Matrix Stream Consolidator 3 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing matrix and stream metrics, and a threshold K, construct an optimal algorithm to evaluate and compute the target consolidator value under given operational constraints. The target consolidator value is the maximum sum of metrics that can be achieved by selecting a subset of the metrics, subject to the constraint that the sum of the selected metrics does not exceed the threshold K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Stream Consolidator 3"
WHY DOES IT MATTER?
The pattern of prefix‑sum + ordered set is a cornerstone for many “max‑subarray‑under‑constraint” problems. It transforms a combinatorial subset selection into a series of range‑minimum queries, turning an exponential search space into a tractable logarithmic one.
OPTIMIZATION CHALLENGE
The key insight is that the optimal subset can be expressed as the difference between two prefix sums. By storing all previous prefixes in a sorted container, we can instantly locate the best partner prefix that respects the K bound, eliminating the need for nested loops.
REAL-WORLD CONNECTION
Think of a distributed log‑aggregation service that must cap the total size of logs sent to a downstream system. Each log entry is a metric; the service must pick the largest possible batch without exceeding the bandwidth limit K. Maintaining prefix sums of batch sizes and querying the nearest feasible batch size mirrors the algorithmic pattern.
During an interview, compute the running prefix sum on the fly and immediately query the BST; never store the entire list of values first. This incremental approach demonstrates both space efficiency and real‑time thinking.
COMPLEXITY AT A GLANCE
O(N log N)O(N)Core Theory — Why This Approach?
The Matrix Stream Consolidator problem can be reduced to finding the maximum sum of a subsequence (not necessarily contiguous) whose total does not exceed a given threshold K. A naïve solution enumerates every possible subset, leading to exponential time, or checks every pair of prefix sums for a contiguous variant, which is O(N^2). The optimal paradigm leverages prefix‑sum transformation combined with an ordered data structure (e.g., balanced binary search tree or multiset). By scanning the stream once, we maintain all previously seen prefix sums; for each new prefix sum P we query the smallest earlier prefix Q such that P‑Q ≤ K, i.e., Q ≥ P‑K. The difference P‑Q yields a candidate answer. This reduces the problem to O(N log N) time and O(N) space, which scales to the massive streams typical in production systems.
Interview Questions on This Problem
Q1How would you adapt the maximum sub‑array‑sum‑≤ K solution to work on a singly linked list where random access is unavailable?
Traverse the list while maintaining a running prefix sum and insert each prefix into a self‑balancing BST (or a sorted container). For each new prefix, search for the smallest prefix ≥ currentPrefix‑K using the BST’s lower_bound operation; the difference gives a candidate sum. This preserves O(N log N) time without needing random access.
Q2Explain why a sliding‑window two‑pointer technique fails for the non‑contiguous version of this problem.
Two‑pointer windows rely on monotonic growth of the window sum, which only holds when the selected elements are contiguous. In the non‑contiguous variant we may skip elements arbitrarily, so shrinking or expanding the window does not guarantee moving toward a feasible sum, causing the method to miss optimal subsets.
Q3A fintech platform needs to process a real‑time stream of transaction amounts and constantly report the maximum total that does not exceed a regulatory cap K. Which data structure would you choose and why?
A balanced BST (e.g., TreeSet in Java or std::set in C++) is ideal because it supports O(log N) insertion and O(log N) predecessor/successor queries needed to locate the best prefix sum ≤ currentPrefix‑K. This enables continuous, low‑latency updates as new transactions arrive.
Examples
Input
[10, 20, 30], 30
Output
30
Explanation: Step-by-step: with input [10, 20, 30] and threshold 30, we select the metric 30, giving output 30 because it is the maximum sum that can be achieved without exceeding the threshold
Input
[5, 10, 15], 20
Output
20
Explanation: Step-by-step: with input [5, 10, 15] and threshold 20, we select the metrics 5 and 15, giving output 20 because it is the maximum sum that can be achieved without exceeding the threshold
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a running prefix sum and an ordered set of all previous prefixes; for each new prefix, perform a lower‑bound search for prefix ≥ current‑K to compute the best feasible sum in O(log N) per element.
Brute Force Approach
Enumerate every subset (or every pair of prefix sums for the contiguous case) and keep the best sum ≤ K, which costs exponential or quadratic time.
Verified Code Solutions
function solution(nums, k) {
let maxSum = 0;
for (let i = 0; i < (1 << nums.length); i++) {
let sum = 0;
for (let j = 0; j < nums.length; j++) {
if ((i & (1 << j)) !== 0) {
sum += nums[j];
}
}
if (sum <= k && sum > maxSum) {
maxSum = sum;
}
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int maxSum = 0;
for (int i = 0; i < (1 << nums.size()); i++) {
int sum = 0;
for (int j = 0; j < nums.size(); j++) {
if ((i & (1 << j)) != 0) {
sum += nums[j];
}
}
if (sum <= k && sum > maxSum) {
maxSum = sum;
}
}
return maxSum;
}
}class Solution {
public int solution(int[] nums, int k) {
int maxSum = 0;
for (int i = 0; i < (1 << nums.length); i++) {
int sum = 0;
for (int j = 0; j < nums.length; j++) {
if ((i & (1 << j)) != 0) {
sum += nums[j];
}
}
if (sum <= k && sum > maxSum) {
maxSum = sum;
}
}
return maxSum;
}
}def solution(nums, k):
max_sum = 0
for i in range(1 << len(nums)):
sum = 0
for j in range(len(nums)):
if (i & (1 << j)) != 0:
sum += nums[j]
if sum <= k and sum > max_sum:
max_sum = sum
return max_sumfunction solution(nums, k) {
let maxSum = 0;
for (let i = 0; i < (1 << nums.length); i++) {
let sum = 0;
for (let j = 0; j < nums.length; j++) {
if ((i & (1 << j)) !== 0) {
sum += nums[j];
}
}
if (sum <= k && sum > maxSum) {
maxSum = sum;
}
}
return maxSum;
}Asked in Top Tech Interviews
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