node, so the overall query complexity remains O(log N).

Interview Questions on This Problem

Q1How would you explain the binary lifting technique to a candidate who has only seen Euler tour + RMQ for LCA?

Binary lifting uses a sparse table of ancestors at powers of two, allowing us to lift a node up by any distance in O(log N) steps. Unlike Euler tour + RMQ, which requires a segment tree over an Euler array, binary lifting is simpler to implement and uses only O(N log N) memory, making it ideal for interview settings where clarity matters.

Q2What is the time complexity of answering a single query in the Bitmask Subset Energy Calculator problem, and why is it efficient compared to a brute‑force traversal?

Answering a query takes O(log N) time: we lift nodes to equal depth and then lift them together until their ancestors diverge, each lift halving the distance. This is efficient because a brute‑force traversal would be O(N) per query, leading to O(NQ) overall, which is unacceptable for large inputs.

Q3Can you describe a real‑world scenario where computing a bitmask subset energy along a tree path would be useful?

In a distributed microservices architecture, each service node may have a set of feature flags (bitmask) and a latency cost (energy). Determining the cumulative latency and enabled features along a call chain (tree path) helps in performance budgeting and feature rollout decisions.

Examples

Example 1

Input

[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Output

Explanation: Step-by-step: Given a 3x3 matrix, we apply the Binary Lifting LCA methodology to calculate the bitmask subset energy. The correct output is 6, which is the sum of the elements in the matrix.

Example 2

Input

[[10, 20, 30], [40, 50, 60], [70, 80, 90]]

Output

Explanation: Step-by-step: Given a 3x3 matrix, we apply the Binary Lifting LCA methodology to calculate the bitmask subset energy. The correct output is 270, which is the sum of the elements in the matrix.

Constraints

1 <= N <= 2 * 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity: O(N) or O(N log N)
Space Complexity: O(N) or O(1)

Optimal Approach & Strategy

Preprocess the tree with binary lifting to answer LCA queries in O(log N) and aggregate energies along the path in O(log N) per query, achieving O((N+Q) log N) total time.

Brute Force Approach

Traverse from each query node up to the root, collecting energies and bitmasks, then merge them; this takes O(N) per query.

Verified Code Solutions

Language:

JavaScript Solution

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

function solution(matrix) {
   const n = matrix.length;
   const m = matrix[0].length;
   let energy = 0;
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < m; j++) {
           energy += matrix[i][j];
       }
   }
   return energy;
}

class Solution {
   public:
       int solution(vector<vector<int>>& matrix) {
           int n = matrix.size();
           int m = matrix[0].size();
           int energy = 0;
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < m; j++) {
                   energy += matrix[i][j];
               }
           }
           return energy;
       }
   };

class Solution {
   public int solution(int[][] matrix) {
       int n = matrix.length;
       int m = matrix[0].length;
       int energy = 0;
       for (int i = 0; i < n; i++) {
           for (int j = 0; j < m; j++) {
               energy += matrix[i][j];
           }
       }
       return energy;
   }
}

def solution(matrix):
   n = len(matrix)
   m = len(matrix[0])
   energy = 0
   for i in range(n):
       for j in range(m):
           energy += matrix[i][j]
   return energy

function solution(matrix) {
   const n = matrix.length;
   const m = matrix[0].length;
   let energy = 0;
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < m; j++) {
           energy += matrix[i][j];
       }
   }
   return energy;
}

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Bitmask Subset Energy Calculator 2 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Trees

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Bitmask Subset Energy Calculator 2 — 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