Vault Buffer Evaluator 31 — 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 evaluator value under given operational constraints. The target evaluator value is the sum of the first 3 values minus the target value.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Buffer Evaluator 31"
WHY DOES IT MATTER?
Extracting a fixed‑size aggregate from a stream avoids unnecessary recomputation.
OPTIMIZATION CHALLENGE
Reduce quadratic scanning to a linear pass with constant extra memory.
REAL-WORLD CONNECTION
Similar to reading sensor headers and a critical alarm value in embedded systems.
Initialize the sum early, then keep the loop tight—avoid extra condition checks inside the main iteration.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a constant‑time arithmetic extraction from a linear data stream. By recognizing that the required value depends only on the first three elements and a single target element, we can avoid any nested iteration and instead use a single pass that maintains a running sum of the first three entries while tracking the target index. Naïve solutions often attempt to recompute sums for every possible sub‑array or use double loops to locate the target, leading to O(n²) time on large inputs, which quickly exceeds time limits. The optimal paradigm leverages prefix‑sum concepts and direct indexing: after reading the first three numbers we store their sum, then continue scanning until the target position is reached, finally returning sum‑target. This yields O(n) time with O(1) auxiliary space, scaling gracefully to massive sequences.
Interview Questions on This Problem
Q1Why does a single‑pass solution suffice for this problem?
Because the answer depends only on the first three values and one target value, both of which can be captured while iterating once.
Q2What edge case must you guard against when n < 3?
If the input length is less than three, the problem is undefined; you should return an error or handle it per specification.
Q3How would you modify the algorithm if the target value were the maximum element in the array?
Maintain the current maximum while scanning, then subtract it from the pre‑computed sum of the first three elements.
Examples
Input
[10, 9, 8, 20, 7, 6, 5]
Output
7
Explanation: Step-by-step: Given the input [10, 9, 8, 20, 7, 6, 5], we first take the sum of the first 3 values: 10 + 9 + 8 = 27. Then, we subtract the target value 20, resulting in 27 - 20 = 7.
Input
[1, 2, 3, 4, 5]
Output
6
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5], we first take the sum of the first 3 values: 1 + 2 + 3 = 6. Since there is no target value, we simply return the sum.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Compute the sum of the first three elements once, then locate the target in a single pass and subtract, achieving O(n) time.
Brute Force Approach
Repeatedly sum the first three elements for each possible target, resulting in O(n²) time.
Verified Code Solutions
function solution(nums) {
if (nums.length < 3) {
return -1; // or throw an error
}
let sum = nums.slice(0, 3).reduce((a, b) => a + b, 0);
return sum - nums[nums.length - 1];
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() < 3) {
return -1; // or throw an error
}
int sum = 0;
for (int i = 0; i < 3; i++) {
sum += nums[i];
}
return sum - nums.back();
}
};class Solution {
public int solution(int[] nums) {
if (nums.length < 3) {
return -1; // or throw an error
}
int sum = 0;
for (int i = 0; i < 3; i++) {
sum += nums[i];
}
return sum - nums[nums.length - 1];
}
}def solution(nums):
if len(nums) < 3:
return -1 # or raise an error
sum = sum(nums[:3])
return sum - nums[-1]function solution(nums) {
if (nums.length < 3) {
return -1; // or throw an error
}
let sum = nums.slice(0, 3).reduce((a, b) => a + b, 0);
return sum - nums[nums.length - 1];
}Asked in Top Tech Interviews
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