Protocol Pipeline Aligner 10 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing protocol and pipeline metrics, construct an optimal algorithm to evaluate and compute the target aligner value under given operational constraints. The input array is of integers and the value of K is provided.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Pipeline Aligner 10"
WHY DOES IT MATTER?
Sliding‑window bit‑frequency patterns turn expensive recomputation into constant‑time updates.
OPTIMIZATION CHALLENGE
The key is reducing per‑window work from O(K) to O(1) by exploiting bit‑level invariants.
REAL-WORLD CONNECTION
Network packet aggregators often need to maintain bit‑flags over moving time windows, mirroring this technique.
Cache the bit counts in a fixed‑size array and update them incrementally; avoid rebuilding the OR from scratch.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to evaluating the bitwise OR over sliding windows of length K. A naive solution recomputes the OR from scratch for each window, leading to O(N·K) time, which quickly becomes prohibitive for N up to 10^5. The optimal paradigm leverages the fact that OR is monotonic: once a bit is set in the window it stays set until all elements containing that bit exit the window. By maintaining a frequency count of each bit (0‑31 for 32‑bit integers) we can update the window in O(1) per step, achieving O(N·B) where B is the number of bits, effectively O(N). This bit‑frequency sliding‑window technique transforms a quadratic‑ish brute force into a linear‑time solution while using only O(B) extra space.
Interview Questions on This Problem
Q1Why can we update a sliding‑window OR in O(1) instead of recomputing it each time?
Because OR is idempotent and monotonic; we only need to track how many numbers in the window contribute each bit.
Q2What is the worst‑case time complexity of the optimized solution and why?
O(N) because each element enters and leaves the window exactly once and each bit update is O(1) for a fixed 32‑bit integer.
Q3How would the algorithm change if the operation were bitwise AND instead of OR?
For AND we must track the count of zeros per bit, because a bit stays 1 only while all elements have it set; the update logic flips accordingly.
Examples
Input
[10, 20, 30, 40, 50, 25, 5]
Output
130
Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50, 25, 5], we need to find the sum of elements greater than K (25). We iterate through the array and add 30, 40, 50 to the sum, giving us a total of 130.
Input
[5, 10, 15, 20, 25]
Output
120
Explanation: Step-by-step: Given the input array [5, 10, 15, 20, 25], we need to find the sum of elements greater than K (15). We iterate through the array and add 20, 25 to the sum, giving us a total of 45. However, the problem statement asks for the sum of elements greater than K, so we should ignore the elements 5, 10, 15. Therefore, the correct output is 45, not 120.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a per‑bit frequency array while sliding the window, updating counters in O(1) per move and recomputing the OR from the counters.
Brute Force Approach
Compute the OR for every possible window of size K by iterating K elements each time, resulting in O(N·K) 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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