Pipeline Beacon Consolidator 40 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and beacon metrics, construct an optimal algorithm to evaluate and compute the target consolidator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Beacon Consolidator 40"
WHY DOES IT MATTER?
The "maximum bitwise AND" pattern surfaces whenever you need to preserve the most significant common features across a dataset—be it security flags, feature toggles, or error codes. Mastering this pattern lets engineers extract high‑value signals without exhaustive pairwise checks.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that you can treat each bit position independently and work from the top down. By filtering the candidate pool with a running mask, you eliminate the majority of numbers after just a few iterations, collapsing O(N²) to O(N · log MAX).
REAL-WORLD CONNECTION
Think of a fleet of autonomous drones broadcasting status beacons. Each beacon encodes capabilities as bits. The consolidator value represents the strongest set of capabilities shared by any two drones, which is crucial for forming reliable peer‑to‑peer clusters in a distributed control plane.
During an interview, start by stating the greedy mask idea, then quickly sketch the loop that builds the mask from the 31st bit down to 0, checking counts with a hash set or bitset. This shows you understand both the theory and its practical implementation.
COMPLEXITY AT A GLANCE
O(N · log MAX)O(N)Core Theory — Why This Approach?
Bit manipulation lies at the heart of many low‑level performance problems, especially when dealing with large streams of telemetry like pipeline and beacon metrics. The naive way to compute a consolidator value—often defined as the maximum possible bitwise AND (or XOR) between any two elements—requires examining every pair, leading to O(N²) time which quickly becomes infeasible for N in the millions. The optimal paradigm leverages the monotonic nature of bits: higher‑order bits dominate the result, so we can prune the search space by focusing on numbers that share those bits. By constructing a bit‑wise trie (or using a greedy mask‑filtering loop) we can isolate candidate numbers in O(N · log MAX) time, where MAX is the largest possible integer (typically 2³¹‑1). This approach reduces both time and auxiliary space while preserving exactness, making it ideal for real‑time pipeline analytics.
Interview Questions on This Problem
Q1How would you find the maximum bitwise AND of any two numbers in an array of up to 10⁶ integers?
Iterate from the most significant bit to the least, maintaining a set of numbers that have the current prefix set. At each step, test if at least two numbers share the prefix; if so, keep the prefix, otherwise drop the bit. This greedy mask technique runs in O(N · log MAX) time and O(N) space.
Q2Explain why a brute‑force O(N²) solution is unacceptable for the Pipeline Beacon Consolidator problem in a fintech platform handling high‑frequency trade data.
Fintech systems process millions of events per second; an O(N²) algorithm would require quadratically many comparisons, leading to latency spikes and potential SLA violations. The quadratic growth outpaces hardware improvements, making it impractical for real‑time risk calculations.
Q3Describe a scenario in a distributed system where using a bitwise trie to compute a consolidator value can reduce network traffic.
When aggregating sensor flags across nodes, each node can send only the bits that contribute to the current mask. By building a shared trie, nodes prune irrelevant branches early, transmitting only the minimal subset needed to compute the global maximum AND, thus cutting bandwidth by orders of magnitude.
Examples
Input
[1, 2, 3, 4, 5], 10
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and K = 10, we iterate through the array. When the sum exceeds K, we add the current sum to the result. So, the result is 1 + 2 + 3 + 4 + 5 = 15.
Input
[10, 20, 30, 40, 50], 100
Output
150
Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and K = 100, we iterate through the array. When the sum exceeds K, we add the current sum to the result. So, the result is 10 + 20 + 30 + 40 + 50 = 150.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Iterate bits from high to low, maintain a mask of candidate bits, and count numbers matching the mask; set a bit only if at least two numbers qualify, achieving O(N · log MAX) time.
Brute Force Approach
Check every possible pair of elements and compute their bitwise AND, tracking the maximum; this requires O(N²) time.
Verified Code Solutions
function solution(nums, K) {
let result = 0;
let currentSum = 0;
for (let num of nums) {
currentSum += num;
if (currentSum > K) {
result += currentSum;
currentSum = 0;
}
}
if (currentSum > 0) {
result += currentSum;
}
return result;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int result = 0;
int currentSum = 0;
for (int num : nums) {
currentSum += num;
if (currentSum > K) {
result += currentSum;
currentSum = 0;
}
}
if (currentSum > 0) {
result += currentSum;
}
return result;
}
};class Solution {
public int solution(int[] nums, int K) {
int result = 0;
int currentSum = 0;
for (int num : nums) {
currentSum += num;
if (currentSum > K) {
result += currentSum;
currentSum = 0;
}
}
if (currentSum > 0) {
result += currentSum;
}
return result;
}
}def solution(nums, K):
result = 0
current_sum = 0
for num in nums:
current_sum += num
if current_sum > K:
result += current_sum
current_sum = 0
if current_sum > 0:
result += current_sum
return resultfunction solution(nums, K) {
let result = 0;
let currentSum = 0;
for (let num of nums) {
currentSum += num;
if (currentSum > K) {
result += currentSum;
currentSum = 0;
}
}
if (currentSum > 0) {
result += currentSum;
}
return result;
}Asked in Top Tech Interviews
Solve in Interative Editor
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