Node Payload Synthesizer 28 — 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 synthesizer value under given operational constraints. The constraints are specified as K, where all elements less than or equal to K should be added to the synthesizer value.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Synthesizer 28"
WHY DOES IT MATTER?
Single-pass aggregation eliminates redundant scans and keeps runtime linear.
OPTIMIZATION CHALLENGE
Reducing from quadratic to linear time by avoiding nested traversals.
REAL-WORLD CONNECTION
Similar to filtering sensor readings in an IoT stream where only values below a threshold are summed.
Always guard against null heads and use a simple accumulator variable to keep the code clean.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to traversing a singly linked list and aggregating values that satisfy a simple predicate (value ≤ K). A naïve solution might repeatedly scan the list for each element or use nested loops, leading to O(N^2) time on large inputs, which quickly exceeds limits. The optimal paradigm leverages a single-pass linear scan, maintaining a running sum while checking each node's payload against K, thus achieving O(N) time and O(1) auxiliary space. This approach aligns with the classic "single traversal" pattern for linked list aggregation, ensuring scalability and cache-friendly access.
Interview Questions on This Problem
Q1How would you handle the case where the linked list is empty?
Return a sum of zero because there are no nodes to evaluate. An empty list should be checked before traversal to avoid null-pointer dereference.
Q2Can you compute the sum without modifying the original list?
Yes, by using only read-only access to each node's value during traversal. No pointer rewiring or node deletion is required.
Q3What is the time and space complexity of the optimal solution?
The algorithm runs in O(N) time, where N is the number of nodes. It uses O(1) extra space beyond the input list.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
Output
350
Explanation: Step-by-step: Given the input [10, 20, 30, 40, 50, 60, 70, 80, 90, 100] and K = 100, we iterate through the array. We add all elements less than or equal to K (10, 20, 30, 40, 50, 60, 70, 80, 90, 100) to the sum, resulting in a total of 350.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100, 200, 300, 400, 500]
Output
1000
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100, 200, 300, 400, 500] and K = 100, we iterate through the array. We add all elements less than or equal to K (1, 2, 3, 4, 5, 6, 7, 8, 9, 10) to the sum, resulting in a total of 1000.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Perform one linear traversal, adding qualifying node values to an accumulator.
Brute Force Approach
Repeatedly scan the list for each node to check the condition, resulting in O(N^2) 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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