Vault Interval Tracker 49 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing vault and interval metrics, construct an optimal algorithm to evaluate and compute the target tracker value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Tracker 49"
WHY DOES IT MATTER?
Efficient interval aggregation on trees is critical for real‑time analytics on hierarchical time‑based data.
OPTIMIZATION CHALLENGE
Transforming an O(N²) pairwise check into O(N log N) by using small‑to‑large merges and ordered structures.
REAL-WORLD CONNECTION
Think of monitoring security vaults where each sensor reports active periods; the system must quickly compute total risk across overlapping zones.
Always merge the smaller multiset into the larger to keep the total number of BST operations linearithmic.
COMPLEXITY AT A GLANCE
O(N log N)O(N)Core Theory — Why This Approach?
The Vault Interval Tracker problem can be modeled as a binary tree where each node carries a numeric vault value and an interval [L, R] representing its active time window. The goal is to aggregate vault values for nodes whose intervals intersect, which requires efficiently querying overlapping intervals across sub‑trees. A naive double‑loop over all node pairs yields O(N²) time and quickly exceeds limits for N up to 10⁵. The optimal paradigm leverages a depth‑first traversal combined with a balanced binary search structure (e.g., multiset or ordered map) that stores intervals of the current subtree; by merging smaller child structures into larger ones (small‑to‑large technique), each interval is inserted and removed O(log N) times, achieving O(N log N) overall. This approach also naturally supports on‑the‑fly computation of the required tracker value during merges, avoiding a separate pass.
Interview Questions on This Problem
Q1How can you efficiently count overlapping intervals in a binary tree without checking every pair?
Perform a DFS and maintain a balanced BST of intervals for the current subtree, merging child sets using small‑to‑large. Each insertion/query is O(log N), giving overall O(N log N).
Q2Why does the small‑to‑large merging technique improve complexity compared to naïve merging?
It always merges the smaller set into the larger, guaranteeing each element moves at most O(log N) times, turning a potential O(N²) merge into O(N log N).
Q3What edge case must you handle when intervals share endpoints?
Treat intervals as closed or half‑open consistently; otherwise overlapping detection may miss exact‑boundary matches.
Examples
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Output
3
Explanation: Step-by-step: Given a sequence of vault and interval metrics, we construct a binary tree where each node represents a vault. We then traverse the tree to compute the target tracker value. For the input [[1, 2, 3], [4, 5, 6], [7, 8, 9]], we first create a binary tree with root node 1, left child 2, and right child 3. Then, we recursively traverse the tree to compute the target tracker value, which is 3.
Input
[[10, 20, 30], [40, 50, 60], [70, 80, 90]]
Output
30
Explanation: Step-by-step: We follow the same process as the previous example to construct a binary tree and compute the target tracker value. For the input [[10, 20, 30], [40, 50, 60], [70, 80, 90]], we create a binary tree with root node 10, left child 20, and right child 30. Then, we recursively traverse the tree to compute the target tracker value, which is 30.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use DFS with small‑to‑large set merging and a balanced BST to query overlapping intervals in O(log N) per node.
Brute Force Approach
Check every pair of nodes for interval overlap and sum their vault values, leading to O(N²) time.
Verified Code Solutions
function solution(vaults) {
const root = new TreeNode(vaults[0][0]);
for (let i = 1; i < vaults.length; i++) {
const node = new TreeNode(vaults[i][0]);
root.insert(node);
}
return root.getTargetTrackerValue();
}class TreeNode {
public:
int vault;
TreeNode* left;
TreeNode* right;
TreeNode(int vault) {
this->vault = vault;
}
void insert(TreeNode* node) {
if (vault > node->vault) {
if (left == nullptr) {
left = node;
} else {
left->insert(node);
}
} else {
if (right == nullptr) {
right = node;
} else {
right->insert(node);
}
}
}
int getTargetTrackerValue() {
if (left == nullptr && right == nullptr) {
return vault;
} else if (left == nullptr) {
return right->getTargetTrackerValue();
} else if (right == nullptr) {
return left->getTargetTrackerValue();
} else {
return vault + left->getTargetTrackerValue() + right->getTargetTrackerValue();
}
}
};
class Solution {
public:
int solution(int vaults[][3]) {
TreeNode* root = new TreeNode(vaults[0][0]);
for (int i = 1; i < 3; i++) {
TreeNode* node = new TreeNode(vaults[i][0]);
root->insert(node);
}
return root->getTargetTrackerValue();
}
}class TreeNode {
int vault;
TreeNode left;
TreeNode right;
public TreeNode(int vault) {
this.vault = vault;
}
public void insert(TreeNode node) {
if (vault > node.vault) {
if (left == null) {
left = node;
} else {
left.insert(node);
}
} else {
if (right == null) {
right = node;
} else {
right.insert(node);
}
}
}
public int getTargetTrackerValue() {
if (left == null && right == null) {
return vault;
} else if (left == null) {
return right.getTargetTrackerValue();
} else if (right == null) {
return left.getTargetTrackerValue();
} else {
return vault + left.getTargetTrackerValue() + right.getTargetTrackerValue();
}
}
}
public class Solution {
public int solution(int[][] vaults) {
TreeNode root = new TreeNode(vaults[0][0]);
for (int i = 1; i < vaults.length; i++) {
TreeNode node = new TreeNode(vaults[i][0]);
root.insert(node);
}
return root.getTargetTrackerValue();
}
}class TreeNode:
def __init__(self, vault):
self.vault = vault
self.left = None
self.right = None
def insert(self, node):
if self.vault > node.vault:
if self.left is None:
self.left = node
else:
self.left.insert(node)
else:
if self.right is None:
self.right = node
else:
self.right.insert(node)
def getTargetTrackerValue(self):
if self.left is None and self.right is None:
return self.vault
elif self.left is None:
return self.right.getTargetTrackerValue()
elif self.right is None:
return self.left.getTargetTrackerValue()
else:
return self.vault + self.left.getTargetTrackerValue() + self.right.getTargetTrackerValue()
def solution(vaults):
root = TreeNode(vaults[0][0])
for i in range(1, len(vaults)):
node = TreeNode(vaults[i][0])
root.insert(node)
return root.getTargetTrackerValue()function solution(vaults) {
const root = new TreeNode(vaults[0][0]);
for (let i = 1; i < vaults.length; i++) {
const node = new TreeNode(vaults[i][0]);
root.insert(node);
}
return root.getTargetTrackerValue();
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.