Network Node Tracker 35 — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums of length *n* and an integer K. Compute the sum of all elements in nums that are strictly greater than K. The algorithm should run in linear time relative to *n* and use only O(1) additional memory beyond the input storage.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Node Tracker 35"
WHY DOES IT MATTER?
Constant‑space linear scans are a core pattern for processing massive streams efficiently.
OPTIMIZATION CHALLENGE
Eliminate any auxiliary containers and avoid multiple passes to keep both time and space minimal.
REAL-WORLD CONNECTION
Think of a network monitor that tallies traffic exceeding a threshold without storing every packet.
Initialize the accumulator outside the loop and update it in‑place; avoid creating temporary lists or using built‑in filter functions that allocate memory.
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 threshold K and, if larger, added to an accumulator. This leverages the principle of in‑place aggregation, which avoids auxiliary data structures and thus respects O(1) extra space. Naïve solutions might sort the array (O(n log n)) or build a filtered list (O(n) space), both unnecessary because the ordering of elements is irrelevant to the sum. The optimal paradigm is a streaming algorithm: maintain a running total while iterating once, guaranteeing linear time and constant auxiliary memory regardless of input size.
Interview Questions on This Problem
Q1Why is sorting the array not an optimal solution for this problem?
Sorting adds O(n log n) time, which is unnecessary because element order doesn't affect the sum. It also uses extra space for the sort algorithm in many implementations.
Q2How would you handle potential integer overflow when summing large numbers?
Use a wider integer type (e.g., long long in C++ or Python's arbitrary‑precision int) for the accumulator. Alternatively, check before addition and handle overflow cases explicitly.
Q3Can this algorithm be parallelized, and what would be the trade‑off?
Yes, by partitioning the array and summing each segment concurrently, then combining partial sums. The trade‑off is added synchronization overhead and loss of strict O(1) extra space per thread.
Examples
Input
5 10 4 12 7 15 9
Output
0
Explanation: Elements greater than 10 are 12 and 15. Their sum is 12 + 15 = 27.
Input
3 -5 -2 -6 0
Output
0
Explanation: Elements greater than -5 are -2 and 0. Their sum is -2 + 0 = -2.
Input
6 100 101 99 150 100 200 50
Output
0
Explanation: Elements greater than 100 are 101, 150 and 200. Their sum is 101 + 150 + 200 = 451.
Constraints
- 1 <= nums.length <= 100000
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
- The answer fits in a 64‑bit signed integer.
Optimal Approach & Strategy
Iterate once, keep a running sum, and add each element directly if it exceeds K, achieving O(n) time and O(1) extra space.
Brute Force Approach
Create a new list of elements > K, then sum that list; this costs O(n) time and O(n) extra space.
Verified Code Solutions
/**
* @param {number[]} nums
* @param {number} K
* @return {number}
*/
function sumGreaterThanK(nums, K) {
let total = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) {
total += nums[i];
}
}
return total;
}
// Example usage
// const n = 5;
// const K = 10;
// const nums = [4, 12, 7, 15, 9];
// console.log(sumGreaterThanK(nums, K));#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
long long sumGreaterThanK(vector<int>& nums, int K) {
long long total = 0;
for (int num : nums) {
if (num > K) {
total += num;
}
}
return total;
}
};
int main() {
int n, K;
cin >> n >> K;
vector<int> nums(n);
for (int i = 0; i < n; i++) {
cin >> nums[i];
}
Solution sol;
cout << sol.sumGreaterThanK(nums, K) << endl;
return 0;
}import java.util.*;
class Solution {
public long sumGreaterThanK(int[] nums, int K) {
long total = 0;
for (int num : nums) {
if (num > K) {
total += num;
}
}
return total;
}
}
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int K = sc.nextInt();
int[] nums = new int[n];
for (int i = 0; i < n; i++) {
nums[i] = sc.nextInt();
}
Solution sol = new Solution();
System.out.println(sol.sumGreaterThanK(nums, K));
}
}def sum_greater_than_k(nums, K):
total = 0
for num in nums:
if num > K:
total += num
return total
# Example usage
# n = 5
# K = 10
# nums = [4, 12, 7, 15, 9]
# print(sum_greater_than_k(nums, K))/**
* @param {number[]} nums
* @param {number} K
* @return {number}
*/
function sumGreaterThanK(nums, K) {
let total = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) {
total += nums[i];
}
}
return total;
}
// Example usage
// const n = 5;
// const K = 10;
// const nums = [4, 12, 7, 15, 9];
// console.log(sumGreaterThanK(nums, K));Asked in Top Tech Interviews
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