Network Network Extractor 28 — Problem Statement & Solution Guide

BacktrackingEasyGreedy Choice

TimeO(n)

SpaceO(1)

Problem Description

You are tasked with optimizing the extraction of data packets from a linear network buffer. The buffer is represented by an array metrics, where each element metrics[i] denotes the signal strength of the packet at index i. The extraction process follows a greedy strategy: you must select a subset of packets such that no two selected packets are adjacent in the buffer. The goal is to maximize the total signal strength of the extracted packets.

Given the array metrics, determine the maximum possible sum of signal strengths achievable by selecting non-adjacent elements. If the array is empty, the result is 0. This problem models a scenario where adjacent sensors interfere with each other, necessitating a selection of non-adjacent nodes to maximize the aggregate reading.

DSA Pattern Breakdown

"Network Network Extractor 28"

easy

WHY DOES IT MATTER?

Selecting non‑adjacent elements maximizes resource usage while respecting interference constraints.

OPTIMIZATION CHALLENGE

Reducing the exponential subset enumeration to a linear scan cuts runtime from O(2^n) to O(n).

REAL-WORLD CONNECTION

Analogous to scheduling non‑overlapping jobs or extracting non‑contiguous memory blocks in a buffer.

Maintain only two rolling variables (prev, curr) to keep the implementation cache‑friendly and O(1) space.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to finding a maximum‑weight independent set on a path graph, which can be solved optimally with dynamic programming. A naive exhaustive search enumerates 2^n subsets, quickly exploding for n > 30, while the DP recurrence dp[i] = max(dp[i‑1], dp[i‑2] + metrics[i]) captures the greedy choice of either skipping or taking the current packet in linear time. This recurrence stems from the optimal substructure: the best solution up to index i either excludes i (leaving dp[i‑1]) or includes i (adding its weight to the best solution up to i‑2). The DP thus transforms the exponential backtracking space into O(n) time and O(1) extra space when using two variables.

Interview Questions on This Problem

Q1Why does the greedy choice of always picking the larger of two adjacent packets fail?

Choosing the larger local value can block a higher‑value packet two steps ahead, breaking optimality. The global optimum depends on future decisions, which greedy local picks ignore.

Q2How does the DP recurrence dp[i] = max(dp[i‑1], dp[i‑2] + metrics[i]) guarantee optimality?

It considers both possibilities for index i—exclude or include—ensuring the best of the two sub‑problems is propagated. By induction, each dp[i] stores the optimal value for the prefix ending at i.

Q3Can this problem be solved with recursion without memoization?

Yes, but it degenerates to exponential time because sub‑problems are recomputed repeatedly. Memoization or iterative DP is required to achieve linear performance.

Examples

Example 1

Input

metrics = [3, 1, 4, 1, 5, 9, 2, 6]

Output

Explanation: We evaluate the maximum sum of non-adjacent elements. 1. Consider index 0 (3) vs index 1 (1). Greedy choice favors 3. 2. Next available is index 2 (4). Sum = 3 + 4 = 7. 3. Next available is index 3 (1) vs index 4 (5). Greedy choice favors 5. Sum = 7 + 5 = 12. 4. Next available is index 5 (9). Sum = 12 + 9 = 21. 5. Next available is index 6 (2) vs index 7 (6). Greedy choice favors 6. Sum = 21 + 6 = 27. Wait, let's re-evaluate using the standard DP/Greedy logic for maximum sum of non-adjacent elements (House Robber style): - dp[0] = 3 - dp[1] = max(3, 1) = 3 - dp[2] = max(3, 4+0) = 4? No, dp[i] = max(dp[i-1], dp[i-2] + nums[i]). - dp[0] = 3 - dp[1] = max(3, 1) = 3 - dp[2] = max(3, 3+4) = 7 - dp[3] = max(7, 3+1) = 7 - dp[4] = max(7, 7+5) = 12 - dp[5] = max(12, 7+9) = 16 - dp[6] = max(16, 12+2) = 16 - dp[7] = max(16, 16+6) = 22 Let's re-read the prompt. It says 'Greedy Choice' pattern. A pure greedy approach (always pick the larger of current and next) is not always optimal for this specific problem (House Robber), but the prompt asks for a 'Greedy Choice' pattern. However, for 'Maximum Sum of Non-Adjacent Elements', the optimal solution is Dynamic Programming. If the problem strictly enforces a *greedy* algorithm that is *optimal*, the problem might be different. Let's adjust the problem statement to fit a true Greedy pattern that is optimal. Alternative Problem: Given an array of intervals, select the maximum number of non-overlapping intervals. This is a classic Greedy problem. Let's pivot the problem to 'Interval Scheduling' to strictly fit the 'Greedy Choice' pattern while keeping the title. Revised Statement: You are given a list of network maintenance windows, each defined by a start and end time. The goal is to schedule the maximum number of non-overlapping maintenance windows. A greedy strategy that always selects the window with the earliest end time is optimal for this problem. Let's re-generate the JSON with this corrected, strictly Greedy problem.

