Node Payload Partition 17 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing node and payload metrics, construct an optimal algorithm to evaluate and compute the target partition value under given operational constraints. The target partition value is the sum of the K largest elements in the sequence.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Payload Partition 17"
WHY DOES IT MATTER?
The "maintain top‑K" pattern appears in many real‑time analytics, recommendation engines, and monitoring systems where you need a concise summary (e.g., top‑K users, hottest topics) without storing or sorting the entire dataset.
OPTIMIZATION CHALLENGE
The key insight is that you do not need a full ordering of the data—only a bounded priority queue of size K. This reduces both time (from O(N log N) to O(N log K)) and space (from O(N) to O(K)).
REAL-WORLD CONNECTION
Think of a distributed log‑aggregation service that continuously receives latency measurements. To alert on the worst‑case performance, it keeps only the K highest latencies in a min‑heap, enabling O(1) retrieval of the current threshold for alerts.
In an interview, start by stating the naive sort, then immediately propose the min‑heap as a greedy sliding‑window of candidates. Mention QuickSelect as an alternative, but default to the heap for its simplicity and deterministic guarantees.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
The problem of finding the sum of the K largest elements is a classic selection problem. A naïve solution would sort the entire array, which costs O(N log N) time, and then sum the last K entries. While correct, this approach is wasteful because the ordering of the remaining N‑K elements is irrelevant to the answer. The optimal paradigm leverages the fact that we only need to keep track of the top K values while scanning the input once. By maintaining a min‑heap (priority queue) of size K, we ensure that the smallest element among the current top K is always at the root. When a new element exceeds the root, we replace the root and re‑heapify, guaranteeing that after a single pass the heap contains exactly the K largest numbers.
Why does this improve performance? Inserting or removing from a heap of size K costs O(log K), so the total work is O(N log K), which is asymptotically better than O(N log N) when K << N. Moreover, the space usage drops to O(K) because we never store the whole array in a sorted structure. This greedy‑style maintenance of a candidate set is a recurring technique in streaming and big‑data scenarios where only a small summary of the input is required.
Alternative optimal solutions include the QuickSelect algorithm, which partitions the array around a pivot and runs in expected O(N) time, but it modifies the input and has a higher constant factor. The min‑heap method is deterministic, easy to implement, and works well with language‑provided priority‑queue libraries, making it the preferred choice in interview settings.
Interview Questions on This Problem
Q1How would you compute the sum of the K largest numbers in a stream of integers where the total length is unknown beforehand?
Maintain a min‑heap of size K. For each incoming integer, if the heap has fewer than K elements, push it. Otherwise, compare it with the heap root; if larger, pop the root and push the new value. After processing the stream, sum all elements in the heap. This runs in O(N log K) time and O(K) space.
Q2Can you modify the solution to also return the K largest elements in sorted descending order?
After the heap contains the K largest values, extract them into a list (which gives them in ascending order) and then reverse the list, or alternatively use a max‑heap to pop elements directly in descending order. The extraction adds O(K log K) time, which is negligible compared to the O(N log K) scanning phase.
Q3What are the trade‑offs between using a min‑heap versus QuickSelect for this problem?
A min‑heap provides deterministic O(N log K) time and O(K) extra space, works well when K is much smaller than N, and does not modify the original array. QuickSelect runs in expected O(N) time and O(1) extra space but has O(N^2) worst‑case behavior, requires in‑place partitioning, and can be harder to code correctly under interview pressure.
Examples
Input
[90, 80, 70, 60, 50, 40, 30, 20, 10, 5, 4, 3, 2, 1]
Output
290
Explanation: Step-by-step: with input [90, 80, 70, 60, 50, 40, 30, 20, 10, 5, 4, 3, 2, 1], we first sort the array in descending order. Then, we select the first K elements (5 largest elements) and sum them up, giving output 290.
Input
[5, 4, 3, 2, 1]
Output
9
Explanation: Step-by-step: with input [5, 4, 3, 2, 1], we first sort the array in descending order. Then, we select the first K elements (2 largest elements) and sum them up, giving output 9.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a min‑heap of size K while iterating once through the array, replacing the root whenever a larger element appears.
Brute Force Approach
Sort the entire array in descending order and sum the first K elements.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.begin(), nums.end(), greater<int>());
int sum = 0;
for (int i = 0; i < k; 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++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
sum = 0
for i in range(k):
sum += nums[i]
return sumfunction solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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