Protocol Sensor Synthesizer 24 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing protocol and sensor metrics, construct an optimal algorithm to evaluate and compute the target synthesizer value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Sensor Synthesizer 24"
WHY DOES IT MATTER?
Bit‑wise aggregation patterns turn exponential combinatorial checks into linear scans.
OPTIMIZATION CHALLENGE
The key is to avoid recomputing the same bit interactions, reducing O(n²) to O(n·log C) or O(n).
REAL-WORLD CONNECTION
Network protocol stacks combine header flags using masks, and sensor fusion pipelines merge bit‑encoded readings in real time.
Pre‑allocate fixed‑size structures (e.g., 32‑node trie) and reuse them across test cases to eliminate allocation overhead.
COMPLEXITY AT A GLANCE
O(n·log C)O(C)Core Theory — Why This Approach?
The problem reduces to extracting a global bit‑wise signature from a sequence of integers, often expressed as the maximum XOR, minimum AND, or a custom combination of bit masks. Naïve pairwise evaluation incurs O(n²) time because each element must be compared with every other, which quickly exceeds limits for n up to 10⁵ or more. The optimal paradigm leverages the linearity of XOR (or AND/OR) and builds a prefix‑based data structure—such as a binary trie for XOR or a running mask for AND/OR—allowing each new element to be merged in O(1) or O(log C) where C is the word size (≤ 32/64). This transforms the problem into a single pass, preserving the exact bit‑wise relationships while dramatically cutting the runtime.
Interview Questions on This Problem
Q1Why does a binary trie enable O(log C) maximum XOR computation instead of O(n²)?
Each number is inserted once, and querying the trie walks the opposite bit at each level to maximize the XOR, giving logarithmic depth per element.
Q2How would you compute the cumulative AND of a sliding window in O(1) per move?
Maintain a count of set bits for each position; when the window slides, decrement counts for outgoing bits and increment for incoming bits, recomputing the AND from the counts.
Q3What edge case breaks a naïve bit‑mask accumulation when the array contains zeros?
Zeros clear all bits in an AND chain, so a simple cumulative mask without reset logic yields an incorrect permanent zero result.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 5
Output
45
Explanation: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 5, we first filter out all values less than or equal to K. This gives us [6, 7, 8, 9, 10]. Then, we sum up all these values to get the target synthesizer value, which is 6 + 7 + 8 + 9 + 10 = 40.
Input
[10, 20, 30, 40, 50], 30
Output
150
Explanation: Given the input array [10, 20, 30, 40, 50] and K = 30, we first filter out all values less than or equal to K. This gives us [40, 50]. Then, we sum up all these values to get the target synthesizer value, which is 40 + 50 = 90.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Insert each element into a binary trie (for XOR) or maintain a running mask (for AND/OR) while scanning once, achieving O(n·log C) or O(n) time.
Brute Force Approach
Iterate over all pairs (or all sub‑segments) and compute the required bit operation directly, leading to O(n²) time.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (num > k) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
}def solution(nums, k):
sum = 0
for num in nums:
if num > k:
sum += num
return sumfunction solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (num > k) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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