Vault Interval Aligner 9 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the data retrieval protocol for a high-security vault system. The vault stores a sequence of integer values representing encrypted data blocks. Your objective is to compute the aggregate checksum of specific blocks based on a modular indexing rule. Given an array of integers nums and a positive integer K, identify all elements whose zero-based index is an exact multiple of K. Return the sum of these selected elements. If no such indices exist, return 0.
The selection process is deterministic: for an array of length n, the valid indices are 0, K, 2K, 3K, ... up to the largest index less than n. This pattern ensures that the retrieval logic scales linearly with the array size, making it suitable for large-scale data processing environments where memory access patterns must be predictable and efficient.
Your solution must handle both positive and negative integers within the array. The summation should be performed using 64-bit integer arithmetic to prevent overflow, although the final result is guaranteed to fit within a standard 32-bit signed integer range for the given constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Aligner 9"
WHY DOES IT MATTER?
Modular grouping converts a combinatorial scan into independent linear streams.
OPTIMIZATION CHALLENGE
The key is to reduce nested iteration by exploiting the periodicity of indices.
REAL-WORLD CONNECTION
Similar to sharding logs by timestamp buckets for parallel processing in distributed systems.
Pre‑allocate a fixed‑size K array and update it in‑place to keep cache locality high.
COMPLEXITY AT A GLANCE
O(N)O(K)Core Theory — Why This Approach?
The problem reduces to grouping array elements by their index modulo K, which creates K independent residue classes. By scanning the array once and accumulating sums per residue, we avoid recomputing overlapping intervals and achieve linear time. A naive double‑loop that checks every possible start index and builds the checksum for each K‑step sequence would be O(N·K) and quickly exceeds limits for N up to 10^6. The optimal paradigm leverages the mathematical property of modular arithmetic to treat each residue class as a separate stream, allowing a single pass aggregation and constant‑time lookup for any remainder.
Interview Questions on This Problem
Q1How does modular indexing turn a seemingly O(N·K) problem into O(N)?
Indices that share the same remainder modulo K form disjoint groups, so each element contributes to exactly one group. Accumulating per‑group sums in one pass eliminates the inner loop.
Q2What edge case must you handle when K > length of nums?
When K exceeds N, each index forms its own residue class, so the algorithm still works because the sum array size is K but only the first N entries are populated. Unused residues remain zero.
Q3Can the solution be extended to compute the maximum checksum among all residues?
Yes, after the single pass you simply scan the K accumulated sums and pick the maximum. This adds O(K) time, still linear overall.
Examples
Input
nums = [10, 20, 30, 40, 50], K = 2
Output
40
Explanation: The array length is 5. The indices that are multiples of 2 are 0 and 2. The element at index 0 is 10. The element at index 2 is 30. The sum is 10 + 30 = 40.
Input
nums = [-5, 15, -25, 35, -45, 55], K = 3
Output
30
Explanation: The array length is 6. The indices that are multiples of 3 are 0 and 3. The element at index 0 is -5. The element at index 3 is 35. The sum is -5 + 35 = 30.
Input
nums = [100, 200, 300], K = 5
Output
100
Explanation: The array length is 3. The only index that is a multiple of 5 and less than 3 is 0. The element at index 0 is 100. The sum is 100.
Input
nums = [7, -7, 7, -7, 7], K = 1
Output
7
Explanation: The array length is 5. Since K is 1, every index (0, 1, 2, 3, 4) is a multiple of 1. The elements are 7, -7, 7, -7, 7. The sum is 7 + (-7) + 7 + (-7) + 7 = 7.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- 1 <= K <= 10^5
Optimal Approach & Strategy
Maintain an array of K sums, add each nums[i] to sums[i % K] in a single pass – O(N) time.
Brute Force Approach
Iterate over every possible start index and step K forward, summing each sequence – O(N·K) time.
Verified Code Solutions
/**
* @param {number[]} nums
* @param {number} K
* @return {number}
*/
var solve = function(nums, K) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (i % K === 0) {
sum += nums[i];
}
}
return sum;
};class Solution {
public:
int solve(vector<int>& nums, int K) {
int n = nums.size();
int sum = 0;
for (int i = 0; i < n; i++) {
if (i % K == 0) {
sum += nums[i];
}
}
return sum;
}
};class Solution {
public int solve(int[] nums, int K) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (i % K == 0) {
sum += nums[i];
}
}
return sum;
}
}class Solution:
def solve(self, nums: List[int], K: int) -> int:
return sum(nums[i] for i in range(len(nums)) if i % K == 0)/**
* @param {number[]} nums
* @param {number} K
* @return {number}
*/
var solve = function(nums, K) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (i % K === 0) {
sum += nums[i];
}
}
return sum;
};Asked in Top Tech Interviews
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