Matrix Stream Tracker 30 — Problem Statement & Solution Guide
Problem Description
You are processing a continuous stream of integer values represented as an array nums. Your task is to identify a specific subset of these values based on a threshold parameter K. Specifically, you must select the K largest elements from nums that strictly satisfy the condition value > K. If fewer than K elements in the array meet this condition, return the sum of all available elements that do. If no elements meet the condition, return 0.
The input consists of an integer array nums and an integer K. The output is a single integer representing the sum of the selected elements. The selection process requires identifying the top K values among those exceeding the threshold K, ensuring that the largest values are prioritized in the summation.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Stream Tracker 30"
WHY DOES IT MATTER?
Selecting a bounded top‑K subset avoids processing the whole dataset, crucial for high‑throughput streams.
OPTIMIZATION CHALLENGE
The key is to reduce O(N log N) sorting to O(N log K) or O(N) by limiting work to K elements.
REAL-WORLD CONNECTION
Similar to maintaining a leaderboard where only the top scores above a threshold are kept in memory.
Initialize the heap only after filtering > K to keep its size minimal and avoid unnecessary heap operations.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
The naive solution sorts the entire array (O(N log N)) and then scans for values > K, which is wasteful when N is huge and only K elements are needed. By leveraging the two‑pointer/selection paradigm—either a min‑heap of size K or QuickSelect partitioning—we can isolate the K largest qualifying elements in linear or near‑linear time, dramatically reducing work on irrelevant data. This approach exploits the fact that we only care about the top K values that also satisfy value > K, allowing us to discard the rest early and keep memory bounded by K. The optimal paradigm thus combines a filter pass (O(N)) with a bounded‑size data structure that maintains the current K best candidates, achieving O(N log K) worst‑case or O(N) expected time while using O(K) extra space.
Interview Questions on This Problem
Q1Why is sorting the entire array suboptimal for this problem?
Sorting costs O(N log N) even though we only need K elements, leading to unnecessary work on the majority of data.
Q2How does a min‑heap of size K help maintain the K largest values > K?
The heap keeps the smallest of the current top K at its root; any new qualifying value larger than the root replaces it, ensuring O(log K) updates.
Q3What edge case must you handle when the number of elements > K is less than K?
You must return all qualifying elements (which may be fewer than K) or indicate that the requirement cannot be met.
Examples
Input
nums = [10, 20, 30, 40, 50], K = 3
Output
120
Explanation: First, filter elements greater than K (3). All elements [10, 20, 30, 40, 50] are greater than 3. Next, select the K (3) largest elements from this filtered set. The largest three are 50, 40, and 30. Sum = 50 + 40 + 30 = 120.
Input
nums = [1, 2, 3, 4, 5], K = 4
Output
9
Explanation: Filter elements greater than K (4). Only the element 5 is greater than 4. The filtered set is [5]. We need the K (4) largest elements, but only 1 is available. Sum the available elements: 5. Wait, the rule says 'If fewer than K elements... return the sum of all available'. So sum is 5. Let me re-read. 'find the sum of the K largest elements... that are greater than K'. If fewer than K exist, sum all that exist. So output is 5. Let me correct the example to be clearer or adjust the numbers. Let's use K=2 for this case to make it a partial match. Revised Example 2: Input: nums = [1, 2, 3, 4, 5], K = 2 Output: 9 Explanation: Filter elements > 2: [3, 4, 5]. Select top 2 largest: 5, 4. Sum = 9.
Input
nums = [1, 2, 3], K = 5
Output
0
Explanation: Filter elements greater than K (5). No elements in [1, 2, 3] are greater than 5. The filtered set is empty. Sum of available elements is 0.
Input
nums = [100, 90, 80, 70, 60], K = 95
Output
100
Explanation: Filter elements greater than K (95). Only 100 is greater than 95. The filtered set is [100]. We need the top K (95) largest, but only 1 is available. Sum the available elements: 100.
Constraints
- 1 <= nums.length <= 10^5
- 1 <= K <= 10^5
- 1 <= nums[i] <= 10^9
Optimal Approach & Strategy
Filter values > K, then use a min‑heap of size K (or QuickSelect) to keep only the K largest qualifying numbers.
Brute Force Approach
Sort the entire array and then pick the K largest elements that are > K.
Verified Code Solutions
function solution(nums, K) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < K; i++) {
if (nums[i] > K) {
sum += nums[i];
} else {
break;
}
}
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++) {
if (nums[i] > K) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < K; i++) {
if (nums[i] > K) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
}def solution(nums, K):
nums.sort(reverse=True)
sum = 0
for i in range(K):
if nums[i] > K:
sum += nums[i]
else:
break
return sumfunction solution(nums, K) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < K; i++) {
if (nums[i] > K) {
sum += nums[i];
} else {
break;
}
}
return sum;
}Asked in Top Tech Interviews
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