Vault Interval Synthesizer 46 — 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 synthesizer value under given operational constraints. The synthesizer value is calculated by summing up the interval values for each pair of vault and interval metrics where the vault value is greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Synthesizer 46"
WHY DOES IT MATTER?
Processing condition‑based aggregates in O(n) is a core skill for high‑throughput systems.
OPTIMIZATION CHALLENGE
The key is reducing pairwise checks to a single linear scan by maintaining state.
REAL-WORLD CONNECTION
Similar to filtering sensor readings in an IoT pipeline where only values above a threshold contribute to analytics.
Prefer a running sum with a lightweight queue over recomputing sums from scratch each iteration.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The naive solution iterates over every possible vault‑interval pair, checking if the vault value exceeds K and, if so, adding the corresponding interval value. This double loop incurs O(n²) time, which quickly becomes infeasible for large streams of data. The optimal paradigm treats the input as a stream and leverages a queue (or simple running sum) to keep only the interval values whose associated vaults satisfy the condition, allowing a single linear pass. By enqueuing interval values when their vaults are > K and dequeuing (or ignoring) when they are not, we maintain the exact contribution to the final synthesizer value without revisiting elements, achieving O(n) time and O(n) auxiliary space in the worst case.
Interview Questions on This Problem
Q1Why does a double‑loop solution exceed time limits for n ≈ 10⁵?
It performs O(n²) comparisons, leading to ~10¹⁰ operations, which far exceeds typical 1‑2 second limits.
Q2How can a queue help compute the sum in a single pass?
The queue stores interval values whose vaults are > K, so we can add or remove contributions in O(1) as we scan the array.
Q3What edge case must be handled when K equals the maximum vault value?
All vaults fail the > K test, so the answer should be zero; forgetting the strict inequality yields an off‑by‑one error.
Examples
Input
[[1, 2], [3, 4], [5, 6]], 3
Output
12
Explanation: Step-by-step: Given the input [[1, 2], [3, 4], [5, 6]], 3, we iterate through each pair of vault and interval metrics. For the first pair [1, 2], the vault value 1 is not greater than K, so we skip it. For the second pair [3, 4], the vault value 3 is greater than K, so we add the interval value 4 to the synthesizer value. For the third pair [5, 6], the vault value 5 is greater than K, so we add the interval value 6 to the synthesizer value. Therefore, the synthesizer value is 2 + 4 + 6 = 12.
Input
[[7, 8], [9, 10], [11, 12]], 10
Output
0
Explanation: Step-by-step: Given the input [[7, 8], [9, 10], [11, 12]], 10, we iterate through each pair of vault and interval metrics. For the first pair [7, 8], the vault value 7 is not greater than K, so we skip it. For the second pair [9, 10], the vault value 9 is not greater than K, so we skip it. For the third pair [11, 12], the vault value 11 is not greater than K, so we skip it. Therefore, the synthesizer value is 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Iterate once, maintain a queue (or running sum) of intervals whose vaults satisfy vault > K, updating the total in constant time per element.
Brute Force Approach
Use two nested loops to examine every vault‑interval pair and sum intervals when vault > K.
Verified Code Solutions
function solution(vaults, k) {
let synthesizerValue = 0;
for (let i = 0; i < vaults.length; i++) {
if (vaults[i][0] > k) {
synthesizerValue += vaults[i][1];
}
}
return synthesizerValue;
}class Solution {
public:
int solution(vector<vector<int>>& vaults, int k) {
int synthesizerValue = 0;
for (int i = 0; i < vaults.size(); i++) {
if (vaults[i][0] > k) {
synthesizerValue += vaults[i][1];
}
}
return synthesizerValue;
}
};class Solution {
public int solution(int[][] vaults, int k) {
int synthesizerValue = 0;
for (int i = 0; i < vaults.length; i++) {
if (vaults[i][0] > k) {
synthesizerValue += vaults[i][1];
}
}
return synthesizerValue;
}
}def solution(vaults, k):
synthesizer_value = 0
for vault in vaults:
if vault[0] > k:
synthesizer_value += vault[1]
return synthesizer_valuefunction solution(vaults, k) {
let synthesizerValue = 0;
for (let i = 0; i < vaults.length; i++) {
if (vaults[i][0] > k) {
synthesizerValue += vaults[i][1];
}
}
return synthesizerValue;
}Asked in Top Tech Interviews
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