Protocol Pipeline Tracker 50 — 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 tracker value under given operational constraints. The algorithm should sum all elements greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Pipeline Tracker 50"
WHY DOES IT MATTER?
This pattern is essential for understanding basic data filtering and aggregation, which are core operations in data processing pipelines. It forms the foundation for more complex algorithms involving conditional sums, threshold-based filtering, and statistical computations.
OPTIMIZATION CHALLENGE
The key insight is recognizing that for unsorted data, a single pass is optimal. However, if the data is static and queries are frequent, preprocessing with sorting and prefix sums can significantly reduce query time. The challenge lies in choosing the right strategy based on the access pattern (one-time vs. repeated queries).
REAL-WORLD CONNECTION
In financial systems, this pattern is used to calculate the total value of transactions exceeding a certain amount for risk assessment. In network monitoring, it helps identify the total bandwidth usage of connections that exceed a specific threshold, aiding in capacity planning.
During an interview, clarify whether the data is static or dynamic and whether the query is one-time or repeated. This determines whether a simple linear scan or a more complex preprocessing approach is appropriate. Always mention the trade-offs between preprocessing time and query time.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem of summing elements greater than a threshold K in an unsorted array is a fundamental linear scan operation. While it appears trivial, it serves as a baseline for understanding data filtering and aggregation in high-throughput systems. The naive approach involves iterating through the entire sequence once, checking each element against K, and accumulating the sum if the condition is met. This approach is optimal for unsorted data because any attempt to skip elements requires prior knowledge of their values, which is not available without sorting or indexing.
Interview Questions on This Problem
Q1How would you modify this algorithm if the input array were sorted in ascending order?
If the array is sorted, you can use binary search to find the first index where the element is greater than K. Then, you only need to sum the elements from that index to the end of the array. This reduces the time complexity from O(N) to O(log N + M), where M is the number of elements greater than K, which is beneficial if M is significantly smaller than N.
Q2In a distributed system where data is partitioned across multiple nodes, how would you parallelize this computation?
You can partition the array into chunks and process each chunk in parallel on different threads or nodes. Each worker computes the local sum of elements greater than K. Finally, a reduction step aggregates the partial sums from all workers. This approach leverages map-reduce patterns and can achieve near-linear speedup with sufficient parallelism.
Q3What if the threshold K changes dynamically during the execution, and you need to query the sum for different K values frequently?
For frequent queries with varying K, you should preprocess the data. Sort the array and compute a prefix sum array. For each query K, use binary search to find the position of K in the sorted array and then calculate the sum of elements from that position to the end using the prefix sums. This reduces each query to O(log N) time after an O(N log N) preprocessing step.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
Output
45
Explanation: Step 1: Initialize two pointers, one at the start and one at the end of the array. Step 2: Iterate through the array, summing all elements. Step 3: Initialize a variable to store the sum of elements less than or equal to K. Step 4: Iterate through the array again, summing all elements less than or equal to K. Step 5: Subtract the sum of elements less than or equal to K from the total sum to get the final answer.
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step 1: Initialize two pointers, one at the start and one at the end of the array. Step 2: Iterate through the array, summing all elements. Step 3: Initialize a variable to store the sum of elements less than or equal to K. Step 4: Iterate through the array again, summing all elements less than or equal to K. Step 5: Subtract the sum of elements less than or equal to K from the total sum to get the final answer.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
For unsorted data, the linear scan is already optimal. For sorted data or frequent queries, preprocess the array by sorting it and computing prefix sums, then use binary search to find the relevant range and calculate the sum in O(log N) time per query.
Brute Force Approach
Iterate through the array, check each element against K, and sum the elements that satisfy the condition. This approach is straightforward and efficient for unsorted data.
Verified Code Solutions
function solution(nums, K) {
let totalSum = 0;
for (let num of nums) {
totalSum += num;
}
let sumLessThanK = 0;
for (let num of nums) {
if (num <= K) {
sumLessThanK += num;
}
}
return totalSum - sumLessThanK;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int totalSum = 0;
for (int num : nums) {
totalSum += num;
}
int sumLessThanK = 0;
for (int num : nums) {
if (num <= K) {
sumLessThanK += num;
}
}
return totalSum - sumLessThanK;
}
};class Solution {
public int solution(int[] nums, int K) {
int totalSum = 0;
for (int num : nums) {
totalSum += num;
}
int sumLessThanK = 0;
for (int num : nums) {
if (num <= K) {
sumLessThanK += num;
}
}
return totalSum - sumLessThanK;
}
}def solution(nums, K):
total_sum = 0
for num in nums:
total_sum += num
sum_less_than_k = 0
for num in nums:
if num <= K:
sum_less_than_k += num
return total_sum - sum_less_than_kfunction solution(nums, K) {
let totalSum = 0;
for (let num of nums) {
totalSum += num;
}
let sumLessThanK = 0;
for (let num of nums) {
if (num <= K) {
sumLessThanK += num;
}
}
return totalSum - sumLessThanK;
}Asked in Top Tech Interviews
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