Node Vault Tracker 7 — 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 tracker value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Vault Tracker 7"
WHY DOES IT MATTER?
Aggregating information in a single pass avoids exponential blow‑up on large trees.
OPTIMIZATION CHALLENGE
The key is to reduce repeated subtree traversals to a single linear pass.
REAL-WORLD CONNECTION
Similar to summarizing sensor data in hierarchical IoT networks where each gateway aggregates child readings.
Cache child results in the call stack and compute parent values immediately to keep memory footprint minimal.
COMPLEXITY AT A GLANCE
O(N)O(H)Core Theory — Why This Approach?
The Node Vault Tracker problem reduces to aggregating values across a binary tree where each node represents a vault with a metric. By performing a single post‑order traversal we can compute the required tracker value using a bottom‑up dynamic programming approach, storing only the necessary intermediate results. Naïve solutions often recompute sub‑tree aggregates for each node, leading to O(N^2) time on skewed trees because each subtree is visited repeatedly. The optimal paradigm leverages the tree’s hierarchical structure: each node’s answer is derived from its children’s answers in constant time, guaranteeing linear O(N) overall complexity.
Interview Questions on This Problem
Q1How does a post‑order traversal help compute aggregate values in a binary tree?
It visits children before the parent, ensuring that all sub‑tree information is available when processing a node. This eliminates redundant recomputation of child results.
Q2Why does a naïve double‑loop over nodes lead to O(N^2) time for this problem?
Each node may trigger a full traversal of its subtree, causing overlapping work across nodes. In the worst case (e.g., a linked‑list shaped tree) the total work sums to a quadratic series.
Q3What is the space complexity of the optimal solution and why?
It is O(H), where H is the tree height, because recursion (or an explicit stack) stores only the call chain. No additional data structures proportional to N are required.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
45
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we first identify the numbers greater than K=5, which are 6, 7, 8, 9, and 10. Then, we sum these numbers to get the target tracker value, which is 45.
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5], we first identify the numbers greater than K=3, which are 4 and 5. Then, we sum these numbers to get the target tracker value, which is 9. However, since 4 and 5 are greater than K=3, we should include them in the sum, giving us a total of 9.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Perform a single DFS that returns the subtree aggregate to its parent, computing each node’s contribution in O(1) time.
Brute Force Approach
Re‑compute the subtree sum for every node by launching a fresh DFS from that node, leading to repeated work.
Verified Code Solutions
function solution(nums, K) {
let tracker = 0;
for (let num of nums) {
if (num > K) {
tracker += num;
}
}
return tracker;
}class Solution {
public:
int solution(vector<int> nums, int K) {
int tracker = 0;
for (int num : nums) {
if (num > K) {
tracker += num;
}
}
return tracker;
}
};class Solution {
public int solution(int[] nums, int K) {
int tracker = 0;
for (int num : nums) {
if (num > K) {
tracker += num;
}
}
return tracker;
}
}def solution(nums, K):
tracker = 0
for num in nums:
if num > K:
tracker += num
return trackerfunction solution(nums, K) {
let tracker = 0;
for (let num of nums) {
if (num > K) {
tracker += num;
}
}
return tracker;
}Asked in Top Tech Interviews
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