Matrix Stream Extractor 11 — 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 operational constraints are that the extractor value should only add the sum of the first two values greater than k.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Stream Extractor 11"
WHY DOES IT MATTER?
Early‑exit selection avoids unnecessary work on massive streams.
OPTIMIZATION CHALLENGE
Reducing a potentially O(n log n) problem to O(n) with constant extra memory.
REAL-WORLD CONNECTION
Similar to network packet filters that stop processing after the first matching packets.
Always track the minimal state needed; a simple counter and accumulator often replace bulky data structures.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a selection task: from a sequential data source (which may be generated by an in‑order traversal of a binary tree representing a matrix) we must locate the first two elements that exceed a threshold k and compute their sum. A naive solution would store all elements, sort them, or repeatedly scan the stream, leading to O(n log n) time or O(n) extra space, which quickly becomes infeasible for large n (e.g., matrices with millions of entries).
The optimal paradigm is a single‑pass, constant‑space filter: as we traverse the stream we maintain a counter of qualifying elements and accumulate their values until two have been found, then terminate early. This leverages the fact that order matters ("first two") and that we only need a fixed‑size state, yielding O(n) worst‑case time and O(1) auxiliary space, which scales gracefully for massive inputs.
Interview Questions on This Problem
Q1How would you modify the algorithm if the requirement changed to the sum of the first three values greater than k?
Increase the counter to three and continue the single pass until three qualifying elements are collected. The time and space complexities remain O(n) and O(1) respectively.
Q2Why is it unsafe to sort the entire stream before extracting the two values?
Sorting incurs O(n log n) time and requires O(n) extra memory, which can exceed limits for large streams. Moreover, sorting destroys the original order, violating the "first two" constraint.
Q3Can this approach be parallelized across multiple cores?
Parallelism is limited because the algorithm depends on the global order of qualifying elements; early termination after two matches prevents independent chunk processing. A possible workaround is to partition the stream, find local candidates, then merge respecting order, but overhead often outweighs benefits.
Examples
Input
[30, 40, 50, 60, 70, 80, 90]
Output
180
Explanation: Step-by-step: Given the input array [30, 40, 50, 60, 70, 80, 90], we need to find the sum of the first two values greater than k. In this case, k is 50. The first two values greater than k are 60 and 70. Therefore, the output is 60 + 70 = 130. However, we need to continue the process until we find the sum of the first two values greater than k. The next two values greater than k are 80 and 90. Therefore, the final output is 130 + 80 + 90 = 300.
Input
[10, 20, 30, 40, 50]
Output
100
Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50], we need to find the sum of the first two values greater than k. In this case, k is 40. The first two values greater than k are 50 and 50. Therefore, the output is 50 + 50 = 100.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Traverse once, keep a count and sum of qualifying elements, stop after two are found.
Brute Force Approach
Collect all elements, sort them, then scan for the first two values > k and sum them.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
let count = 0;
for (let num of nums) {
if (num > k) {
count++;
if (count === 1) {
sum += num;
} else if (count === 2) {
sum += num;
break;
}
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sum = 0;
int count = 0;
for (int num : nums) {
if (num > k) {
count++;
if (count == 1) {
sum += num;
} else if (count == 2) {
sum += num;
break;
}
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
int count = 0;
for (int num : nums) {
if (num > k) {
count++;
if (count == 1) {
sum += num;
} else if (count == 2) {
sum += num;
break;
}
}
}
return sum;
}
}def solution(nums, k):
sum = 0
count = 0
for num in nums:
if num > k:
count += 1
if count == 1:
sum += num
elif count == 2:
sum += num
break
return sumfunction solution(nums, k) {
let sum = 0;
let count = 0;
for (let num of nums) {
if (num > k) {
count++;
if (count === 1) {
sum += num;
} else if (count === 2) {
sum += num;
break;
}
}
}
return sum;
}Asked in Top Tech Interviews
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