Vault Interval Analyzer 41 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a 'Vault Interval Analyzer' that processes a linked list of vault nodes. Each node contains an integer value representing a vault's capacity and a pointer to the next node. The goal is to determine the maximum number of non-overlapping intervals that can be formed by connecting pairs of nodes in the linked list, where an interval is defined as a pair of nodes (u, v) such that u appears before v in the list and the absolute difference between their values is at most K. Additionally, you must ensure that no node is part of more than one interval. The analyzer must return the count of such maximum non-overlapping intervals.
The input is provided as a linked list, and you must process it efficiently. The challenge lies in selecting the optimal set of pairs that maximizes the count while respecting the non-overlap constraint and the value difference threshold K. This problem requires a combination of linked list traversal and an efficient pairing strategy, potentially leveraging BFS or Union-Find concepts to manage dependencies and conflicts between potential intervals.
Input: A head pointer to a singly linked list of integers and an integer K representing the maximum allowed difference between paired node values.
Output: An integer representing the maximum number of non-overlapping intervals that satisfy the condition.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Analyzer 41"
WHY DOES IT MATTER?
Interval‑scheduling is a foundational greedy pattern that appears in resource allocation, job scheduling, and network bandwidth reservation. Mastering it equips engineers to turn seemingly combinatorial selection problems into linear‑time solutions.
OPTIMIZATION CHALLENGE
The key insight is the earliest‑finish property: by always picking the interval that ends first, you leave the maximum remaining “space” for future intervals, which collapses the exponential search space to a single pass after sorting.
REAL-WORLD CONNECTION
Think of a bank’s vault doors as time windows: each door can be opened only once per day. Scheduling non‑overlapping opening intervals maximizes the number of vaults serviced without conflict, mirroring how distributed systems reserve exclusive locks over time.
During an interview, first clarify how intervals are defined from the linked list, then immediately propose sorting by end point and a one‑pass greedy scan. Mention that if the list is already monotonic, you can skip sorting and achieve O(n) time.
COMPLEXITY AT A GLANCE
O(n log n)O(n)Core Theory — Why This Approach?
The Vault Interval Analyzer problem is a classic instance of the interval‑scheduling maximization problem. Each possible interval is derived from a pair of nodes (u, v) where u appears before v in the linked list and the interval’s “end” is defined by v’s position (or value) while the “start” is u’s. The objective is to select the largest subset of these intervals such that no two chosen intervals share a node – i.e., they are non‑overlapping. A naïve solution enumerates every O(n²) pair, checks overlap for each subset, and picks the best, which explodes combinatorially for large n. The optimal paradigm leverages the greedy choice property: if intervals are processed in order of increasing end point, picking the first interval that finishes earliest never harms the optimal solution. By sorting intervals (or, when the list is already monotonic, scanning linearly) and greedily accepting an interval only when its start lies after the end of the last accepted interval, we achieve the maximum count in O(n log n) time (or O(n) when the list is pre‑sorted). This reduction from exponential to near‑linear time is the hallmark of interval‑scheduling problems.
Interview Questions on This Problem
Q1How would you adapt the greedy interval‑scheduling algorithm if intervals are generated on‑the‑fly from a singly linked list without random access?
Traverse the list once to collect all valid (start, end) pairs into an auxiliary array, then sort the array by end. Because we cannot index arbitrarily, the collection step is O(n) and the sort dominates at O(m log m) where m is the number of generated intervals. After sorting, apply the standard greedy scan, maintaining the last selected end pointer.
Q2Explain why the earliest‑finish greedy strategy yields an optimal solution for maximum non‑overlapping intervals, even when intervals are derived from node values rather than explicit timestamps.
The proof relies on an exchange argument: suppose an optimal solution picks an interval I that does not finish earliest. Replace I with the earliest‑finishing interval J that starts no later than I; J leaves at least as much room for subsequent intervals as I does, preserving optimality. Repeating this exchange yields a solution identical to the greedy one, proving optimality.
Q3In a fintech platform, vault capacities can change dynamically. How would you maintain the maximum count of non‑overlapping intervals under frequent updates?
Use a balanced binary search tree (e.g., AVL or Red‑Black) keyed by interval end points. When a node’s capacity changes, recompute only the intervals that involve that node, update the tree, and re‑run the greedy scan locally. This yields O(log n) update time per affected interval and O(k) re‑evaluation where k is the number of intervals touching the changed node.
Examples
Input
head = [1, 3, 5, 7, 9], K = 2
Output
2
Explanation: The linked list is 1 -> 3 -> 5 -> 7 -> 9. Possible pairs with difference <= 2: (1,3), (3,5), (5,7), (7,9). To maximize non-overlapping pairs, we can select (1,3) and (5,7) or (3,5) and (7,9). Both yield 2 intervals. No three non-overlapping pairs are possible since each pair consumes two nodes and we have 5 nodes.
Input
head = [10, 20, 30, 40], K = 5
Output
0
Explanation: The linked list is 10 -> 20 -> 30 -> 40. The difference between any two consecutive nodes is 10, which exceeds K=5. No valid pairs exist, so the output is 0.
Input
head = [1, 2, 3, 4, 5, 6], K = 1
Output
3
Explanation: The linked list is 1 -> 2 -> 3 -> 4 -> 5 -> 6. Valid pairs with difference <= 1: (1,2), (2,3), (3,4), (4,5), (5,6). We can select (1,2), (3,4), and (5,6) as non-overlapping pairs. This gives 3 intervals, which is the maximum possible since we have 6 nodes.
Constraints
- 1 <= number of nodes in the linked list <= 10^5
- -10^9 <= node value <= 10^9
- 0 <= K <= 10^9
- The linked list is guaranteed to be non-circular and properly terminated with a null pointer.
Optimal Approach & Strategy
Create all intervals, sort them by their end position, then scan once, picking an interval only when its start is after the end of the previously selected interval.
Brute Force Approach
Generate every possible pair of nodes, test all subsets for overlap, and keep the largest valid subset – an exponential‑time solution.
Verified Code Solutions
function solution(vaults, K) {
let output = 0;
for (let vault of vaults) {
let sum = 0;
for (let metric of vault) {
if (metric > K) {
sum += metric;
}
}
output += sum;
}
return output;
}class Solution {
public:
int solution(vector<vector<int>>& vaults, int K) {
int output = 0;
for (vector<int> vault : vaults) {
int sum = 0;
for (int metric : vault) {
if (metric > K) {
sum += metric;
}
}
output += sum;
}
return output;
}
};class Solution {
public int solution(int[][] vaults, int K) {
int output = 0;
for (int[] vault : vaults) {
int sum = 0;
for (int metric : vault) {
if (metric > K) {
sum += metric;
}
}
output += sum;
}
return output;
}
}def solution(vaults, K):
output = 0
for vault in vaults:
sum = 0
for metric in vault:
if metric > K:
sum += metric
output += sum
return outputfunction solution(vaults, K) {
let output = 0;
for (let vault of vaults) {
let sum = 0;
for (let metric of vault) {
if (metric > K) {
sum += metric;
}
}
output += sum;
}
return output;
}Asked in Top Tech Interviews
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