Node Vault Consolidator 16 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing node and vault metrics, construct an optimal algorithm to evaluate and compute the target consolidator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Vault Consolidator 16"
WHY DOES IT MATTER?
Graph traversal is the foundational pattern for any problem involving relationships, dependencies, or networks. Mastering it allows engineers to solve a vast array of problems from social network analysis to route finding and resource allocation.
OPTIMIZATION CHALLENGE
The key insight is maintaining a 'visited' state to prevent infinite loops in cyclic graphs and to ensure O(V+E) time complexity by avoiding redundant processing of nodes.
REAL-WORLD CONNECTION
This pattern directly mirrors how data centers manage load balancing across server clusters or how logistics companies optimize delivery routes across a network of warehouses and distribution hubs.
Always clarify the graph structure (directed/undirected, cyclic/acyclic) before coding. For easy problems, a recursive DFS is often cleaner, but iterative BFS with a queue is safer for avoiding stack overflow on deep graphs.
COMPLEXITY AT A GLANCE
O(V + E)O(V)Core Theory — Why This Approach?
The 'Node Vault Consolidator' problem fundamentally revolves around graph traversal and aggregation, specifically utilizing Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse a connected component or the entire graph. The core theoretical challenge lies in efficiently accumulating metrics (such as node weights or vault capacities) while respecting operational constraints, which often manifest as cycle detection or specific path requirements. In an easy difficulty context, this typically implies a tree structure or a simple connected graph where the goal is to sum or count properties of nodes reachable from a specific source or across the entire structure.
Interview Questions on This Problem
Q1At a fintech platform, how would you design a system to aggregate transaction volumes across a network of regional banks (nodes) connected by settlement channels (edges) without double-counting shared settlements?
Model the banks as nodes and settlement channels as edges. Use a visited set during graph traversal (DFS/BFS) to ensure each node's metric is processed exactly once, preventing double-counting in cyclic settlement networks.
Q2For a high-growth startup's microservices architecture, how do you calculate the total latency impact of a dependency chain if services form a directed acyclic graph (DAG)?
Perform a topological sort or DFS on the DAG. Accumulate the latency metrics as you traverse from source to sink, ensuring that shared dependencies are only added to the total path cost once per unique path or aggregated globally depending on the specific SLA requirement.
Q3In a global product company's CDN network, how do you determine the total storage capacity available in a specific geographic region given that some servers are shared across regions?
Represent servers as nodes and regional boundaries as constraints. Use a graph traversal algorithm to visit all nodes within the region's subgraph, summing their storage capacities while using a global visited set to exclude servers that belong to multiple regions if the requirement is exclusive ownership.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5], 10
Output
400
Explanation: Step-by-step: The input array is [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5] and the target number K is 10. We iterate through the array and sum all the numbers: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 1 + 2 + 3 + 4 + 5 = 60 + 70 + 80 + 90 + 100 + 60 + 70 + 80 + 90 + 100 + 60 + 70 + 80 + 90 + 100 = 400.
Input
[1, 2, 3, 4, 5], 10
Output
0
Explanation: Step-by-step: The input array is [1, 2, 3, 4, 5] and the target number K is 10. We iterate through the array and sum all the numbers: 1 + 2 + 3 + 4 + 5 = 15. Since 15 is less than 10, we return 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a single pass graph traversal (DFS or BFS) with a visited set to ensure each node is processed exactly once. Accumulate the node metrics during the traversal to compute the total consolidator value in linear time relative to the graph size.
Brute Force Approach
Recursively explore all possible paths from the start node, summing metrics for every path, which leads to exponential time complexity due to revisiting nodes in cyclic graphs. This approach fails on large inputs because it does not track visited states, causing redundant calculations and potential stack overflows.
Verified Code Solutions
function solution(nums, K) {
if (nums.length === 0 || K < 0) return 0;
let sum = 0;
for (let num of nums) {
if (typeof num !== 'number') continue;
sum += num;
}
return sum > K ? sum : 0;
}class Solution {
public:
int solution(vector<int> nums, int K) {
if (nums.size() == 0 || K < 0) return 0;
int sum = 0;
for (int num : nums) {
if (std::isnan(num)) continue;
sum += num;
}
return sum > K ? sum : 0;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length == 0 || K < 0) return 0;
int sum = 0;
for (int num : nums) {
if (Double.isNaN(num)) continue;
sum += num;
}
return sum > K ? sum : 0;
}
}def solution(nums, K):
if not nums or K < 0:
return 0
sum = 0
for num in nums:
if not isinstance(num, (int, float)):
continue
sum += num
return sum if sum > K else 0function solution(nums, K) {
if (nums.length === 0 || K < 0) return 0;
let sum = 0;
for (let num of nums) {
if (typeof num !== 'number') continue;
sum += num;
}
return sum > K ? sum : 0;
}Asked in Top Tech Interviews
Solve in Interative Editor
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