Payload Token Synthesizer 42 — 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 synthesizer value under given operational constraints. The function should return the sum of all numbers in the array greater than a given threshold K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Synthesizer 42"
WHY DOES IT MATTER?
Sliding‑window transforms repeated scans into a single linear pass, crucial for high‑throughput data streams.
OPTIMIZATION CHALLENGE
The key is to update the aggregate incrementally instead of recomputing from scratch for each window shift.
REAL-WORLD CONNECTION
Network routers aggregate packet sizes above a threshold in real time using a moving window buffer.
Cache the contribution of each element (whether it qualifies) and adjust the sum only when the element enters or exits the window.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The naive solution iterates over the array and adds each element that exceeds K, which is already O(N) time but can become inefficient when combined with additional constraints such as moving windows, multiple queries, or streaming data where repeated scans are required. In large‑scale systems, repeatedly scanning the entire dataset for each query leads to prohibitive latency and CPU usage.
The optimal paradigm leverages a single‑pass sliding‑window or prefix‑sum technique: maintain a running total while traversing the array once, updating the sum only when the current element satisfies the >K condition. This reduces the overall work to linear time with constant auxiliary space, and it easily extends to sliding‑window variants where the window bounds shift dynamically, preserving O(N) complexity across all operations.
Interview Questions on This Problem
Q1How would you modify the solution if you needed the sum of elements > K for every subarray of size W?
Maintain a sliding window of size W and a running sum, adding the new element if it > K and subtracting the outgoing element if it > K. This yields O(N) time for all windows.
Q2What is the time‑space trade‑off when precomputing prefix sums for multiple K queries?
Prefix sums give O(1) query time but require O(N) extra space; each query then becomes O(log N) if you binary‑search the threshold index.
Q3Why is a single pass preferable to sorting the array before summing elements > K?
Sorting costs O(N log N) and destroys original order, while a single pass achieves the result in O(N) without extra memory.
Examples
Input
[2, 3, 4, 5], 1
Output
14
Explanation: Step-by-step: with input [2, 3, 4, 5] and K = 1, we sum all numbers greater than 1, which are 2, 3, 4, and 5. So, the output is 2 + 3 + 4 + 5 = 14
Input
[], 1
Output
0
Explanation: Step-by-step: with input [] and K = 1, the array is empty, so the function should return 0 as per the problem statement
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a single traversal (or sliding window) to maintain the sum, achieving O(N) total time and O(1) auxiliary space.
Brute Force Approach
For each query, scan the entire array and sum qualifying elements, leading to O(N) per query and O(1) extra space.
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):
return sum(num for num in nums if num > k)function 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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