Sensor Checkpoint Extractor 41 — Problem Statement & Solution Guide
Problem Description
Sensor Checkpoint Extractor 41
Given an integer array nums and a positive integer k, compute the sum of the k greatest elements contained in nums. The order of elements in the array is irrelevant; only their values matter. Return the resulting sum as a 64‑bit signed integer.
Input: The function receives two arguments – an array of integers nums and an integer k (1 ≤ k ≤ nums.length).
Output: A single integer representing the sum of the k largest values in nums.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Checkpoint Extractor 41"
WHY DOES IT MATTER?
Selecting the top‑k elements is a fundamental reduction step in analytics and ranking pipelines.
OPTIMIZATION CHALLENGE
Turning an O(n log n) full sort into O(n log k) or expected O(n) cuts runtime dramatically for large n.
REAL-WORLD CONNECTION
Databases use similar logic for LIMIT k queries and search engines rank the highest‑scoring documents.
When k is known and small, a pre‑allocated min‑heap avoids reallocations and yields predictable performance.
COMPLEXITY AT A GLANCE
O(n log k)O(k)Core Theory — Why This Approach?
The problem is a classic selection task: we need the sum of the k largest values in an unsorted array. A naïve solution sorts the entire array (O(n log n)) and then adds the last k elements, which is wasteful when k is much smaller than n because sorting imposes a global order that we never fully use. The optimal paradigm leverages partial ordering – either a min‑heap of size k (O(n log k) time, O(k) space) or a quick‑select algorithm that partitions around the kth largest element in expected O(n) time – to isolate only the needed top‑k values. Both approaches avoid the full sort cost and keep the memory footprint low, while a 64‑bit accumulator prevents overflow when the individual numbers or their sum exceed 32‑bit limits.
Interview Questions on This Problem
Q1What data structure gives the most efficient way to keep track of the k largest elements while scanning the array once?
A min‑heap of capacity k. It lets you insert each element in O(log k) and discard the smallest when the heap exceeds k.
Q2How does the time complexity of the heap‑based solution compare to sorting the entire array?
Sorting costs O(n log n). The heap solution runs in O(n log k), which is asymptotically better when k << n. It therefore scales better for large inputs with a small k.
Q3Why must the accumulator be a 64‑bit integer, and how do you ensure correct arithmetic in code?
Individual elements may be up to 32‑bit, but their sum can exceed that range. Declaring the sum variable as long long (or int64) and casting each addition prevents overflow.
Examples
Input
[4, 7, 1, 3, 9], k=3
Output
20
Explanation: The three largest numbers are 9, 7 and 4. Their sum is 9 + 7 + 4 = 20.
Input
[-5, -2, -8, -1], k=2
Output
-3
Explanation: Sorting descending gives [-1, -2, -5, -8]. The top two are -1 and -2, and -1 + (-2) = -3.
Input
[10, 10, 10, 10], k=4
Output
40
Explanation: All elements are equal and each equals 10. Selecting all four yields 10 + 10 + 10 + 10 = 40.
Constraints
- 1 <= nums.length <= 200000
- 1 <= k <= nums.length
- -10^9 <= nums[i] <= 10^9
- The answer fits in a 64‑bit signed integer.
Optimal Approach & Strategy
Use a min‑heap of size k while scanning, or apply quick‑select to partition around the kth largest element and then sum the top k.
Brute Force Approach
Sort the entire array and sum the last k elements; simple but O(n log n) time.
Verified Code Solutions
function solution(nums, k) {
if (k > nums.length) {
k = nums.length;
}
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) {
if (k > nums.size()) {
k = nums.size();
}
sort(nums.rbegin(), nums.rend());
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (k > nums.length) {
k = nums.length;
}
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
if k > len(nums):
k = len(nums)
nums.sort(reverse=True)
return sum(nums[:k])function solution(nums, k) {
if (k > nums.length) {
k = nums.length;
}
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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