Node Payload Resolver 42 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing node and payload metrics, and an integer K, construct an optimal algorithm to evaluate and compute the sum of all elements greater than the Kth largest element.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Resolver 42"
WHY DOES IT MATTER?
This pattern is essential because many real-world problems involve finding order statistics (Kth largest/smallest) and aggregating data relative to that threshold. It tests the candidate's ability to move beyond sorting and utilize selection algorithms or heaps for more efficient solutions.
OPTIMIZATION CHALLENGE
The key insight is recognizing that you do not need to sort the entire array. You only need to identify the Kth largest element and sum the elements that are strictly greater than it. Using a min-heap of size K allows you to keep track of the top K elements efficiently, and the sum can be computed either during the heap construction or by iterating through the heap after construction.
REAL-WORLD CONNECTION
This is directly analogous to calculating tax brackets or percentile-based bonuses in HR systems, where you need to sum up values that exceed a certain threshold (the Kth largest salary) to determine total payout or tax liability.
In an interview, explicitly state the trade-offs. If K is close to N, sorting might be simpler and competitive. If K is small, a min-heap of size K is optimal. Mentioning Quickselect shows depth, but the heap approach is often easier to implement correctly under time pressure and handles the 'sum' part more naturally if you maintain the sum incrementally.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
The problem of finding the sum of elements greater than the Kth largest element is fundamentally a selection problem combined with an aggregation task. Naively, one might sort the array in descending order, identify the Kth element, and sum the preceding elements. While this approach is straightforward, it incurs a time complexity of O(N log N) due to the sorting step. For large datasets where N can be in the millions, this logarithmic factor becomes a significant bottleneck, especially when the goal is merely to find a specific order statistic and aggregate values relative to it, rather than fully ordering the dataset.
Interview Questions on This Problem
Q1At a high-growth fintech platform, you need to calculate the total value of transactions exceeding the 95th percentile threshold in real-time. How would you optimize this calculation if the data stream is continuous and memory is constrained?
Use a min-heap of size K (where K corresponds to the percentile count) to maintain the top K largest elements. As new elements arrive, if they are larger than the heap's minimum, replace the minimum and push the new element. The sum can be maintained incrementally by subtracting the removed element and adding the new one. This ensures O(log K) per insertion and O(K) space, which is optimal for streaming percentiles.
Q2In a distributed systems interview at a major tech company, you are asked to find the sum of all values greater than the Kth largest in an array of 10^7 integers. Why is using Quickselect preferred over sorting, and what is the worst-case scenario for Quickselect?
Quickselect has an average time complexity of O(N), which is superior to O(N log N) for sorting. It works by partitioning the array around a pivot until the Kth element is in its correct position. The worst-case scenario is O(N^2) if the pivot selection is consistently poor (e.g., always picking the smallest or largest element), but this can be mitigated by using randomized pivot selection or the Median-of-Medians algorithm to guarantee O(N).
Q3For a startup building a recommendation engine, you need to identify the top K items by score and sum their scores. If K is very small compared to N (e.g., K=10, N=1,000,000), which data structure is more efficient: a max-heap of size N or a min-heap of size K?
A min-heap of size K is more efficient. Building a max-heap of size N takes O(N) time, but maintaining it for dynamic updates or if we only care about the top K is less optimal. A min-heap of size K allows us to process each element in O(log K) time. Since K << N, log K is much smaller than log N, and the space complexity is reduced from O(N) to O(K), making it ideal for memory-constrained environments.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100], 3
Output
250
Explanation: Step-by-step: First, we sort the array in descending order. Then, we find the Kth largest element, which is 40 in this case. Next, we filter out all elements that are not greater than 40 and sum them up. Finally, we return the sum, which is 250.
Input
[15, 20, 25, 30, 35, 40, 45, 50, 55, 60], 4
Output
155
Explanation: Step-by-step: First, we sort the array in descending order. Then, we find the Kth largest element, which is 30 in this case. Next, we filter out all elements that are not greater than 30 and sum them up. Finally, we return the sum, which is 155.
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 maintain the K largest elements. Iterate through the array, updating the heap and a running sum. The Kth largest is the heap's minimum, and the sum of elements greater than it is derived from the heap's contents.
Brute Force Approach
Sort the array in descending order. Identify the Kth element and sum all elements from index 0 to K-1 (or K-2 if strictly greater).
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let kthLargest = nums[k - 1];
return nums.filter(x => x > kthLargest).reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.rbegin(), nums.rend());
int kthLargest = nums[k - 1];
int sum = 0;
for (int x : nums) {
if (x > kthLargest) {
sum += x;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int kthLargest = nums[k - 1];
int sum = 0;
for (int x : nums) {
if (x > kthLargest) {
sum += x;
}
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
kthLargest = nums[k - 1]
return sum(x for x in nums if x > kthLargest)function solution(nums, k) {
nums.sort((a, b) => b - a);
let kthLargest = nums[k - 1];
return nums.filter(x => x > kthLargest).reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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