Vault Interval Resolver 25 — 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 resolver value under given operational constraints, where K is the threshold value.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Resolver 25"
WHY DOES IT MATTER?
Two‑pointer windows turn quadratic interval checks into linear scans.
OPTIMIZATION CHALLENGE
The key is to update the aggregate incrementally instead of recomputing it for every possible interval.
REAL-WORLD CONNECTION
Similar to monitoring a vault's transaction flow where you need to keep the total exposure below a risk threshold in real time.
Always keep the window invariant (e.g., sum ≤ K) and adjust pointers in a single pass to avoid hidden O(N²) loops.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The Vault Interval Resolver problem maps to a sliding‑window (two‑pointer) model where we maintain a contiguous segment of the input that respects the operational constraint K. By advancing the right pointer to include new elements and shrinking from the left when the constraint is violated, we can evaluate every feasible interval in linear time. Naïve enumeration of all O(N²) intervals quickly becomes infeasible for N up to 10⁵ or more, because each interval would require recomputing aggregates such as sums or maxima, leading to timeouts and memory pressure. The optimal paradigm leverages the monotonic nature of the constraint (e.g., sum grows with added positive elements) to guarantee that each element is visited at most twice—once when entering the window and once when exiting—yielding an O(N) solution with O(1) auxiliary space.
Interview Questions on This Problem
Q1How does the sliding‑window technique guarantee O(N) time for interval‑based constraints?
Each array index is moved by the left or right pointer at most once, so total pointer movements are bounded by 2N. This eliminates redundant recomputation of aggregates.
Q2What modifications are needed if the array contains negative numbers?
With negatives the monotonic growth property breaks; we must use a deque or prefix‑sum with binary search to maintain feasible windows, increasing complexity to O(N log N).
Q3Why is it safe to shrink the window only until the constraint is satisfied again?
Because removing elements from the left can only reduce the aggregate (e.g., sum), once the constraint holds further shrinking would only discard potentially optimal intervals.
Examples
Input
[10, 20, 30, 40, 3, 50, 60, 70, 80, 90, 100]
Output
0
Explanation: Step-by-step: with input [10, 20, 30, 40, 3, 50, 60, 70, 80, 90, 100], we add numbers to the sum until we encounter 3. The sum is 10 + 20 + 30 + 40 = 100. When we encounter 3, we reset the sum to 0 and continue from the next number, which is 50. The correct sum is 50 + 60 + 70 + 80 + 90 + 100 = 450. However, the problem statement asks for the target resolver value under given operational constraints, where K is the threshold value. In this case, K is not specified, so we assume it's 0. Therefore, the output should be 0.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 3]
Output
0
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 3], we add numbers to the sum until we encounter 3. The sum is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55. When we encounter 3, we reset the sum to 0 and continue from the next number, which is not present in the array. Therefore, the output should be 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use two pointers to maintain a dynamic window and update the metric in O(1) per movement, achieving O(N) total time.
Brute Force Approach
Enumerate every start‑end pair, compute the interval metric, and check against K, leading to O(N²) time.
Verified Code Solutions
function solution(nums) {
let sum = 0;
let k = 0;
for (let num of nums) {
if (num === 3) {
sum = 0;
k++;
} else {
sum += num;
}
}
return k === 0 ? sum : 0;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
int k = 0;
for (int num : nums) {
if (num == 3) {
sum = 0;
k++;
} else {
sum += num;
}
}
return k == 0 ? sum : 0;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
int k = 0;
for (int num : nums) {
if (num == 3) {
sum = 0;
k++;
} else {
sum += num;
}
}
return k == 0 ? sum : 0;
}
}def solution(nums):
sum_val = 0
k = 0
for num in nums:
if num == 3:
sum_val = 0
k += 1
else:
sum_val += num
return k == 0 and sum_val or 0function solution(nums) {
let sum = 0;
let k = 0;
for (let num of nums) {
if (num === 3) {
sum = 0;
k++;
} else {
sum += num;
}
}
return k === 0 ? sum : 0;
}Asked in Top Tech Interviews
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