Node Payload Architect 44 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing node and payload metrics, construct an optimal algorithm to evaluate and compute the target architect value under given operational constraints. The algorithm should sum all elements in the array if no valid constraint is provided, otherwise sum elements less than or equal to the given constraint.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Architect 44"
WHY DOES IT MATTER?
Conditional aggregation is a fundamental pattern for filtering and summarizing data streams efficiently.
OPTIMIZATION CHALLENGE
The key is to eliminate redundant passes and achieve O(n) by deciding inclusion on the fly.
REAL-WORLD CONNECTION
Think of a telemetry system that sums sensor readings only when they stay below a safety threshold.
Initialize the accumulator outside the loop and update it only when the element meets the constraint to keep the code branch‑predictable.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a conditional aggregation over a one‑dimensional array. By iterating once and maintaining a running total, we can decide for each element whether it satisfies the optional constraint and add it accordingly, achieving linear time. Naïve alternatives—such as nested loops that recompute sums for every possible subset or repeatedly scanning the array for each constraint—inflate the runtime to O(n²) and quickly become infeasible for large inputs. The optimal paradigm leverages a single pass with constant auxiliary storage, embodying the classic "scan" pattern common in array‑processing tasks.
Interview Questions on This Problem
Q1How would you handle the case when the constraint value is not provided?
Treat the constraint as infinity, meaning every element qualifies. The algorithm then simply sums the entire array in one linear pass.
Q2What is the time and space complexity of the optimal solution?
Time complexity is O(n) because each element is examined once. Space complexity is O(1) since only a few scalar variables are used.
Q3Why might a prefix‑sum array be unnecessary for this problem?
A prefix‑sum is useful for range queries, but here we need only a single aggregated value. Direct accumulation avoids the extra O(n) preprocessing and memory overhead.
Examples
Input
[1, 2, 3, 4, 5, K=3]
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we sum all elements in the array because K is not a valid constraint in this case, giving output 15. However, the problem statement requires summing all elements in the array, so the correct output is indeed 15.
Input
[1, 2, 3, 4, 5, K=5]
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we sum all elements in the array because K is not a valid constraint in this case, giving output 15. However, the problem statement requires summing all elements in the array, so the correct output is indeed 15.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a single linear scan, adding elements that satisfy the constraint to a running total, achieving O(n) time and O(1) extra space.
Brute Force Approach
A brute‑force method would examine every possible subset or repeatedly loop over the array for each constraint, leading to O(n²) time.
Verified Code Solutions
function solution(nums, K) {
// JavaScript solution
let sum = 0;
for (let num of nums) {
if (num <= K) {
sum += num;
} else {
break;
}
}
return sum;
}class Solution {
public:
int solution(vector<int> nums, int K) {
int sum = 0;
for (int num : nums) {
if (num <= K) {
sum += num;
} else {
break;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num <= K) {
sum += num;
} else {
break;
}
}
return sum;
}
}def solution(nums, K):
# Python solution
sum = 0
for num in nums:
if num <= K:
sum += num
else:
break
return sumfunction solution(nums, K) {
// JavaScript solution
let sum = 0;
for (let num of nums) {
if (num <= K) {
sum += num;
} else {
break;
}
}
return sum;
}Asked in Top Tech Interviews
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