Protocol Sensor Validator 11 — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums and a non‑negative integer K. Compute the sum of the first K elements of nums (using 0‑based indexing, i.e., elements nums[0] through nums[K‑1]). If K equals 0, the result is 0. The input guarantees 0 ≤ K ≤ nums.length. Output the computed sum as a 64‑bit signed integer.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Sensor Validator 11"
WHY DOES IT MATTER?
This pattern demonstrates the importance of linear-time aggregation and constant space usage, which are essential for performance-critical applications such as real-time analytics, financial tick processing, and large-scale data pipelines. By focusing on the minimal subset of data required, developers can avoid unnecessary work and reduce latency.
OPTIMIZATION CHALLENGE
The key insight is that you only need to iterate over the first K elements, not the entire array. This reduces the time complexity from O(n) to O(K) and eliminates the need for auxiliary data structures, keeping space usage at O(1).
REAL-WORLD CONNECTION
Consider a stock trading platform that needs to compute the cumulative volume of trades for the first K seconds of a trading day. Rather than reprocessing all trades each time, the platform maintains a running sum of volumes as trades arrive, enabling instant responses to user queries about early-day activity.
When explaining this to an interviewer, emphasize that the algorithm is optimal because each element in the prefix must be examined at least once, and that no faster algorithm exists for a static array. Highlight the constant space advantage and the simplicity of the implementation.
COMPLEXITY AT A GLANCE
O(K)O(1)Core Theory — Why This Approach?
The problem of computing the sum of the first K elements of an array is a classic example of a linear-time prefix aggregation. In the naive approach, one might consider iterating over the entire array and conditionally adding elements based on their index, which would still be O(n) but with unnecessary overhead. However, the optimal strategy leverages the fact that we only need the first K elements; thus, a single pass over those K elements suffices, yielding an O(K) time complexity. This linear scan is optimal because each element must be examined at least once to determine its contribution to the sum, and no sub-linear algorithm can avoid inspecting each of the K items. The space complexity remains O(1) since we only maintain an accumulator variable, regardless of the array size. This pattern is fundamental in many streaming and real‑time data processing scenarios where partial aggregates are required without full traversal.
In large-scale systems, such as real-time analytics or financial tick processing, computing partial sums efficiently is critical. A naive approach that recomputes the sum from scratch for each query would lead to quadratic time complexity when repeated, causing unacceptable latency. By contrast, maintaining a running prefix sum or simply iterating over the required slice ensures that each query is answered in linear time relative to K, which is often far smaller than the total dataset size. This principle extends to more complex data structures like Fenwick trees or segment trees, where prefix sums can be updated and queried in logarithmic time, but for a static array the simple linear scan remains the most straightforward and performant solution.
The key insight is that the problem reduces to a single accumulation loop, and that the constraints guarantee K is within bounds, eliminating the need for bounds checking beyond the loop condition. This guarantees that the algorithm is both time‑efficient and easy to reason about, making it an ideal teaching example for beginners and a reliable baseline for interviewers to assess a candidate’s understanding of basic algorithmic optimization.
Interview Questions on This Problem
Q1How would you modify this algorithm to handle a dynamic array where elements can be inserted or removed, and you still need to answer prefix sum queries efficiently?
You would replace the simple linear scan with a data structure that supports dynamic updates and prefix queries, such as a Binary Indexed Tree (Fenwick Tree) or a Segment Tree. These structures allow updates in O(log n) time and prefix sum queries in O(log n) time, enabling efficient handling of insertions, deletions, and sum queries on the fly.
Q2In a distributed system processing a stream of integers, how can you compute the sum of the first K elements without storing the entire stream?
You can maintain a running accumulator and a counter. As each element arrives, increment the counter and add the element to the accumulator until the counter reaches K. After that point, you can discard subsequent elements or continue processing them separately, ensuring constant memory usage.
Q3What potential pitfalls should you watch for when implementing this sum in a language with 32-bit integer types, given the problem requires a 64-bit signed integer output?
You must cast or use a 64-bit integer type (e.g., long long in C++/Java, long in Java, or int64 in Python) for the accumulator to avoid overflow. If you use a 32-bit type, large sums will wrap around, producing incorrect results. Additionally, ensure that the input array elements are also treated as 64-bit values if they can exceed 32-bit limits.
Examples
Input
5 3 2 -1 4 7 0
Output
5
Explanation: The first three elements are 2, -1, and 4. Their sum is 2 + (-1) + 4 = 5.
Input
4 4 10 -5 3 2
Output
10
Explanation: All four elements are included: 10 + (-5) + 3 + 2 = 10.
Input
6 0 1 2 3 4 5 6
Output
0
Explanation: K is 0, so no elements are taken. The sum of an empty prefix is defined as 0.
Constraints
- 1 <= nums.length <= 100000
- 0 <= K <= nums.length
- -1000000000 <= nums[i] <= 1000000000
- The answer fits in a signed 64‑bit integer
Optimal Approach & Strategy
Iterate only over the first K elements, adding each to an accumulator. This yields O(K) time and O(1) space, which is optimal for this problem.
Brute Force Approach
Loop through the entire array, adding each element to a sum only if its index is less than K. This approach still scans all n elements, resulting in O(n) time even when K is small.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] <= k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}class Solution {
public:
int solution(vector<int> nums, int k) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] <= k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] <= k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
}def solution(nums, k):
sum = 0
for i in range(len(nums)):
if nums[i] <= k:
sum += nums[i]
else:
break
return sumfunction solution(nums, k) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] <= k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}Asked in Top Tech Interviews
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