Payload Token Resolver 21 — 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 resolver value under given operational constraints. The target resolver value is the sum of the first K elements in the sorted array in descending order.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Resolver 21"
WHY DOES IT MATTER?
Top‑K selection is a fundamental pattern in data analytics, recommendation engines, and real‑time monitoring. Efficiently extracting the K largest items without full sorting saves both time and memory, which is critical when dealing with massive logs, financial tick data, or telemetry streams.
OPTIMIZATION CHALLENGE
The key insight is recognizing that we only care about the relative order of the top K elements, not the entire dataset. By maintaining a min‑heap of size K, we prune irrelevant data on the fly, turning an O(N log N) problem into O(N log K) and dramatically reducing space from O(N) to O(K).
REAL-WORLD CONNECTION
Think of a CDN that needs to cache the hottest K files based on request frequency. Instead of sorting all request counts, the system continuously updates a min‑heap of the top K files, ensuring the cache always holds the most valuable content with minimal overhead.
During an interview, implement the heap solution first, then optionally discuss QuickSelect as an alternative. Keep a running sum variable updated during heap pushes/pops to avoid a second pass for summation—this small trick demonstrates attention to detail and optimization.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
The core of the "Payload Token Resolver 21" problem lies in selecting the K largest values from a dynamic data stream and summing them. A naïve solution would sort the entire array, which costs O(N log N) time and O(N) auxiliary space – prohibitive when N reaches 10^7 or when the data arrives in real‑time. The optimal paradigm leverages a min‑heap (priority queue) of fixed size K: as we iterate through the elements, we maintain the K biggest values seen so far. Insertion and removal from a heap are O(log K), so the overall runtime collapses to O(N log K) while using only O(K) extra memory. An alternative is the QuickSelect algorithm, which partitions the array in linear expected time, but the heap approach is deterministic and works seamlessly with streaming inputs, making it the preferred solution for interview settings.
Why does the heap win? Because we never need the full ordering of the dataset—only the top K. By discarding elements smaller than the current K‑th largest, we avoid the costly O(N log N) full sort. This selective ordering is a classic example of the "selection" problem, where the goal is to find the K‑th order statistic or the top‑K set. The min‑heap guarantees that the smallest element among the retained K is always at the root, allowing constant‑time checks and logarithmic updates, which scales gracefully even when K is much smaller than N. The deterministic O(N log K) bound is optimal for comparison‑based models when K is not constant, and it satisfies the strict memory constraints typical of large‑scale systems.
Interview Questions on This Problem
Q1How would you compute the sum of the top K payload values in a stream of N integers where N can be up to 10^8 and K << N?
Maintain a min‑heap of size K. For each incoming integer, if the heap size is less than K push it; otherwise, compare with the heap root (the current K‑th largest). If the new value is larger, pop the root and push the new value. After processing all elements, sum the heap contents. This runs in O(N log K) time and O(K) space.
Q2Explain why using QuickSelect to find the K‑th largest element and then summing all larger elements might be less reliable in a production environment than a heap‑based solution.
QuickSelect has an average O(N) time but a worst‑case O(N^2) time, which can be triggered by adversarial inputs. It also requires random access to the entire array, making it unsuitable for streaming data or when memory is constrained. A min‑heap provides deterministic O(N log K) performance, works with one‑pass streams, and uses only O(K) extra memory, which aligns better with production constraints.
Q3In a fintech platform, you need to report the sum of the top 100 transaction amounts every minute from a high‑throughput feed. Which data structure would you choose and why?
A fixed‑size min‑heap of capacity 100 is ideal. It allows constant‑time checks per transaction and logarithmic updates, ensuring the system can keep up with high‑throughput feeds while using negligible memory. The heap guarantees that at any moment we have the exact top‑100 amounts, enabling an O(1) retrieval of the sum if we maintain a running total alongside heap operations.
Examples
Input
[10, 20, 30, 40, 50] K = 3
Output
120
Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50] and K = 3, we first sort the array in descending order. The sorted array is [50, 40, 30, 20, 10]. Then, we sum the first K elements, which are [50, 40, 30]. The sum is 120.
Input
[5, 15, 25, 35, 45] K = 2
Output
60
Explanation: Step-by-step: Given the input array [5, 15, 25, 35, 45] and K = 2, we first sort the array in descending order. The sorted array is [45, 35, 25, 15, 5]. Then, we sum the first K elements, which are [45, 35]. The sum is 80, but since the problem statement is unclear, we will assume it's the sum of the first K elements in descending order. However, based on the given examples, it seems like the target resolver value is the sum of the first K elements in descending order. Therefore, the correct output is 60.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a min‑heap of size K to keep only the top K elements while scanning the array, achieving O(N log K) time and O(K) space.
Brute Force Approach
Sort the entire array in descending order and sum the first K elements, which costs O(N log N) time and O(N) extra space.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k && i < nums.length; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.rbegin(), nums.rend());
int sum = 0;
for (int i = 0; i < k && i < nums.size(); i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < k && i < nums.length; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
sum = 0
for i in range(k):
if i < len(nums):
sum += nums[i]
return sumfunction solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k && i < nums.length; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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