Pipeline Grid Optimizer 45 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and grid metrics, and an integer K, construct an optimal algorithm to compute the sum of numbers greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Grid Optimizer 45"
WHY DOES IT MATTER?
Processing streams with a queue avoids repeated passes, crucial for high‑throughput systems.
OPTIMIZATION CHALLENGE
The key is reducing per‑element work to constant time while maintaining only necessary state.
REAL-WORLD CONNECTION
Similar to filtering packets in a network router, where only packets exceeding a threshold are aggregated.
Keep the queue lightweight; often you can replace it with index pointers to avoid actual data movement.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The naive solution iterates through the entire sequence, checking each element against K and accumulating the sum, which runs in O(N) time but can become a bottleneck when the data stream is massive or arrives in real‑time, requiring repeated passes for multiple queries. By treating the input as a queue, we can process elements in a single pass, maintaining a running sum of qualifying values and discarding elements as they exit the window, thus preserving linear time while using constant auxiliary space.
When multiple K‑based queries are required, a pre‑processing step such as prefix sums or a balanced binary indexed tree can answer each query in O(log N) or O(1) after O(N) setup, dramatically reducing total runtime. This shift from repeated scanning to incremental aggregation exemplifies the optimal paradigm of streaming algorithms: compute results on the fly with minimal state, leveraging the FIFO nature of queues to avoid redundant work.
Interview Questions on This Problem
Q1Why is a single linear scan sufficient to compute the sum of elements greater than K?
Because each element is examined exactly once and we can update the sum on the spot. No element needs to be revisited, guaranteeing O(N) time.
Q2How would you adapt the solution if you needed to answer multiple K queries efficiently?
Pre‑compute a sorted list of elements with their prefix sums or use a Fenwick tree. Each query then becomes a binary search plus O(log N) sum retrieval.
Q3What edge case must you handle when the input sequence is empty or all elements are ≤ K?
The algorithm should return 0 for the sum. Ensure the loop handles zero‑length input without errors.
Examples
Input
[1, 2, 3, 4, 5]
Output
9
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we filter numbers greater than K (let's say K = 3), giving output 9. We sum 4 and 5 which are greater than 3.
Input
[10, 20, 30, 40, 50]
Output
150
Explanation: Step-by-step: with input [10, 20, 30, 40, 50], we filter numbers greater than K (let's say K = 30), giving output 150. We sum 40 and 50 which are greater than 30.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Perform a single pass, updating a running sum for elements > K, achieving O(N) total time and O(1) extra space.
Brute Force Approach
Loop through the array for each query, checking every element against K and summing matches, leading to O(N) per query.
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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