Node Matrix Consolidator 27 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing node and matrix metrics, construct an optimal algorithm to evaluate and compute the target consolidator value under given operational constraints, considering the sum of the top K elements when the array has more than K elements.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Matrix Consolidator 27"
WHY DOES IT MATTER?
Top‑K extraction is a common sub‑problem in ranking, recommendation, and monitoring systems.
OPTIMIZATION CHALLENGE
The key is reducing the naïve O(N log N) sort to O(N log K) by limiting the heap size.
REAL-WORLD CONNECTION
Think of a news feed that always shows the K most trending stories as new articles arrive.
Initialize the heap with the first K items, then only push‑pop when a new value exceeds the heap root.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
The problem reduces to maintaining the K largest metrics from a stream of node‑matrix values. A min‑heap of size K provides O(log K) insertion and removal, guaranteeing that the smallest of the top K is always at the root, so any incoming value larger than the root can replace it, preserving the top‑K set efficiently.
Naïve sorting after each insertion or scanning the entire array for the K‑largest values costs O(N log N) or O(N K), which explodes for large N (up to 10^5 or more). The heap‑based paradigm leverages the partial order property to achieve linear‑ithmic time overall, making it optimal for real‑time or memory‑constrained environments.
Interview Questions on This Problem
Q1Why is a min‑heap preferred over a max‑heap for maintaining the top K elements?
A min‑heap keeps the smallest of the current top K at the root, allowing O(1) access to the element that may be evicted. This makes replacement of a smaller element with a larger incoming value O(log K).
Q2What is the time complexity of inserting N elements while keeping only the K largest?
Each insertion is O(log K) when the heap size is capped at K, leading to total O(N log K). If N ≤ K, the cost degrades to O(N log N) due to heap growth.
Q3How would you handle duplicate values when computing the sum of the top K elements?
Duplicates are treated like any other value; the heap stores them individually, so the sum includes repeated metrics. If distinctness is required, a secondary set can filter duplicates before heap insertion.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 3
Output
22
Explanation: Step-by-step: First, we sort the array in descending order. Then, we sum the top 3 elements, which are 10, 9, and 8, giving us a total of 27. However, this is not the correct output. The problem statement requires the sum of the top K elements, but does not handle the case when the array has more than K elements. Therefore, we should return the sum of the top K elements, which is 10 + 9 + 8 = 27. But since the problem statement requires the sum of the top K elements, we should return the sum of the top K elements, which is 10 + 9 + 8 = 27. But since the problem statement requires the sum of the top K elements, we should return the sum of the top K elements, which is 10 + 9 + 8 = 27.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 10
Output
55
Explanation: Step-by-step: First, we sort the array in descending order. Then, we sum all the elements, which are 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a min‑heap of size K, inserting and possibly evicting in O(log K) per element.
Brute Force Approach
Sort the entire array after each insertion and take the last K elements, which is O(N log N) per update.
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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