Matrix Stream Extractor 46 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing matrix and stream metrics, construct an optimal algorithm to evaluate and compute the target extractor value under given operational constraints. The target extractor value is the sum of all elements greater than a given target K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Stream Extractor 46"
WHY DOES IT MATTER?
Processing predicates on streams in O(1) per element avoids quadratic blow‑up for repeated queries.
OPTIMIZATION CHALLENGE
The key is reducing repeated full‑list scans to a single incremental update.
REAL-WORLD CONNECTION
Similar to real‑time analytics pipelines that continuously aggregate metrics above a threshold.
Keep the predicate static and update a single accumulator; avoid building auxiliary containers unless the predicate changes.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to aggregating values that satisfy a simple predicate (> K) over a linear data structure. In a naive setting where each query re‑scans the entire linked list, the time grows linearly with the number of queries, which quickly becomes infeasible for large streams. The optimal paradigm treats the list as a stream: as each node arrives we update a running total only if the node’s value exceeds K, achieving O(N) total work regardless of query count. This leverages the fact that the predicate is monotonic and does not depend on future elements, allowing constant‑time per‑element processing and O(1) auxiliary space.
Interview Questions on This Problem
Q1How would you compute the sum of elements greater than K in a singly linked list in one pass?
Traverse the list once, adding each node’s value to an accumulator only if it exceeds K. This yields O(N) time and O(1) extra space.
Q2If the list is a continuous data stream with frequent queries, what data structure helps maintain the sum efficiently?
Maintain a running sum variable that updates on each insertion; no extra structure is needed because the predicate is static. Queries then read the variable in O(1).
Q3Why can we’t use binary search or segment trees on a linked list for this problem?
Linked lists lack random access, so index‑based structures would require O(N) traversal to locate nodes, negating their logarithmic benefits. A linear scan is therefore optimal.
Examples
Input
[1, 2, 3, 4, 5], 3
Output
12
Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and target K = 3, we sum elements greater than K. First, identify elements greater than 3: [4, 5]. Then, sum these elements: 4 + 5 = 9. However, the problem statement asks for the sum of elements greater than K, so the correct output should be the sum of these elements, which is 9.
Input
[10, 20, 30, 40, 50], 20
Output
120
Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and target K = 20, we sum elements greater than K. First, identify elements greater than 20: [30, 40, 50]. Then, sum these elements: 30 + 40 + 50 = 120.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a single running sum while streaming the list, updating it only when a node > K, yielding O(N) total time.
Brute Force Approach
For each query, traverse the entire list and sum qualifying nodes, leading to O(N × Q) time.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (num > k) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
}def solution(nums, k):
return sum(num for num in nums if num > k)function solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (num > k) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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