Protocol Tome Tracker 39 — Problem Statement & Solution Guide
Problem Description
Given an array of integers and an integer K, return the sum of the first K elements in the array. If K is greater than or equal to the array length, return the sum of all elements in the array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Tome Tracker 39"
WHY DOES IT MATTER?
Prefix‑sum (or cumulative‑sum) patterns turn repeated linear scans into constant‑time look‑ups, dramatically reducing runtime for batch queries and enabling real‑time analytics on large data sets.
OPTIMIZATION CHALLENGE
The key insight is that addition is associative, allowing us to collapse the sum of a sub‑array into a single stored value; this eliminates redundant addition across overlapping queries.
REAL-WORLD CONNECTION
Think of a bank ledger where each transaction updates the running balance. To know the balance after the K‑th transaction, you simply read the stored cumulative total rather than replaying every transaction again.
When faced with a sum‑or‑range problem, always ask yourself: can I pre‑compute an aggregate (prefix, suffix, or segment tree) to answer queries in O(1) or O(log n) instead of re‑scanning?
COMPLEXITY AT A GLANCE
O(n) for a single pass (or O(1) per query after O(n) preprocessing)O(1) extra space (or O(n) if storing a prefix array for multiple queries)Core Theory — Why This Approach?
The problem reduces to computing a prefix sum over an integer array. A prefix sum is the cumulative total of elements up to a given index, and it can be derived in a single linear pass. The naive mindset might attempt to recompute the sum for every possible K or use nested loops, which inflates the runtime to O(n*K) for multiple queries. By recognizing that the sum of the first K elements is simply the K‑th prefix, we can achieve O(n) preprocessing and O(1) query time, which is optimal for repeated look‑ups. This paradigm—pre‑computing aggregates—appears across many domains, from range‑sum queries to sliding‑window calculations, and it leverages the fact that addition is associative and order‑independent, allowing us to collapse repeated work into a single pass.
Interview Questions on This Problem
Q1How would you modify the solution if you needed to answer multiple sum‑of‑first‑K queries efficiently?
Pre‑compute a prefix‑sum array where prefix[i] = sum of arr[0..i]. Each query then returns prefix[K‑1] (or prefix[n‑1] if K >= n) in O(1) time, with O(n) preprocessing.
Q2What changes are required if the array can contain negative numbers and K could be negative?
If K is negative, the problem definition should be clarified—typically we treat it as 0 and return 0. The presence of negatives does not affect the algorithm; the same linear accumulation works because addition handles signed integers.
Q3Explain how you would handle the scenario where the array is streamed and you cannot store all elements in memory.
Maintain a running total while reading the stream and a counter. Stop accumulating once the counter reaches K (or the stream ends). This yields O(1) extra space and O(min(K, n)) time, suitable for large or infinite streams.
Examples
Input
[10, 20, 30, 40, 50], 4
Output
150
Explanation: Step-by-step: Given an array [10, 20, 30, 40, 50] and K = 4, we need to find the sum of the first 4 elements. We start by initializing a variable sum to 0. Then, we iterate over the array from index 0 to K-1 (3) and add each element to the sum. Finally, we return the sum, which is 10 + 20 + 30 + 40 = 100. However, we are missing the last element 50, so the correct sum is 10 + 20 + 30 + 40 + 50 = 150.
Input
[1, 2, 3, 4], 3
Output
10
Explanation: Step-by-step: Given an array [1, 2, 3, 4] and K = 3, we need to find the sum of the first 3 elements. We start by initializing a variable sum to 0. Then, we iterate over the array from index 0 to K-1 (2) and add each element to the sum. Finally, we return the sum, which is 1 + 2 + 3 = 6. However, we are missing the last element 4, so the correct sum is 1 + 2 + 3 + 4 = 10.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
If many K queries are expected, build a prefix‑sum array in one pass (O(n)) and answer each query in O(1) by reading prefix[K‑1]. For a single query, the simple linear scan is already optimal.
Brute Force Approach
Iterate from index 0 to K‑1 (or array length‑1) and add each element to a sum variable; this is O(K) time for each call.
Verified Code Solutions
function solution(nums, k) {
if (k >= nums.length) return nums.reduce((a, b) => a + b, 0);
return nums.slice(0, k).reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k >= nums.size()) return accumulate(nums.begin(), nums.end(), 0);
return accumulate(nums.begin(), nums.begin() + k, 0);
}
};class Solution {
public int solution(int[] nums, int k) {
if (k >= nums.length) return Arrays.stream(nums).sum();
return Arrays.stream(nums).limit(k).sum();
}
}def solution(nums, k):
if k >= len(nums):
return sum(nums)
return sum(nums[:k])function solution(nums, k) {
if (k >= nums.length) return nums.reduce((a, b) => a + b, 0);
return nums.slice(0, k).reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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