Payload Token Detector 13 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and token metrics, construct an optimal algorithm to evaluate and compute the target detector value under given operational constraints. The target detector value is the sum of all elements greater than K, excluding the last element and elements not greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Detector 13"
WHY DOES IT MATTER?
Filtering and aggregating streams in order is a fundamental queue pattern for real‑time analytics.
OPTIMIZATION CHALLENGE
Eliminate redundant passes by maintaining a running total, reducing time from quadratic to linear.
REAL-WORLD CONNECTION
Network routers sum packet sizes above a threshold while discarding the tail packet for latency calculations.
When using a queue, dequeue only when you need to drop the last element; otherwise, just stop reading before it.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a single-pass filter over a linear data stream, a classic use‑case for the queue abstraction where elements are processed in arrival order and the tail (last element) is ignored. A naive double‑loop that recomputes sums for each prefix would be O(N²) and quickly exceeds time limits for large N, because each element would be revisited many times. The optimal paradigm leverages the fact that the sum of qualifying elements can be accumulated incrementally: as each element arrives (except the final one), we test it against K and, if larger, add it to a running total. This yields a linear‑time solution with O(1) auxiliary space, which is the hallmark of efficient streaming algorithms and aligns with queue‑based processing where only the current element matters.
Interview Questions on This Problem
Q1Why does a nested loop approach lead to O(N²) time for this problem?
Each inner loop re‑examines previously seen elements, causing repeated work. The total number of comparisons grows quadratically with N.
Q2How can you handle the “exclude last element” rule while still using a single pass?
Process elements up to index N‑2 during the scan and simply ignore the final element. This can be done by looping to length‑1 or by checking the index inside the loop.
Q3What is the space complexity of the optimal solution and why?
It uses O(1) extra space because only a running sum and a few counters are stored. No additional data structures proportional to N are required.
Examples
Input
[10, 20, 30, 40, 50] and K = 35
Output
100
Explanation: Step-by-step: 1. Initialize sum to 0. 2. Iterate through the array from the start. 3. For each element, check if it is greater than K. 4. If it is, add it to the sum. 5. After the loop, return the sum. In this case, the sum of elements greater than 35 is 10 + 20 + 30 + 40 = 100.
Input
[10, 20, 30, 40, 50] and K = 60
Output
0
Explanation: Step-by-step: 1. Initialize sum to 0. 2. Iterate through the array from the start. 3. For each element, check if it is greater than K. 4. If it is, add it to the sum. 5. After the loop, return the sum. In this case, there are no elements greater than 60, so the sum is 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a single running sum while iterating once over the array, skipping the last element.
Brute Force Approach
Use two nested loops to recompute the sum for every prefix, which leads to O(N²) time.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let i = 0; i < nums.length - 1; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int i = 0; i < nums.size() - 1; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int i = 0; i < nums.length - 1; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
return sum;
}
}def solution(nums, K):
sum = 0
for i in range(len(nums) - 1):
if nums[i] > K:
sum += nums[i]
return sumfunction solution(nums, K) {
let sum = 0;
for (let i = 0; i < nums.length - 1; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
return sum;
}Asked in Top Tech Interviews
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