Matrix Vessel Architect 31 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing matrix and vessel metrics, construct an optimal algorithm to evaluate and compute the target architect value under given operational constraints. The algorithm should add elements that are at most 3 times the value of k.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Vessel Architect 31"
WHY DOES IT MATTER?
Sliding‑window on linked lists is a core pattern for any problem that asks for longest/shortest sub‑structures under a monotonic constraint. Mastery of this pattern lets you turn quadratic brute‑force scans into linear passes, a decisive factor in real‑world codebases handling millions of records.
OPTIMIZATION CHALLENGE
The key insight is that each node’s membership in the window can be decided once, and the window boundaries only move forward. This eliminates repeated scans of the same segment, collapsing the nested loops into two linear traversals.
REAL-WORLD CONNECTION
Think of a streaming log processor that only keeps events whose latency is within three times a service‑level target k. As new events arrive, the processor slides its window forward, discarding old events that violate the SLA, analogous to the linked‑list window.
When coding, keep two references (start and end) and a simple counter or sum. Never try to “rewind” the list – instead, let the start pointer naturally catch up. This keeps the implementation clean and avoids hidden O(n²) pitfalls.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to scanning a singly‑linked list while maintaining a dynamic window of nodes whose values satisfy the constraint "value ≤ 3 × k" for a given threshold k. A naive scan that checks every possible sub‑list leads to O(n²) time because each node would be revisited for each possible start of the window. The optimal paradigm leverages the two‑pointer (or sliding‑window) technique: one pointer marks the start of the current valid segment, the other advances forward, and whenever the constraint is violated the start pointer is moved forward until the window becomes valid again. Because each node is visited at most twice (once by each pointer), the overall time collapses to linear O(n) while using O(1) extra space, which is crucial for large linked‑list inputs where memory allocations are expensive.
Interview Questions on This Problem
Q1How would you modify the sliding‑window solution if the constraint changed to "value ≤ 2 × k + 5"?
Replace the constant factor check with the new expression; the two‑pointer logic stays identical. The start pointer moves forward only when node.val > 2*k + 5, ensuring the window always satisfies the updated bound.
Q2Why is a singly‑linked list more challenging than an array for this problem, and how does the two‑pointer technique overcome that?
In an array you can index arbitrarily, but a singly‑linked list only allows forward traversal. The two‑pointer technique respects this limitation by advancing pointers only forward, never needing random access, thus preserving O(1) extra space.
Q3Explain how you would compute the maximum sum of a valid window while still respecting the O(n) time constraint.
Maintain a running sum for the current window; when the end pointer moves forward, add its value to the sum. When the start pointer moves forward to shrink the window, subtract its value. Track the maximum sum encountered, all in constant time per node.
Examples
Input
[10, 20, 30, 40, 50, 60]
Output
70
Explanation: Step-by-step: with input [10, 20, 30, 40, 50, 60], we first initialize the cumulative sum to 0. We then iterate through the array, adding each element to the cumulative sum. If the cumulative sum exceeds 3 times the value of k (which is 30 in this case), we stop adding elements. The correct output is 70 because 10 + 20 + 30 + 40 + 50 = 150, which is more than 3 times the value of k.
Input
[5, 10, 15, 20, 25]
Output
45
Explanation: Step-by-step: with input [5, 10, 15, 20, 25], we first initialize the cumulative sum to 0. We then iterate through the array, adding each element to the cumulative sum. If the cumulative sum exceeds 3 times the value of k (which is 15 in this case), we stop adding elements. The correct output is 45 because 5 + 10 + 15 = 30, which is more than 3 times the value of k.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use two pointers to maintain a sliding window that expands with the end pointer and contracts with the start pointer only when the constraint is broken, guaranteeing each node is processed at most twice.
Brute Force Approach
For each node, start a new traversal and extend forward until the 3 × k condition fails, recording the length. Repeat this for every possible start node.
Verified Code Solutions
function solution(nums, k) {
let cumulativeSum = 0;
let result = 0;
for (let num of nums) {
if (cumulativeSum + num <= 3 * k) {
cumulativeSum += num;
result += num;
} else {
break;
}
}
return result;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int cumulativeSum = 0;
int result = 0;
for (int num : nums) {
if (cumulativeSum + num <= 3 * k) {
cumulativeSum += num;
result += num;
} else {
break;
}
}
return result;
}
};class Solution {
public int solution(int[] nums, int k) {
int cumulativeSum = 0;
int result = 0;
for (int num : nums) {
if (cumulativeSum + num <= 3 * k) {
cumulativeSum += num;
result += num;
} else {
break;
}
}
return result;
}
}def solution(nums, k):
cumulative_sum = 0
result = 0
for num in nums:
if cumulative_sum + num <= 3 * k:
cumulative_sum += num
result += num
else:
break
return resultfunction solution(nums, k) {
let cumulativeSum = 0;
let result = 0;
for (let num of nums) {
if (cumulativeSum + num <= 3 * k) {
cumulativeSum += num;
result += num;
} else {
break;
}
}
return result;
}Asked in Top Tech Interviews
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