Network Node Resolver 46 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing network and node metrics, construct an optimal algorithm to evaluate and compute the target resolver value under given operational constraints. The algorithm should handle the case where K is greater than all elements in the array and the case where the input array is empty.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Node Resolver 46"
WHY DOES IT MATTER?
Two‑pointer patterns turn quadratic scans into linear traversals.
OPTIMIZATION CHALLENGE
The key is to reduce the search space by moving pointers based on monotonic comparisons.
REAL-WORLD CONNECTION
Similar to sliding a window over a network packet stream to locate the first packet meeting a threshold.
Initialize pointers at opposite ends and always advance the one that cannot satisfy the condition.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The two‑pointer technique exploits a monotonic property—usually sorted order—to traverse an array with a pair of indices that move in a coordinated fashion. By advancing the pointers based on the comparison with the target K, we can locate the required resolver value in a single linear pass, eliminating the need for nested loops.
A naïve solution would scan every element for each possible condition, yielding O(N²) time on large inputs and quickly exhausting time limits. The optimal paradigm leverages the sorted nature (or can sort first) and uses two pointers to achieve O(N) time, while keeping auxiliary space constant, which scales gracefully for massive datasets.
Interview Questions on This Problem
Q1How does the two‑pointer method achieve linear time on a sorted array?
Both pointers only move forward, each element is examined at most once. This eliminates redundant comparisons inherent in brute‑force nested loops.
Q2What edge case must you handle when K exceeds all array elements?
The algorithm should return a sentinel (e.g., -1) indicating no valid resolver exists. Failing to check this leads to out‑of‑bounds access.
Q3Why is constant extra space important for this pattern?
It ensures the solution scales without additional memory overhead, crucial for embedded or high‑throughput systems. Two pointers use only a few integer variables.
Examples
Input
[4, 5, 6, 7, 8, 9, 10], 3
Output
49
Explanation: Step-by-step: with input [4, 5, 6, 7, 8, 9, 10] and K = 3, we sum all elements greater than 3, which are 4, 5, 6, 7, 8, 9, 10. The sum is 4 + 5 + 6 + 7 + 8 + 9 + 10 = 49.
Input
[1, 2, 3], 5
Output
0
Explanation: Step-by-step: with input [1, 2, 3] and K = 5, we sum all elements greater than 5. Since there are no elements greater than 5, the sum is 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a single forward pointer on the sorted array; stop at the first element > K, achieving O(N) time and O(1) space.
Brute Force Approach
Iterate over every element and check if it exceeds K, returning the first match; worst‑case O(N) but with extra checks for each element.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (typeof num !== 'number') {
throw new Error('Input array contains non-numeric values');
}
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):
if not all(isinstance(num, (int, float)) for num in nums):
raise ValueError('Input array contains non-numeric values')
return sum(num for num in nums if num > k)function solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (typeof num !== 'number') {
throw new Error('Input array contains non-numeric values');
}
if (num > k) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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