Network Protocol Consolidator 31 — Problem Statement & Solution Guide

Two PointersEasyFrequency Hash Map

TimeO(n log n)

SpaceO(1)

Problem Description

You are tasked with optimizing the throughput of a legacy network switch that processes a stream of protocol identifiers. The switch receives an array of integers, where each integer represents a unique protocol ID. To prevent buffer overflow, the system must identify the 'consolidator value,' defined as the count of the most frequently occurring protocol ID in the stream. If multiple IDs share the highest frequency, the consolidator value remains that shared maximum count.

Your objective is to design an efficient algorithm that computes this maximum frequency in a single pass through the data. The solution should leverage a frequency mapping strategy to track occurrences and determine the peak load on the switch's processing units. Return the integer representing this maximum frequency.

Input: An array of integers protocols representing the sequence of protocol IDs.

Output: An integer representing the maximum number of times any single protocol ID appears in the array.

DSA Pattern Breakdown

"Network Protocol Consolidator 31"

easy

WHY DOES IT MATTER?

Identifying the most frequent element is a common reduction step in many analytics pipelines.

OPTIMIZATION CHALLENGE

The key is collapsing repeated values into a single pass, turning quadratic work into linear or near‑linear.

REAL-WORLD CONNECTION

Network devices often need to prioritize the protocol that dominates traffic to allocate buffers efficiently.

Prefer in‑place sorting when memory is tight; otherwise, a hash map is simpler and faster for unsorted data.

COMPLEXITY AT A GLANCE

⏱ Time:O(n log n)

💾 Space:O(1)

Core Theory — Why This Approach?

The consolidator value is essentially the mode frequency of an integer array. A naive O(n²) scan compares each element with every other, which quickly becomes infeasible for large streams because the number of comparisons grows quadratically. The optimal paradigm leverages either a hash map for linear counting or a sort‑then‑two‑pointer sweep: sorting clusters identical IDs together, allowing a single linear pass to count consecutive runs and track the maximum count. This reduces the dominant work to O(n log n) time with O(1) extra space (aside from the sort) or O(n) time with O(n) auxiliary space using a hash map, both dramatically faster than the brute force method.

Interview Questions on This Problem

Q1How would you compute the consolidator value without extra space?

Sort the array in‑place, then iterate with two pointers to count the length of each block of equal values. Update the maximum count whenever a block ends.

Q2What is the time‑space trade‑off between using a hash map versus sorting?

A hash map gives O(n) time but requires O(n) additional space for the frequency table. Sorting costs O(n log n) time but only O(1) extra space if the sort is in‑place.

Q3Why does a naive double loop fail on the maximum input size?

It performs O(n²) comparisons, leading to billions of operations for typical constraints (e.g., n = 10⁵). Such workload exceeds time limits and memory caches, causing time‑outs.

Examples

Example 1

Input

protocols = [101, 202, 101, 303, 202, 101]

Output

Explanation: Initialize an empty frequency map. Process 101: map becomes {101: 1}. Process 202: map becomes {101: 1, 202: 1}. Process 101: map becomes {101: 2, 202: 1}. Process 303: map becomes {101: 2, 202: 1, 303: 1}. Process 202: map becomes {101: 2, 202: 2, 303: 1}. Process 101: map becomes {101: 3, 202: 2, 303: 1}. The maximum value in the map is 3 (for ID 101). Return 3.

Example 2

Input

protocols = [5, 5, 5, 5]

Output

Explanation: All elements are identical. The frequency map will contain a single entry {5: 4}. The maximum frequency is 4. Return 4.

Example 3

Input

protocols = [1, 2, 3, 4, 5]

Output

Explanation: Each protocol ID appears exactly once. The frequency map is {1: 1, 2: 1, 3: 1, 4: 1, 5: 1}. The maximum frequency is 1. Return 1.

Example 4

Input

protocols = [7, 7, 8, 8, 9, 9, 10]

Output

Explanation: ID 7 appears twice, ID 8 appears twice, ID 9 appears twice, and ID 10 appears once. The frequency map is {7: 2, 8: 2, 9: 2, 10: 1}. The maximum frequency is 2. Return 2.

Constraints

1 <= protocols.length <= 10^5
1 <= protocols[i] <= 10^9
The sum of frequencies for all unique IDs equals protocols.length

Optimal Approach & Strategy

Sort the array and use a sliding window/two‑pointer technique to count consecutive equal values in one pass.

Brute Force Approach

Use two nested loops to count occurrences of each element, updating the maximum each time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n log n)

/**
 * @param {number[]} protocols
 * @return {number}
 */
var consolidateProtocols = function(protocols) {
    const freq = {};
    let maxCount = 0;
    for (let id of protocols) {
        freq[id] = (freq[id] || 0) + 1;
        if (freq[id] > maxCount) {
            maxCount = freq[id];
        }
    }
    return maxCount;
};

console.log(consolidateProtocols([101, 202, 101, 303, 202, 101]));

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

class Solution {
public:
    int consolidateProtocols(vector<int>& protocols) {
        unordered_map<int, int> freq;
        int maxCount = 0;
        for (int id : protocols) {
            freq[id]++;
            if (freq[id] > maxCount) {
                maxCount = freq[id];
            }
        }
        return maxCount;
    }
};

int main() {
    vector<int> protocols = {101, 202, 101, 303, 202, 101};
    Solution sol;
    cout << sol.consolidateProtocols(protocols) << endl;
    return 0;
}

import java.util.*;

class Solution {
    public int consolidateProtocols(int[] protocols) {
        Map<Integer, Integer> freq = new HashMap<>();
        int maxCount = 0;
        for (int id : protocols) {
            int count = freq.getOrDefault(id, 0) + 1;
            freq.put(id, count);
            if (count > maxCount) {
                maxCount = count;
            }
        }
        return maxCount;
    }

    public static void main(String[] args) {
        int[] protocols = {101, 202, 101, 303, 202, 101};
        Solution sol = new Solution();
        System.out.println(sol.consolidateProtocols(protocols));
    }
}

from typing import List

class Solution:
    def consolidate_protocols(self, protocols: List[int]) -> int:
        freq = {}
        max_count = 0
        for pid in protocols:
            freq[pid] = freq.get(pid, 0) + 1
            if freq[pid] > max_count:
                max_count = freq[pid]
        return max_count

if __name__ == "__main__":
    sol = Solution()
    print(sol.consolidate_protocols([101, 202, 101, 303, 202, 101]))

/**
 * @param {number[]} protocols
 * @return {number}
 */
var consolidateProtocols = function(protocols) {
    const freq = {};
    let maxCount = 0;
    for (let id of protocols) {
        freq[id] = (freq[id] || 0) + 1;
        if (freq[id] > maxCount) {
            maxCount = freq[id];
        }
    }
    return maxCount;
};

console.log(consolidateProtocols([101, 202, 101, 303, 202, 101]));

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Network Protocol Consolidator 31 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Two Pointers

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Network Protocol Consolidator 31 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor