Network Protocol Analyzer 15 — Problem Statement & Solution Guide
Problem Description
You are tasked with analyzing a network of nodes represented by an array of integers, where each integer denotes the unique identifier of a node. The network is initially fragmented, and you must determine the minimum number of connection operations required to ensure that all nodes with identifiers less than or equal to a given threshold K are part of a single connected component. A connection operation allows you to link any two nodes directly. If no nodes satisfy the threshold condition, the network is considered trivially connected, and the cost is zero. Your goal is to compute this minimum connection count efficiently.
Input: An array nums of integers representing node identifiers and an integer K representing the threshold. Output: An integer representing the minimum number of connections needed to unify all valid nodes (those with value <= K) into one component. If there are zero or one valid nodes, return 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Protocol Analyzer 15"
WHY DOES IT MATTER?
Transforming segment counting into a constant‑time update eliminates nested loops.
OPTIMIZATION CHALLENGE
The key is reducing the problem from O(N²) pair checks to a single O(N) scan.
REAL-WORLD CONNECTION
It mirrors merging isolated subnetworks in a data center with the fewest cable runs.
Maintain a simple flag for “inside a good segment” to avoid extra state and keep the code branch‑light.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to identifying contiguous segments of nodes whose identifiers are ≤ K. Each segment is already internally connected, and any two separate segments require exactly one connection operation to merge, so the minimal number of operations equals the number of such segments minus one. A naive solution would examine every pair of nodes or simulate all possible connections, leading to O(N²) time, which fails for N up to 10⁵ or higher. The optimal paradigm leverages a single linear pass, counting transitions between “good” (≤ K) and “bad” (> K) elements, achieving O(N) time and O(1) extra space.
Interview Questions on This Problem
Q1How do you compute the minimum connections needed for nodes ≤ K in one pass?
Count the number of contiguous blocks of elements ≤ K while scanning the array. The answer is blocks − 1.
Q2Why does the naive O(N²) approach break on large inputs?
It examines every possible pair or simulates each connection, leading to quadratic time. For N = 10⁵ this exceeds typical time limits.
Q3Can the solution be adapted if connections have different costs?
Yes, by weighting each merge with its cost and using a greedy or union‑find approach to pick the cheapest merges. The linear‑scan trick no longer applies directly.
Examples
Input
nums = [1, 2, 3, 4, 5], K = 3
Output
2
Explanation: The valid nodes are 1, 2, and 3 (since they are <= 3). These three nodes are initially disconnected. To connect 3 nodes into a single component, you need exactly 2 edges (connections). Thus, the answer is 2.
Input
nums = [10, 20, 30], K = 5
Output
0
Explanation: No node in the array has a value less than or equal to 5. Therefore, there are zero valid nodes. By definition, a set of zero nodes requires no connections. The answer is 0.
Input
nums = [7, 7, 7, 7], K = 7
Output
3
Explanation: All four nodes have the value 7, which is <= K. There are 4 valid nodes. To connect 4 distinct nodes into a single connected component, you need 4 - 1 = 3 connections. The answer is 3.
Input
nums = [1, 100, 2, 100, 3], K = 10
Output
2
Explanation: The valid nodes are 1, 2, and 3 (values <= 10). The nodes 100 are ignored. There are 3 valid nodes. Connecting 3 nodes requires 2 edges. The answer is 2.
Constraints
- 1 <= nums.length <= 10^5
- 1 <= nums[i] <= 10^9
- 1 <= K <= 10^9
Optimal Approach & Strategy
Perform a single linear scan, count contiguous ≤ K blocks, and return blockCount − 1.
Brute Force Approach
Check every pair of nodes ≤ K and simulate connecting them, counting needed operations, which is O(N²).
Verified Code Solutions
function minConnections(nums, K) {
let count = 0;
for (const id of nums) {
if (id <= K) count++;
}
return Math.max(0, count - 1);
}#include <bits/stdc++.h>
using namespace std;
int main() {
vector<int> nums = {1, 2, 3, 4, 5};
int K = 3;
int count = 0;
for (int id : nums) {
if (id <= K) count++;
}
int result = max(0, count - 1);
cout << result << endl;
return 0;
}public class Solution {
public int minConnections(int[] nums, int K) {
int count = 0;
for (int id : nums) {
if (id <= K) count++;
}
return Math.max(0, count - 1);
}
}def min_connections(nums, K):
count = sum(1 for id in nums if id <= K)
return max(0, count - 1)function minConnections(nums, K) {
let count = 0;
for (const id of nums) {
if (id <= K) count++;
}
return Math.max(0, count - 1);
}Asked in Top Tech Interviews
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