Network Network Evaluator 17 — Problem Statement & Solution Guide
Problem Description
You are provided with a sequence of integers arr and a threshold integer threshold. Your objective is to calculate the total sum of all elements in arr that are strictly greater than threshold. Elements that are equal to or less than threshold must be ignored in the computation. If no elements in the sequence exceed the threshold, the result should be 0.
This task requires a single linear pass through the data structure, applying a conditional check at each position to determine whether the current element contributes to the cumulative total. The solution must operate in O(n) time complexity and O(1) auxiliary space, leveraging the properties of sequential access and immediate aggregation.
The input will always be a valid array of integers and a valid integer threshold. The output is a single integer representing the computed sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Network Evaluator 17"
WHY DOES IT MATTER?
Two‑pointer scanning turns a potentially quadratic check into a guaranteed linear pass.
OPTIMIZATION CHALLENGE
The key is to avoid extra passes or sorting, reducing time from O(n log n) to O(n).
REAL-WORLD CONNECTION
It's akin to filtering sensor readings in real time, discarding noise below a safety threshold.
Keep the pointer logic simple: advance the read pointer, and only update the accumulator when the condition holds.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a linear aggregation where each element must be examined exactly once to decide if it contributes to the sum. A naive double‑loop or repeated sorting would inflate the runtime to O(n²) or O(n log n), which is unnecessary because the decision criterion (greater than a static threshold) is independent of other elements. The optimal paradigm leverages the two‑pointer (or fast‑slow) technique: one pointer iterates through the array while the other maintains the running total, allowing a single pass with O(1) auxiliary space. This approach exploits the fact that the predicate is monotonic with respect to each element, eliminating any need for backtracking or re‑evaluation, thus achieving linear time complexity.
Interview Questions on This Problem
Q1How would you modify the solution if the threshold could change after each query?
Maintain a prefix sum array and perform binary search to locate the first element > threshold, then compute the sum of the suffix. This yields O(log n) per query after O(n) preprocessing.
Q2Can this problem be solved using a divide‑and‑conquer approach?
Yes, recursively compute sums for left and right halves and combine, but the overhead matches O(n) time without any advantage over the linear scan. The recursion adds O(log n) stack space.
Q3What edge cases must you handle when implementing the algorithm?
Empty array, all elements ≤ threshold, and integer overflow when summing large values. Guard against these by early returns and using a wider integer type if needed.
Examples
Input
arr = [12, 5, 18, 7, 22], threshold = 10
Output
52
Explanation: Traverse the array: 12 > 10 (add 12, sum=12); 5 <= 10 (skip); 18 > 10 (add 18, sum=30); 7 <= 10 (skip); 22 > 10 (add 22, sum=52). Final result is 52.
Input
arr = [3, 3, 3, 3], threshold = 3
Output
0
Explanation: Traverse the array: 3 <= 3 (skip); 3 <= 3 (skip); 3 <= 3 (skip); 3 <= 3 (skip). No elements strictly exceed the threshold. Final result is 0.
Input
arr = [-5, 0, 100, -10, 50], threshold = 0
Output
150
Explanation: Traverse the array: -5 <= 0 (skip); 0 <= 0 (skip); 100 > 0 (add 100, sum=100); -10 <= 0 (skip); 50 > 0 (add 50, sum=150). Final result is 150.
Input
arr = [1, 2, 3, 4, 5], threshold = 0
Output
15
Explanation: Traverse the array: 1 > 0 (add 1, sum=1); 2 > 0 (add 2, sum=3); 3 > 0 (add 3, sum=6); 4 > 0 (add 4, sum=10); 5 > 0 (add 5, sum=15). Final result is 15.
Constraints
- 1 <= arr.length <= 10^5
- -10^9 <= arr[i] <= 10^9
- -10^9 <= threshold <= 10^9
Optimal Approach & Strategy
Use a single pass with a fast pointer to evaluate each element and accumulate the sum when the condition holds.
Brute Force Approach
A brute force might sort the array then scan from the end, incurring O(n log n) time unnecessarily.
Verified Code Solutions
function sumAboveThreshold(nums, threshold) {
let sum = 0;
for (const num of nums) {
if (num > threshold) sum += num;
}
return sum;
}
const nums = [12,5,8,20,3];
const threshold = 10;
console.log(sumAboveThreshold(nums, threshold));#include <bits/stdc++.h>
using namespace std;
int sumAboveThreshold(const vector<int>& nums, int threshold) {
int sum = 0;
for (int num : nums) {
if (num > threshold) sum += num;
}
return sum;
}
int main() {
vector<int> nums = {12,5,8,20,3};
int threshold = 10;
cout << sumAboveThreshold(nums, threshold) << endl;
return 0;
}public class Solution {
public int sumAboveThreshold(int[] nums, int threshold) {
int sum = 0;
for (int num : nums) {
if (num > threshold) sum += num;
}
return sum;
}
public static void main(String[] args) {
int[] nums = {12,5,8,20,3};
int threshold = 10;
System.out.println(new Solution().sumAboveThreshold(nums, threshold));
}
}def sum_above_threshold(nums, threshold):
total = 0
for num in nums:
if num > threshold:
total += num
return total
nums = [12,5,8,20,3]
threshold = 10
print(sum_above_threshold(nums, threshold))function sumAboveThreshold(nums, threshold) {
let sum = 0;
for (const num of nums) {
if (num > threshold) sum += num;
}
return sum;
}
const nums = [12,5,8,20,3];
const threshold = 10;
console.log(sumAboveThreshold(nums, threshold));Asked in Top Tech Interviews
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