Bitmask Subset Energy Calculator 3 — Problem Statement & Solution Guide
Problem Description
Given a complex dataset of length N representing system constraints and values, calculate the bitmask subset energy using the Parenthesis Score Calculator methodology.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Bitmask Subset Energy Calculator 3"
WHY DOES IT MATTER?
Bitmask DP is a cornerstone technique for exponential‑size state spaces where the problem exhibits optimal substructure across subsets. It turns otherwise intractable brute‑force enumeration into a systematic, memoized process, enabling interviewers to assess a candidate's ability to reason about state compression and transition design.
OPTIMIZATION CHALLENGE
The key insight is that the parentheses score of a superset can be derived by adjusting the score of its immediate sub‑mask with the contribution of the newly added character, which can be pre‑computed using a single stack pass. This eliminates the need for a full re‑scan of the string for each subset.
REAL-WORLD CONNECTION
Think of a distributed microservice architecture where each service can be toggled on/off. The overall system latency (energy) depends on which services are active and how they call each other. Computing the latency for every configuration efficiently mirrors the bitmask subset energy calculation.
When implementing, first run a linear stack pass to record for each position the index of its matching partner and the base score of that pair. Store these in arrays; then the DP transition becomes a simple lookup, drastically reducing constant factors.
COMPLEXITY AT A GLANCE
O(N·2^N)O(2^N)Core Theory — Why This Approach?
The problem blends two classic algorithmic ideas: a stack‑based evaluation of well‑formed parentheses (the "Parenthesis Score Calculator") and a bitmask DP that enumerates subsets of the input sequence. The stack computes the local contribution of each matched pair – typically 1 for "()" and 2 × innerScore for nested structures – in linear time. Naïvely, to obtain the energy of every possible subset you would generate all 2^N subsets and, for each, re‑run the stack algorithm, leading to O(N·2^N) time but with a huge constant factor and repeated work. The optimal paradigm observes that the score of a superset can be built incrementally from its sub‑masks: when a new element is added, only the interactions that involve the newly added position need to be recomputed. By storing the score of each mask in a DP array and updating it using the pre‑computed contribution of the added character (derived from the stack), we achieve O(N·2^N) overall time with a single pass per mask and O(2^N) space. This reduction eliminates the inner O(N) recomputation for each subset, turning an exponential‑times‑exponential blow‑up into a manageable exponential DP suitable for N ≤ 20‑22.
Interview Questions on This Problem
Q1How would you compute the score of a parentheses string using a stack in O(N) time?
Traverse the string, push a sentinel for '('; when encountering ')', pop the top. If the popped element is a sentinel, push 1 (score of "()"); otherwise, sum all popped scores until a sentinel, double the sum (for nesting), and push the result back. The final stack sum is the total score.
Q2Explain how bitmask DP can be used to evaluate a function over all subsets of a sequence.
Represent each subset as an integer mask where bit i indicates inclusion of element i. Initialize DP[0] = base case. For each mask, iterate over its set bits; for each bit i, transition from mask ^ (1<<i) to mask by adding the contribution of element i, possibly using pre‑computed interaction data. This builds solutions for larger masks from smaller ones in O(N·2^N) time.
Q3Why is it unsafe to recompute the parentheses score from scratch for every subset, and how does the incremental DP avoid this pitfall?
Recomputing from scratch repeats the same stack work for overlapping subsets, leading to O(N·2^N) with a large hidden constant and potential TLE for N≈20. The incremental DP reuses previously computed scores: when adding a new character to a subset, only the local effect of that character (and its immediate matches) changes, so we update the stored score in O(1) or O(depth) instead of O(N).
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5], we calculate the sum of its elements: 1 + 2 + 3 + 4 + 5 = 15.
Input
[10, 20, 30, 40, 50]
Output
150
Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50], we calculate the sum of its elements: 10 + 20 + 30 + 40 + 50 = 150.
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
Pre‑compute matching indices and base scores with one stack pass, then fill a DP array over bitmask subsets, updating each mask from its sub‑mask in O(1) per added element, achieving O(N·2^N) overall.
Brute Force Approach
Generate every subset (2^N) and for each run the full stack‑based score computation, resulting in O(N·2^N) time with a large hidden constant.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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