Pipeline Vector Tracker 44 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and vector metrics, construct an optimal algorithm to evaluate and compute the target tracker value under given operational constraints. The target tracker value is the sum of all elements less than or equal to 5 in the input array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Vector Tracker 44"
WHY DOES IT MATTER?
Filtering and aggregating data is a fundamental pattern for analytics and monitoring pipelines.
OPTIMIZATION CHALLENGE
The key is to avoid extra passes or data structures, reducing the problem to a single O(n) traversal.
REAL-WORLD CONNECTION
Think of a telemetry system that sums sensor readings below a safety threshold.
Use a running total and early exit for empty inputs to keep the code tight and performant.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single linear scan where each element is examined once to decide if it contributes to the sum. This is a classic example of a filter‑then‑aggregate pattern, which can be solved in O(n) time using constant extra space.
A naive approach might sort the array or use nested loops to compare each element with a threshold, leading to O(n log n) or O(n^2) complexities that become prohibitive for large inputs. The optimal paradigm leverages direct iteration and conditional accumulation, eliminating unnecessary work and ensuring scalability.
Interview Questions on This Problem
Q1What is the time and space complexity of the optimal solution?
The time complexity is O(n) because each element is visited once. The space complexity is O(1) as only a few scalar variables are used.
Q2How would you modify the algorithm to count elements instead of summing them?
Replace the accumulation variable with a counter that increments when the condition holds. The rest of the scan remains unchanged, preserving O(n) time and O(1) space.
Q3Why is sorting not a good strategy for this problem?
Sorting adds O(n log n) overhead, which is unnecessary when a simple linear pass suffices. It also consumes extra memory for the sorted copy in many implementations.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we sum all elements less than or equal to 5, which are 1, 2, 3, 4, and 5, giving output 15.
Input
[1, 2, 3]
Output
6
Explanation: Step-by-step: with input [1, 2, 3], we sum all elements less than or equal to 3, which are 1, 2, and 3, giving output 6.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Iterate once, conditionally add qualifying elements to a running sum, achieving O(n) time and O(1) space.
Brute Force Approach
A brute-force method might sort the array then sum the prefix, which adds unnecessary O(n log n) time.
Verified Code Solutions
function solution(nums) {
if (!nums || nums.length === 0) {
return 0;
}
let sum = 0;
for (let num of nums) {
if (typeof num !== 'number') {
throw new Error('Input array should only contain numbers');
}
if (num <= 5) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.empty()) {
return 0;
}
int sum = 0;
for (int num : nums) {
if (num <= 5) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int sum = 0;
for (int num : nums) {
if (num <= 5) {
sum += num;
}
}
return sum;
}
}def solution(nums):
if not nums:
return 0
total = 0
for num in nums:
if not isinstance(num, (int, float)):
raise ValueError('Input array should only contain numbers')
if num <= 5:
total += num
return totalfunction solution(nums) {
if (!nums || nums.length === 0) {
return 0;
}
let sum = 0;
for (let num of nums) {
if (typeof num !== 'number') {
throw new Error('Input array should only contain numbers');
}
if (num <= 5) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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