Node Payload Analyzer 23 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a diagnostic routine for a distributed data pipeline. The system receives a sequence of integer metrics, metrics, and a threshold parameter K. The goal is to compute the 'Analyzer Value' based on the following strict operational logic:
1. If any single metric in the sequence exceeds the threshold K, the system flags a critical overflow, and the Analyzer Value must be exactly 0.
2. If and only if every metric in the sequence is less than or equal to K, the Analyzer Value is defined as the sum of all metrics in the sequence.
Your function must efficiently determine this value. Note that the problem is framed within a backtracking context to encourage exploring state transitions, although for this specific 'easy' difficulty, a linear scan suffices to verify the global constraint before computing the aggregate sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Analyzer 23"
WHY DOES IT MATTER?
Backtracking with pruning turns an exponential brute‑force problem into a tractable solution for moderate input sizes.
OPTIMIZATION CHALLENGE
The key is to cut branches early by tracking the current sum and comparing it to K.
REAL-WORLD CONNECTION
It mirrors resource allocation in distributed systems where tasks must fit within a capacity budget.
Always sort descending and stop recursion as soon as the partial sum exceeds K to maximize pruning.
COMPLEXITY AT A GLANCE
O(2^n) in the worst case, much less with pruningO(n) recursion stackCore Theory — Why This Approach?
The problem is a classic subset‑sum variant where we must select a subset of the given metrics whose total does not exceed the threshold K and is as large as possible. A naive exhaustive scan of all 2^n subsets quickly becomes infeasible for n > 30 because the time grows exponentially, leading to time‑outs on large inputs. By applying backtracking we can prune the search space: sorting the metrics in descending order lets us stop exploring a branch as soon as the running sum exceeds K, and we can also skip identical values to avoid duplicate work. This yields an optimal exponential‑time algorithm that runs fast enough for the intended constraints while using only O(n) auxiliary space for the recursion stack.
Interview Questions on This Problem
Q1How does sorting the input before backtracking improve performance?
Sorting enables early pruning because larger numbers are tried first, causing the running sum to exceed K sooner and cutting off deeper recursion. It also helps skip duplicate values, reducing redundant branches.
Q2What is the difference between backtracking and dynamic programming for subset‑sum?
Backtracking explores combinations recursively with pruning, suitable when n is small or when we need the actual subset. DP builds a table of reachable sums, using O(n·K) time and space, which can be overkill for large K.
Q3Why is it safe to ignore a metric that is greater than K?
Any subset containing a metric > K would already violate the threshold, so such metrics can never be part of a valid solution. Removing them reduces the search space without affecting the answer.
Examples
Input
metrics = [1, 2, 3], K = 5
Output
6
Explanation: Step 1: Check each element against K=5. 1 <= 5, 2 <= 5, 3 <= 5. All elements are within the threshold. Step 2: Since no element exceeds K, compute the sum: 1 + 2 + 3 = 6. Output: 6.
Input
metrics = [1, 6, 3], K = 5
Output
0
Explanation: Step 1: Check each element against K=5. 1 <= 5, but 6 > 5. Step 2: Since an element (6) exceeds the threshold, the critical overflow condition is met. Output: 0.
Input
metrics = [5, 5, 5], K = 5
Output
15
Explanation: Step 1: Check each element against K=5. 5 <= 5, 5 <= 5, 5 <= 5. All elements are within the threshold (equality is allowed). Step 2: Compute the sum: 5 + 5 + 5 = 15. Output: 15.
Input
metrics = [-1, 0, 1], K = 0
Output
0
Explanation: Step 1: Check each element against K=0. -1 <= 0, 0 <= 0, but 1 > 0. Step 2: Since 1 exceeds the threshold, the overflow condition triggers. Output: 0.
Constraints
- 1 <= metrics.length <= 10^5
- -10^9 <= metrics[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Sort the array, then recursively include or exclude each metric while pruning any branch whose running sum exceeds K.
Brute Force Approach
Generate every possible subset, compute its sum, and keep the best valid one.
Verified Code Solutions
/**
* @param {number[]} metrics
* @param {number} K
* @return {number}
*/
var analyzerValue = function(metrics, K) {
let sum = 0;
for (let m of metrics) {
if (m > K) return 0;
sum += m;
}
return sum;
};class Solution {
public:
int analyzerValue(vector<int>& metrics, int K) {
int sum = 0;
for (int m : metrics) {
if (m > K) return 0;
sum += m;
}
return sum;
}
};class Solution {
public int analyzerValue(int[] metrics, int K) {
int sum = 0;
for (int m : metrics) {
if (m > K) return 0;
sum += m;
}
return sum;
}
}class Solution:
def analyzerValue(self, metrics: List[int], K: int) -> int:
total = 0
for m in metrics:
if m > K:
return 0
total += m
return total/**
* @param {number[]} metrics
* @param {number} K
* @return {number}
*/
var analyzerValue = function(metrics, K) {
let sum = 0;
for (let m of metrics) {
if (m > K) return 0;
sum += m;
}
return sum;
};Asked in Top Tech Interviews
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