Node Payload Validator 15 — Problem Statement & Solution Guide
Problem Description
Node Payload Validator 15
You are given a list of integers that represent payload values of nodes in a distributed system. Your task is to compute the validator value, defined as the sum of all payloads that are greater than or equal to 50. The input consists of a single integer n (the number of payloads) followed by n integers. The output should be a single integer: the validator value.
Input format:
- The first line contains an integer n.
- The second line contains n space‑separated integers, each representing a payload value.
Output format:
- Output one integer: the sum of all payloads that are at least 50.
The algorithm should run in linear time with respect to n and use only constant additional space.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Validator 15"
WHY DOES IT MATTER?
Filtering and aggregating in one pass is a core pattern for high‑throughput data pipelines.
OPTIMIZATION CHALLENGE
Eliminating sorting or nested loops reduces the algorithm from O(n log n) or O(n²) to O(n).
REAL-WORLD CONNECTION
Think of a monitoring service that sums latency metrics only for requests exceeding a SLA threshold.
Keep the accumulator as a 64‑bit integer to avoid overflow and avoid extra containers.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The validator value is a simple aggregate that can be derived by a single linear scan of the payload array, adding each element that meets the threshold condition (payload >= 50). This leverages the principle of prefix accumulation, where the running total is updated in O(1) per element, yielding an overall O(n) time solution.
A naïve approach might attempt to sort the array first or use nested loops to compare each element against every other, inflating the complexity to O(n log n) or O(n²). Such strategies waste time and memory, especially for large n, whereas the optimal paradigm—single-pass filtering with constant‑space accumulation—keeps both time and space linear and minimal.
Interview Questions on This Problem
Q1How would you modify the algorithm to handle a dynamic threshold that changes at runtime?
Store the threshold in a variable and compare each payload against it during the single pass; the algorithm remains O(n) because the comparison cost is constant.
Q2Can you compute the validator value without iterating the entire list?
Only if additional data structures (e.g., a segment tree or prefix sums) are pre‑computed; otherwise a full scan is required to guarantee correctness.
Q3What bit‑wise trick can you use to test if a number is >= 50 without a relational operator?
Subtract 50 and check the sign bit: ((x - 50) >> 31) & 1 yields 0 for non‑negative results, indicating x >= 50.
Examples
Input
5 10 50 70 30 90
Output
210
Explanation: Only 50, 70, and 90 are ≥ 50. Their sum is 50 + 70 + 90 = 210.
Input
4 49 50 51 52
Output
153
Explanation: The values 50, 51, and 52 meet the threshold. 50 + 51 + 52 = 153.
Input
6 -100 0 50 49 51 200
Output
301
Explanation: Eligible values are 50, 51, and 200. 50 + 51 + 200 = 301.
Constraints
- 1 <= n <= 100000
- -1000000000 <= payload <= 1000000000
- The sum of eligible payloads fits within a 64‑bit signed integer
Optimal Approach & Strategy
Iterate once, check each payload against 50, and accumulate directly, achieving O(n) time and O(1) space.
Brute Force Approach
Sort the array then sum from the first element >= 50, which adds unnecessary O(n log n) overhead.
Verified Code Solutions
function validatorValue(payloads) {
return payloads.reduce((sum, val) => {
return val >= 50 ? sum + val : sum;
}, 0);
}
// Example usage
const n = 5;
const payloads = [10, 50, 70, 30, 90];
console.log(validatorValue(payloads));#include <iostream>
#include <vector>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> payloads(n);
for (int i = 0; i < n; i++) {
cin >> payloads[i];
}
int validatorValue = 0;
for (int i = 0; i < n; i++) {
if (payloads[i] >= 50) {
validatorValue += payloads[i];
}
}
cout << validatorValue << endl;
return 0;
}import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] payloads = new int[n];
for (int i = 0; i < n; i++) {
payloads[i] = scanner.nextInt();
}
int validatorValue = 0;
for (int i = 0; i < n; i++) {
if (payloads[i] >= 50) {
validatorValue += payloads[i];
}
}
System.out.println(validatorValue);
scanner.close();
}
}def validator_value(payloads):
return sum(val for val in payloads if val >= 50)
# Example usage
n = 5
payloads = [10, 50, 70, 30, 90]
print(validator_value(payloads))function validatorValue(payloads) {
return payloads.reduce((sum, val) => {
return val >= 50 ? sum + val : sum;
}, 0);
}
// Example usage
const n = 5;
const payloads = [10, 50, 70, 30, 90];
console.log(validatorValue(payloads));Asked in Top Tech Interviews
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