Network Protocol Partition 45 — Problem Statement & Solution Guide
Problem Description
You are given an array of integers and a threshold value. Your task is to calculate the sum of all elements that are strictly greater than the threshold. The input consists of the number of elements, the threshold, and the array itself. The output is a single integer representing the required sum. The algorithm should run in linear time relative to the array size and use constant additional space.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Protocol Partition 45"
WHY DOES IT MATTER?
Linear‑time aggregation patterns are foundational for processing massive streams efficiently.
OPTIMIZATION CHALLENGE
Eliminating unnecessary sorting or nested loops cuts the complexity from O(n log n) or O(n^2) to O(n).
REAL-WORLD CONNECTION
Network routers often sum packet sizes above a threshold to trigger alerts or rate limiting.
Keep the accumulator in a primitive type and avoid extra containers to minimize cache misses.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single linear scan where each element is compared against a fixed threshold and, if larger, added to an accumulator. This leverages the additive property of sums and the fact that each element's contribution is independent, allowing O(n) time without extra data structures. Naïve approaches might attempt sorting or nested loops, which inflate time complexity to O(n log n) or O(n^2) and are unnecessary because ordering provides no benefit for a simple threshold check. The optimal paradigm is the greedy single-pass algorithm, which makes an irrevocable decision for each element as it is encountered, guaranteeing both correctness and minimal work.
Interview Questions on This Problem
Q1Why is sorting the array before summing elements > threshold suboptimal?
Sorting adds O(n log n) overhead while the ordering does not affect the sum. A single pass achieves the same result in linear time.
Q2How would you handle potential integer overflow when summing large values?
Use a wider numeric type (e.g., 64‑bit long) or language‑specific big integer support. Validate input ranges if constraints are known.
Q3Can this algorithm be parallelized, and what would be the trade‑off?
Yes, by partitioning the array and summing locally before a final reduction. The overhead of thread management may outweigh benefits for small n.
Examples
Input
5 3 1 4 5 2 6
Output
15
Explanation: Elements greater than 3 are 4, 5, and 6. Their sum is 4 + 5 + 6 = 15.
Input
4 10 12 9 10 11
Output
23
Explanation: Only 12 and 11 exceed 10. Their sum is 12 + 11 = 23.
Input
6 -2 -5 -1 0 3 -2 4
Output
6
Explanation: Numbers greater than -2 are -1, 0, 3, and 4. Their sum is -1 + 0 + 3 + 4 = 6.
Constraints
- 1 <= n <= 100000
- -1000000000 <= nums[i] <= 1000000000
- -1000000000 <= K <= 1000000000
- The sum of qualifying elements fits within a 64‑bit signed integer
- Time limit: 1 second; Memory limit: 256 MB
Optimal Approach & Strategy
Simply iterate once, compare each element to the threshold, and accumulate when larger, achieving O(n) time and O(1) space.
Brute Force Approach
You could sort the array then sum from the first element exceeding the threshold, which costs O(n log n).
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):
sum = 0
for num in nums:
if num > K:
sum += num
return sumfunction 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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