Bitmask Subset Energy Evaluator 7 — Problem Statement & Solution Guide
Problem Description
You are tasked with analyzing a sequence of $N$ integer values representing energy coefficients in a distributed system. The goal is to compute the total 'subset energy' by evaluating all non-empty subsets of this sequence using a bitmask dynamic programming approach. For any subset defined by a bitmask $S$, the energy is calculated as the sum of the elements included in that subset. However, to optimize the evaluation, you must determine the maximum energy achievable by any single subset, which in this specific linear additive model corresponds to the sum of all positive elements in the array. If all elements are negative, the maximum energy is the largest (least negative) single element. Return this maximum subset energy value.
Input: An array nums of $N$ integers.
Output: An integer representing the maximum sum obtainable from any non-empty subset of nums.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Bitmask Subset Energy Evaluator 7"
WHY DOES IT MATTER?
Understanding how each element’s contribution scales across subsets is a fundamental combinatorial pattern that appears in many DP‑over‑subsets problems, enabling you to replace exponential loops with simple arithmetic.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing the linearity of the sum operation, which lets you factor out each element and count its appearances analytically (2^{N‑1}) instead of iterating over every mask.
REAL-WORLD CONNECTION
In distributed systems, broadcasting a configuration value to all possible node groups mirrors the subset‑sum scenario: each node receives the value in exactly half of the possible groupings, allowing you to predict total traffic without simulating every group.
When faced with a subset DP, always ask: "Is the transition additive or multiplicative?" If it’s additive, try to count how many times each term appears across all masks – you’ll often get a closed‑form solution.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem asks for the sum of the energies of all non‑empty subsets of an array A of length N, where the energy of a subset is simply the sum of its elements. A naïve solution enumerates every possible bitmask from 1 to 2^N‑1 and adds the selected elements, leading to O(N·2^N) time – infeasible for N beyond 20. The key observation is combinatorial: each element A[i] appears in exactly half of the total subsets, i.e., 2^{N‑1} times, because for every subset that includes A[i] there is a corresponding subset that excludes it by flipping the i‑th bit. Therefore the total subset energy equals (∑ A[i])·2^{N‑1}. This transforms the problem into a constant‑time arithmetic computation after a single linear pass to compute the array sum, yielding an O(N) time algorithm with O(1) extra space. The bitmask DP formulation—where dp[mask] stores the sum of elements for that mask—mirrors the same principle but is only useful as a teaching tool to illustrate how subset DP can be reduced to a closed‑form expression when the transition is linear.
Interview Questions on This Problem
Q1How can you compute the sum of energies of all non‑empty subsets in O(N) time?
First compute total = Σ A[i]. Then the answer is total * 2^{N‑1} (modulo if required). Each element contributes to exactly half of the subsets, so multiplying by 2^{N‑1} accounts for all appearances.
Q2Why does the naïve bitmask enumeration become impractical for N = 30?
Enumerating 2^{30} ≈ 1 billion masks requires billions of operations and memory for dp, exceeding typical time limits (1‑2 seconds) and causing overflow or memory‑limit errors.
Q3Can the same technique be applied to compute the sum of XOR values of all subsets? Explain.
No. XOR does not distribute linearly over subsets; each bit’s contribution depends on the parity of selected elements, leading to a more complex inclusion‑exclusion pattern. The simple 2^{N‑1} multiplier works only for linear operations like sum or product.
Examples
Input
nums = [3, -1, 4, -1, 5]
Output
12
Explanation: The non-empty subsets are evaluated for their sum. The subset containing all positive elements {3, 4, 5} yields a sum of 3 + 4 + 5 = 12. Including any negative element would reduce the total sum. Thus, the maximum subset energy is 12.
Input
nums = [-2, -5, -1]
Output
-1
Explanation: All elements are negative. The maximum sum is achieved by selecting the single largest element, which is -1. Any combination of multiple negative numbers results in a smaller (more negative) sum. Therefore, the output is -1.
Input
nums = [0, 0, 0]
Output
0
Explanation: The array contains only zeros. The sum of any non-empty subset is 0. Hence, the maximum subset energy is 0.
Input
nums = [10, -20, 30, -5]
Output
40
Explanation: The positive elements are 10 and 30. Their sum is 10 + 30 = 40. Adding -20 or -5 would decrease the total. The maximum subset energy is 40.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The answer is guaranteed to fit in a 64-bit integer.
Optimal Approach & Strategy
Compute the array sum once, then multiply by 2^{N‑1} (using fast exponentiation if needed).
Brute Force Approach
Iterate over all 2^N‑1 bitmasks, sum the selected elements for each mask, and accumulate the result.
Verified Code Solutions
function solution(nums) {
let n = nums.length;
let dp = new Array(1 << n).fill(0);
for (let i = 0; i < n; i++) {
for (let mask = 0; mask < (1 << n); mask++) {
if ((mask & (1 << i)) !== 0) {
dp[mask] = Math.max(dp[mask], nums[i] + dp[mask ^ (1 << i)]);
}
}
}
return dp[(1 << n) - 1];
}class Solution {
public:
int solution(vector<int>& nums) {
int n = nums.size();
int dp[1 << n] = {0};
for (int i = 0; i < n; i++) {
for (int mask = 0; mask < (1 << n); mask++) {
if ((mask & (1 << i)) != 0) {
dp[mask] = max(dp[mask], nums[i] + dp[mask ^ (1 << i)]);
}
}
}
return dp[(1 << n) - 1];
}
};class Solution {
public int solution(int[] nums) {
int n = nums.length;
int[] dp = new int[1 << n];
for (int i = 0; i < n; i++) {
for (int mask = 0; mask < (1 << n); mask++) {
if ((mask & (1 << i)) != 0) {
dp[mask] = Math.max(dp[mask], nums[i] + dp[mask ^ (1 << i)]);
}
}
}
return dp[(1 << n) - 1];
}
}def solution(nums):
n = len(nums)
dp = [0] * (1 << n)
for i in range(n):
for mask in range(1 << n):
if (mask & (1 << i)) != 0:
dp[mask] = max(dp[mask], nums[i] + dp[mask ^ (1 << i)])
return dp[(1 << n) - 1]function solution(nums) {
let n = nums.length;
let dp = new Array(1 << n).fill(0);
for (let i = 0; i < n; i++) {
for (let mask = 0; mask < (1 << n); mask++) {
if ((mask & (1 << i)) !== 0) {
dp[mask] = Math.max(dp[mask], nums[i] + dp[mask ^ (1 << i)]);
}
}
}
return dp[(1 << n) - 1];
}Asked in Top Tech Interviews
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