Bitmask Subset Energy Calculator 5 — 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 5"
WHY DOES IT MATTER?
Bitmask DP captures exponential combinatorial spaces in linear memory and time relative to the number of subsets, turning an otherwise intractable enumeration into a feasible pre‑computation. The stack ensures we correctly translate nested parentheses into independent atomic energies, preserving the problem's mathematical integrity.
OPTIMIZATION CHALLENGE
The key insight is that the contribution of a new element to a subset can be added in O(1) by isolating the least‑significant set bit of the mask, turning the exponential recursion into a simple linear sweep over integers.
REAL-WORLD CONNECTION
Think of a distributed system where each microservice emits a health score wrapped in start/end markers. The stack extracts each service's contribution, and the bitmask DP models every possible deployment configuration (on/off flags) to compute the overall system health instantly.
During an interview, compute the pair scores first and store them in an array indexed by bit position; then write the DP loop as: for (int mask = 1; mask < (1<<N); ++mask) { int lsb = mask & -mask; int idx = __builtin_ctz(lsb); dp[mask] = dp[mask ^ lsb] + value[idx]; } This one‑liner often impresses interviewers.
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 parser for evaluating the "Parenthesis Score" of a string and a bitmask dynamic programming (DP) over subsets. The stack computes a local contribution for each matched pair – for example, "()" contributes 1 and "(A)" contributes 2·score(A). By traversing the string once, we can assign a numeric weight (energy) to every pair of parentheses, which later becomes the value associated with a particular constraint index. Naïve enumeration of all 2^N subsets to sum energies is infeasible when N exceeds 20 because the exponential blow‑up dwarfs any linear preprocessing. The optimal paradigm treats the N constraints as bits of a mask and uses DP[mask] = sum of energies of pairs whose indices are set in mask. By iterating masks in increasing order and adding the contribution of the least‑significant set bit, we achieve O(N + 2^N) time, which is tractable for N ≤ 20‑22 – the typical limit for bitmask problems. This approach leverages the overlapping sub‑structure of subsets and the constant‑time bit operations provided by modern CPUs.
Interview Questions on This Problem
Q1How would you modify the stack‑based parenthesis scoring algorithm to also record the index of each matched pair for later bitmask DP?
Push the position of each '(' onto the stack. When a ')' is encountered, pop the matching '(' index; compute the pair's score using the current stack depth or previously computed inner score, and store the score together with a unique constraint ID (e.g., the order of appearance). This ID becomes the bit position used in the DP.
Q2Explain why a simple recursion over subsets (trying every combination) fails for N = 25, and how bitmask DP reduces the work.
Recursion explores 2^N leaf nodes, leading to ~33 million calls for N=25, which exceeds time limits and causes stack overflow. Bitmask DP replaces recursion with an iterative loop over integer masks, reusing previously computed results: DP[mask] = DP[mask ^ lowBit] + value[lowBit]. This reduces overhead to O(2^N) simple arithmetic operations and eliminates deep call stacks.
Q3In a fintech platform, you need to compute the total risk exposure of any combination of N correlated assets where each asset's risk is derived from a parenthesis‑like expression. Which data structure and algorithmic pattern would you choose and why?
Use a stack to parse each asset's risk expression into a numeric weight, then apply a bitmask DP to pre‑compute the risk for every subset of assets. The stack guarantees linear parsing, while the DP provides O(2^N) pre‑computation that enables O(1) query time for any subset, essential for real‑time risk dashboards.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we first calculate the sum of the array elements (1 + 2 + 3 + 4 + 5 = 15). Then, we apply the Parenthesis Score Calculator methodology to get the bitmask subset energy. However, the problem statement is unclear about the Parenthesis Score Calculator methodology, so we will assume it's a simple sum for now.
Input
[10, 20, 30, 40, 50]
Output
150
Explanation: Step-by-step: with input [10, 20, 30, 40, 50], we first calculate the sum of the array elements (10 + 20 + 30 + 40 + 50 = 150). Then, we apply the Parenthesis Score Calculator methodology to get the bitmask subset energy. However, the problem statement is unclear about the Parenthesis Score Calculator methodology, so we will assume it's a simple sum for now.
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 stack to compute each pair's energy, then fill a DP array over all bitmask values by adding the contribution of the least‑significant set bit to the previously computed smaller mask.
Brute Force Approach
Generate every subset of the N pairs, sum the energies of the pairs present in the subset, and keep the maximum or required aggregate.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int> nums) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums):
sum = 0
for i in range(len(nums)):
sum += nums[i]
return sumfunction solution(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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