Matrix Stream Optimizer 19 — Problem Statement & Solution Guide
Problem Description
You are given a sequence of integers and a positive integer k. Your task is to compute the sum of the k largest values in the sequence. The input consists of two lines: the first line contains two integers n and k, where n is the number of elements in the sequence and k is the number of top elements to consider. The second line contains n space‑separated integers representing the sequence. The output should be a single integer: the sum of the k greatest elements. If the sequence contains duplicate values, each occurrence is treated independently when selecting the top k elements.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Stream Optimizer 19"
WHY DOES IT MATTER?
This pattern is essential for top-k problems, which are ubiquitous in data analytics, recommendation systems, and real-time monitoring. It teaches the trade-off between sorting (O(n log n)) and partial selection (O(n log k)), a critical concept for optimizing performance in large-scale systems.
OPTIMIZATION CHALLENGE
The key insight is that we do not need a fully sorted array; we only need the boundary between the top k and the rest. By maintaining a Min-Heap of size k, the root always represents the smallest element in our 'top k' set. Any new element larger than this root displaces it, ensuring the heap always contains the k largest elements seen so far.
REAL-WORLD CONNECTION
Consider a stock trading platform that needs to display the top 10 most volatile stocks out of 5,000 listed stocks in real-time. Sorting all 5,000 stocks every second is wasteful. Using a Min-Heap of size 10 allows the system to update the top 10 list in O(log 10) time per new data point, ensuring low latency.
In an interview, explicitly state the constraints. If k is small relative to n, the heap is optimal. If k is large, consider Quickselect. Always mention the space complexity: the heap uses O(k) space, which is often a significant advantage over sorting in-place if memory is constrained.
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 in a sequence is a classic selection problem that challenges candidates to move beyond simple sorting. The naive approach involves sorting the entire array in descending order and summing the first k elements, which results in a time complexity of O(n log n). While this is acceptable for small datasets, it is inefficient for large-scale systems where n can be in the millions and k is relatively small (e.g., k << n). Sorting the entire array performs unnecessary work on the n-k elements that are not part of the final result.
Interview Questions on This Problem
Q1How would you optimize the solution if k is very close to n, such that k = n - 1?
If k is close to n, finding the k largest elements is equivalent to finding the smallest element and subtracting it from the total sum. In this specific edge case, a linear scan to find the minimum element O(n) is more efficient than using a heap of size k, which would be O(n log k) ≈ O(n log n). However, the general heap-based solution remains robust and is the standard expected answer unless the interviewer specifically probes for this edge-case optimization.
Q2Can you solve this problem in O(n) average time complexity?
Yes, using the Quickselect algorithm. By partitioning the array around a pivot, we can isolate the k largest elements in O(n) average time. However, the worst-case time complexity is O(n^2). In interviews, it is safer to propose the Min-Heap solution O(n log k) as it guarantees consistent performance, but mentioning Quickselect demonstrates deep algorithmic knowledge.
Q3How would you handle this problem if the input is a stream of data rather than a static array?
For a streaming input, we cannot store all elements if the stream is infinite. We must maintain a Min-Heap of size k. As each new element arrives, if the heap is not full, we add it. If the heap is full and the new element is larger than the root (the smallest of the current top k), we pop the root and push the new element. This allows us to compute the sum of the top k elements at any point in time with O(log k) update time per element.
Examples
Input
5 3 3 1 5 2 4
Output
12
Explanation: The sorted sequence in descending order is [5,4,3,2,1]. The top 3 elements are 5, 4, and 3. Their sum is 5+4+3=12.
Input
4 2 -1 -5 0 2
Output
2
Explanation: Descending order: [2,0,-1,-5]. The two largest are 2 and 0, summing to 2.
Input
3 2 10 10 10
Output
20
Explanation: All elements are equal to 10. The two largest are 10 and 10, giving a sum of 20.
Constraints
- 1 <= n <= 100000
- -1000000000 <= nums[i] <= 1000000000
- 1 <= k <= n
- The result fits in a 64‑bit signed integer
Optimal Approach & Strategy
Use a Min-Heap of size k to keep track of the k largest elements seen so far. Iterate through the array, replacing the smallest element in the heap if a larger one is found, then sum the heap contents at the end.
Brute Force Approach
Sort the entire array in descending order and sum the first k elements. This approach is simple but inefficient for large n because it processes all elements unnecessarily.
Verified Code Solutions
const fs=require('fs');
const input=fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let idx=0;
const n=input[idx++];
const k=input[idx++];
const arr=input.slice(idx, idx+n);
arr.sort((a,b)=>b-a);
let sum=0;
for(let i=0;i<k && i<arr.length;i++) sum+=arr[i];
console.log(sum);#include <bits/stdc++.h>
using namespace std;
int main(){ios::sync_with_stdio(false);cin.tie(nullptr);
int n,k; if(!(cin>>n>>k)) return 0; vector<long long>a(n); for(int i=0;i<n;i++)cin>>a[i];
sort(a.begin(), a.end(), greater<long long>());
long long sum=0; for(int i=0;i<k && i<n;i++) sum+=a[i];
cout<<sum; return 0;}
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
st = new StringTokenizer(br.readLine());
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = Long.parseLong(st.nextToken());
}
Arrays.sort(arr);
long sum = 0;
for (int i = n - 1; i >= n - k && i >= 0; i--) {
sum += arr[i];
}
System.out.println(sum);
}
}
import sys
def main():
data=sys.stdin.read().strip().split()
if not data:
return
n=int(data[0]); k=int(data[1])
arr=list(map(int, data[2:2+n]))
arr.sort(reverse=True)
print(sum(arr[:k]))
if __name__=='__main__':
main()
const fs=require('fs');
const input=fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let idx=0;
const n=input[idx++];
const k=input[idx++];
const arr=input.slice(idx, idx+n);
arr.sort((a,b)=>b-a);
let sum=0;
for(let i=0;i<k && i<arr.length;i++) sum+=arr[i];
console.log(sum);
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