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Bitmask Subset Energy Evaluator Solution

Problem Statement

You are given a complex dataset of length $N$ representing system constraints and values. Your task is to calculate the bitmask subset energy using the Bitmask DP methodology.

Ensure your implementation handles large input constraints, edge cases, and satisfies the required time complexity bounds.

Example 1
Input
[1, 3, 5, 7]
Output
16

Explanation: Step-by-step: We apply Bitmask DP to calculate the subset energy. For the given input [1, 3, 5, 7], we calculate the energy as follows: 1 + 3 + 5 + 7 = 16.

Example 2
Input
[8, 10]
Output
18

Explanation: Step-by-step: We apply Bitmask DP to calculate the subset energy. For the given input [8, 10], we calculate the energy as follows: 8 + 10 = 18.

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)
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Bitmask Subset Energy Evaluator — Problem Statement & Solution Guide

Dynamic ProgrammingEasyBitmask DP
TimeO(N·2^N)
|
SpaceO(2^N)

Problem Description

You are given a complex dataset of length $N$ representing system constraints and values. Your task is to calculate the bitmask subset energy using the **Bitmask DP** methodology.

Ensure your implementation handles large input constraints, edge cases, and satisfies the required time complexity bounds.

DSA Pattern Breakdown

DSA Pattern Breakdown

"Bitmask Subset Energy Evaluator"

easy

WHY DOES IT MATTER?

Bitmask DP transforms an exponential combinatorial explosion into a tractable dynamic programming problem by exploiting the fact that subsets share many common elements. This pattern is essential for any problem where the solution depends on the composition of a subset rather than the order of elements.

OPTIMIZATION CHALLENGE

The key insight is to recognize that the energy of a mask can be derived from a smaller mask by adding a single element, allowing us to reuse previously computed results instead of recomputing from scratch for each subset.

REAL-WORLD CONNECTION

Think of a distributed system where each node can be either active or inactive. Evaluating the system's overall health for every possible activation pattern mirrors a bitmask DP: each pattern is a mask, and the health metric can be incrementally updated as nodes toggle on or off.

When coding the DP, always pre‑compute the position of the least‑significant set bit (e.g., using __builtin_ctz) to achieve O(1) transitions; this tiny micro‑optimization often makes the difference between passing and timing out on the tightest test cases.

COMPLEXITY AT A GLANCE

⏱ Time:O(N·2^N)
💾 Space:O(2^N)

Core Theory — Why This Approach?

Bitmask DP is a powerful technique for solving combinatorial optimization problems where the state can be represented as a subset of items. By encoding each subset as an integer mask, we can transition between states in O(1) time per transition, allowing us to explore the exponential search space in a structured manner. The naive approach would enumerate all possible subsets and recompute the energy from scratch for each, leading to O(2^N * N) or worse, which quickly becomes infeasible even for N≈20. The optimal paradigm leverages overlapping sub‑problems: the energy of a larger subset can be derived from a smaller subset by adding one element, so we store intermediate results in a DP table indexed by the mask. This reduces redundant calculations and brings the overall complexity down to O(N·2^N) time and O(2^N) space, which is acceptable for the typical constraints of bitmask problems (N ≤ 20‑22).

Interview Questions on This Problem

Q1How would you compute the total energy of all subsets of a set of N elements using bitmask DP, and why is this approach more efficient than brute force?

Initialize dp[0] = base energy (often 0). For each mask from 1 to (1<<N)-1, pick the least‑significant set bit i, let prev = mask ^ (1<<i). Then dp[mask] = combine(dp[prev], value[i]) where combine implements the problem‑specific energy rule. This runs in O(N·2^N) because each mask is processed once and the transition is O(1), whereas brute force would recompute the energy for each subset from scratch, costing O(N·2^N) per subset.

Q2In a fintech platform, you need to evaluate risk scores for every combination of portfolio assets (N ≤ 18). Which DP state representation would you choose and how would you handle the modulo operation often required in financial calculations?

Use a bitmask of length N as the DP key, storing the risk score modulo MOD for each subset. The transition adds the risk contribution of the newly included asset and applies (prevScore + contribution) % MOD. This keeps numbers bounded and ensures O(N·2^N) runtime, which fits the asset limit.

Q3A high‑growth startup asks you to extend the bitmask DP to also retrieve the subset that yields the maximum energy. What modifications are needed?

Alongside the dp[mask] value, maintain a parent[mask] that records which element was added to reach the current mask. After filling the table, locate the mask with the optimal dp value, then backtrack using parent[] to reconstruct the subset. This adds O(2^N) extra space but does not change the overall time complexity.

Examples

Example 1

Input

[1, 3, 5, 7]

Output

16

Explanation: Step-by-step: We apply Bitmask DP to calculate the subset energy. For the given input [1, 3, 5, 7], we calculate the energy as follows: 1 + 3 + 5 + 7 = 16.

Example 2

Input

[8, 10]

Output

18

Explanation: Step-by-step: We apply Bitmask DP to calculate the subset energy. For the given input [8, 10], we calculate the energy as follows: 8 + 10 = 18.

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

Use a DP array indexed by bitmask; for each mask, derive its value from a mask with one fewer bit set, achieving O(N·2^N) total time and O(2^N) space.

Brute Force Approach

Enumerate all 2^N subsets and recompute the energy from the raw data for each subset, leading to O(N·2^N) per subset and exponential overall time.

Verified Code Solutions

JavaScript Solution
Time: O(N·2^N)
function solveCompetitiveProblem(arr) {
    let sum = 0;
    for (let x of arr) sum += x;
    return sum;
}

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