Vault Buffer Optimizer 15 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing vault and buffer metrics, construct an optimal algorithm to evaluate and compute the target optimizer value under given operational constraints. The target optimizer value is the sum of the first K elements in the array, where K is a positive integer less than or equal to the length of the array. If the input array contains non-numeric values or K is less than 1 or greater than the length of the array, return -1.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Buffer Optimizer 15"
WHY DOES IT MATTER?
Binary search on derived monotonic arrays turns expensive linear scans into logarithmic lookups.
OPTIMIZATION CHALLENGE
The key is reducing repeated sum calculations from O(N^2) to O(N) by exploiting prefix‑sum monotonicity.
REAL-WORLD CONNECTION
Think of a bank vault where cumulative deposits grow; finding the day when a balance threshold is crossed mirrors this pattern.
Always pre‑compute reusable aggregates (like prefix sums) before applying binary search to avoid hidden O(N) loops inside the log factor.
COMPLEXITY AT A GLANCE
O(N + log N) ≈ O(N)O(N)Core Theory — Why This Approach?
Binary search leverages the monotonic property of prefix sums: as K increases, the sum of the first K elements never decreases. By pre‑computing a prefix‑sum array, we transform the problem of finding the optimal K into searching for a target value in a sorted list, enabling O(log N) queries instead of linear scans. Naïve approaches recompute sums for each K, leading to O(N^2) time on large inputs, which quickly exceeds limits. The optimal paradigm combines prefix sums (O(N) preprocessing) with binary search (O(log N) query) to achieve overall O(N + log N) ≈ O(N) time while using O(N) auxiliary space.
Interview Questions on This Problem
Q1Why does binary search require a monotonic condition, and how does the prefix‑sum array guarantee it for this problem?
Binary search works only when the predicate changes at most once (false→true). Prefix sums are cumulative, so the sum grows monotonically with K, satisfying the condition.
Q2What is the time‑space trade‑off when using a prefix‑sum array versus recomputing sums on the fly?
Prefix sums use O(N) extra space but reduce query time from O(N) to O(1), turning an overall O(N^2) solution into O(N). Without it, you save space but pay quadratic time.
Q3How would you adapt the solution if the array could contain negative numbers?
Negative values break monotonicity; you would need a different structure (e.g., segment tree) or a two‑pointer sliding window instead of binary search.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100] K = 5
Output
150
Explanation: Step-by-step: Given an array of vault and buffer metrics, we need to find the sum of the first K elements. In this case, K is 5. So, we sum the first 5 elements: 10 + 20 + 30 + 40 + 50 = 150.
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100] K = 3
Output
90
Explanation: Step-by-step: Given an array of vault and buffer metrics, we need to find the sum of the first K elements. In this case, K is 3. So, we sum the first 3 elements: 10 + 20 + 30 = 60. However, the problem seems to be asking for the sum of the first K elements, not the sum of the first 3 elements. Therefore, we need to clarify the problem statement.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Compute prefix sums once, then binary‑search the prefix array for the target K.
Brute Force Approach
Iterate K from 1 to N, recompute the sum each time, and stop when the condition holds.
Verified Code Solutions
function solution(nums, k) {
if (k < 1 || k > nums.length) {
return -1;
}
let sum = 0;
for (let i = 0; i < k; i++) {
if (typeof nums[i] !== 'number') {
return -1;
}
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int> nums, int k) {
if (k < 1 || k > nums.size()) {
return -1;
}
int sum = 0;
for (int i = 0; i < k; i++) {
if (nums[i] == NULL || !std::is_integral<decltype(nums[i])>::value) {
return -1;
}
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (k < 1 || k > nums.length) {
return -1;
}
int sum = 0;
for (int i = 0; i < k; i++) {
if (nums[i] == null || !(nums[i] instanceof Integer)) {
return -1;
}
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
if k < 1 or k > len(nums):
return -1
sum = 0
for i in range(k):
if not isinstance(nums[i], (int, float)):
return -1
sum += nums[i]
return sumfunction solution(nums, k) {
if (k < 1 || k > nums.length) {
return -1;
}
let sum = 0;
for (let i = 0; i < k; i++) {
if (typeof nums[i] !== 'number') {
return -1;
}
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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