Vault Interval Tracker 14 — 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. The algorithm should add up the first k elements and return the sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Tracker 14"
WHY DOES IT MATTER?
Prefix‑sum patterns turn repeated range‑sum queries into constant‑time operations.
OPTIMIZATION CHALLENGE
The key is reducing repeated addition by reusing previously computed partial sums.
REAL-WORLD CONNECTION
Databases use cumulative aggregates to quickly compute totals over sliding windows of transactions.
Cache the running total in a local variable and avoid extra array allocations for maximum cache friendliness.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to computing the sum of the first k elements of an array, a classic prefix‑sum scenario. A naive double loop would recompute overlapping sub‑sums, leading to O(n·k) time, which explodes for large n and k. The optimal paradigm leverages a single pass accumulating a running total, achieving O(n) time and O(1) extra space, because each element is visited exactly once and the partial sum is updated incrementally.
Interview Questions on This Problem
Q1How would you compute the sum of the first k elements in a single pass?
Initialize a variable sum = 0 and iterate i from 0 to k‑1, adding arr[i] to sum each step. Return sum after the loop.
Q2What edge cases must you guard against when k may exceed the array length?
If k > n, treat k as n (sum the whole array) or return an error per spec. Always check bounds before accessing elements.
Q3Why is a prefix‑sum array useful for multiple queries of different k values?
A prefix‑sum array stores cumulative sums so any query sum(0..k‑1) is answered in O(1) by prefix[k‑1]. It trades O(n) preprocessing for constant‑time queries.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150]
Output
150
Explanation: Step-by-step: Given the input [10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150], we need to add up the first k elements. Here, k is 5. So, we add 10 + 20 + 30 + 40 + 50 = 150.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
Output
15
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], we need to add up the first k elements. Here, k is 5. So, we add 1 + 2 + 3 + 4 + 5 = 15.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a running sum while scanning the array once, stopping after k elements.
Brute Force Approach
Use a nested loop that recomputes sums for each element, resulting in O(n·k) time.
Verified Code Solutions
function solution(nums, k) {
// JavaScript solution
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
# Python solution
sum = 0
for i in range(k):
sum += nums[i]
return sumfunction solution(nums, k) {
// JavaScript solution
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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