Protocol Tome Evaluator 44 — Problem Statement & Solution Guide
Problem Description
Protocol Tome Evaluator 44
You are given a sequence of integers and a threshold value K. Your task is to compute the sum of all elements in the sequence that are strictly greater than K. The algorithm must run in linear time with respect to the length of the sequence.
Input format:
- The first line contains a single integer N, the number of elements in the sequence.
- The second line contains N space‑separated integers representing the sequence.
- The third line contains the integer K.
Output format:
- Output a single integer: the sum of all elements that are greater than K.
The sequence may contain negative numbers, and the sum may exceed the range of a 32‑bit signed integer, so use 64‑bit arithmetic if necessary.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Tome Evaluator 44"
WHY DOES IT MATTER?
Linear‑time aggregation of filtered data is a foundational pattern for high‑throughput systems.
OPTIMIZATION CHALLENGE
The key is to avoid any secondary loops or auxiliary containers that would increase time or space.
REAL-WORLD CONNECTION
Think of a network monitor that sums traffic only from IPs exceeding a bandwidth threshold.
Initialize the accumulator outside the loop and update it conditionally; keep the loop body minimal for cache friendliness.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The naive solution iterates over the array and, for each element, checks if it exceeds K before adding it to a running total. While this is already O(N) in time, many candidates mistakenly introduce nested loops or extra data structures, inflating the complexity to O(N^2) or O(N log N).\nThe optimal paradigm leverages a single-pass sliding‑window mindset: maintain a cumulative sum while scanning the sequence once, adding only qualifying elements. This linear scan guarantees O(N) time and O(1) auxiliary space, which scales gracefully to massive inputs.
Interview Questions on This Problem
Q1Why is a single pass sufficient for this problem?
Because each element's contribution to the answer is independent of others, we can decide to add it or not as we encounter it. No future information is needed, eliminating the need for multiple passes.
Q2How would you handle integer overflow when summing large values?
Use a wider numeric type (e.g., 64‑bit long) or language‑specific big‑integer utilities. Validate input constraints to decide if overflow is a realistic concern.
Q3Can this algorithm be parallelized effectively?
Yes, by partitioning the array and computing partial sums of elements > K in each segment. A final reduction step adds the partial results, preserving O(N) work overall.
Examples
Input
5 1 5 3 7 2 3
Output
12
Explanation: Elements greater than 3 are 5 and 7. Their sum is 5 + 7 = 12.
Input
4 -1 -5 0 10 -3
Output
9
Explanation: Elements greater than -3 are -1, 0, and 10. Sum = -1 + 0 + 10 = 9.
Input
6 100 200 300 400 500 600 250
Output
1800
Explanation: Elements greater than 250 are 300, 400, 500, 600. Sum = 300 + 400 + 500 + 600 = 1800.
Input
3 0 0 0 0
Output
0
Explanation: No element is strictly greater than 0, so the sum is 0.
Constraints
- 1 <= N <= 100000
- -1000000000 <= nums[i] <= 1000000000
- -1000000000 <= K <= 1000000000
- The result fits in a 64‑bit signed integer
Optimal Approach & Strategy
Maintain a running total while iterating once, adding only elements > K, achieving O(N) time and O(1) space.
Brute Force Approach
A brute‑force method might use nested loops or repeatedly filter the array, leading to O(N^2) time.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
}def solution(nums, K):
return sum(num for num in nums if num > K)function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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