Example 2

Input

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

Output

Explanation: Selecting indices 0, 2, 4 gives 1 + 3 + 5 = 9. Selecting indices 1, 3 gives 2 + 4 = 6. The maximum is 9. Wait, my previous calculation for the first example was 22. Let's stick to the House Robber logic as it's a common 'easy' backtracking/DP problem, but the prompt says 'Greedy Choice'. Actually, the prompt says 'Pattern: Greedy Choice'. I must provide a problem where Greedy is the correct and optimal approach. Problem: Coin Change (Min Coins) is Greedy only for specific coin systems. Problem: Activity Selection is Greedy. Let's use Activity Selection (Interval Scheduling). Input: List of intervals [start, end]. Output: Max number of non-overlapping intervals. Example 1: [[1,3], [2,4], [3,5], [6,7]] Sort by end time: [1,3], [2,4], [3,5], [6,7]. Pick [1,3]. Next must start >= 3. [3,5] starts at 3. Pick [3,5]. Next must start >= 5. [6,7] starts at 6. Pick [6,7]. Total 3. Example 2: [[1,2], [2,3], [3,4]] Sort by end: [1,2], [2,3], [3,4]. Pick [1,2]. Next start >= 2. [2,3] starts at 2. Pick [2,3]. Next start >= 3. [3,4] starts at 3. Pick [3,4]. Total 3. Example 3: [[1,10], [2,3], [4,5], [6,7]] Sort by end: [2,3], [4,5], [6,7], [1,10]. Pick [2,3]. Next start >= 3. [4,5] starts at 4. Pick [4,5]. Next start >= 5. [6,7] starts at 6. Pick [6,7]. Next start >= 7. [1,10] starts at 1 (invalid). Total 3. This fits the 'Greedy Choice' pattern perfectly.

Constraints

1 <= N <= 10^5
-10^4 <= metrics[i] <= 10^4
1 <= K <= N

Optimal Approach & Strategy

Iterate once, updating two variables that store the best sums for the previous two positions using the DP recurrence.

Brute Force Approach

Enumerate every subset, filter those with no adjacent indices, and track the maximum sum; runs in O(2^n) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

/**
 * @param {number[]} metrics
 * @return {number}
 */
var extractMaxSignal = function(metrics) {
    const n = metrics.length;
    if (n === 0) return 0;
    if (n === 1) return metrics[0];
    
    let prev2 = metrics[0];
    let prev1 = Math.max(metrics[0], metrics[1]);
    
    for (let i = 2; i < n; i++) {
        const current = Math.max(prev1, prev2 + metrics[i]);
        prev2 = prev1;
        prev1 = current;
    }
    
    return prev1;
};

#include <vector>
using namespace std;

class Solution {
public:
    int extractMaxSignal(vector<int>& metrics) {
        int n = metrics.size();
        if (n == 0) return 0;
        if (n == 1) return metrics[0];
        
        vector<int> dp(n, 0);
        dp[0] = metrics[0];
        dp[1] = max(metrics[0], metrics[1]);
        
        for (int i = 2; i < n; i++) {
            dp[i] = max(dp[i-1], dp[i-2] + metrics[i]);
        }
        
        return dp[n-1];
    }
};

import java.util.*;

class Solution {
    public int extractMaxSignal(int[] metrics) {
        int n = metrics.length;
        if (n == 0) return 0;
        if (n == 1) return metrics[0];
        
        int prev2 = metrics[0];
        int prev1 = Math.max(metrics[0], metrics[1]);
        
        for (int i = 2; i < n; i++) {
            int current = Math.max(prev1, prev2 + metrics[i]);
            prev2 = prev1;
            prev1 = current;
        }
        
        return prev1;
    }
}

from typing import List

class Solution:
    def extractMaxSignal(self, metrics: List[int]) -> int:
        n = len(metrics)
        if n == 0:
            return 0
        if n == 1:
            return metrics[0]
        
        prev2 = metrics[0]
        prev1 = max(metrics[0], metrics[1])
        
        for i in range(2, n):
            current = max(prev1, prev2 + metrics[i])
            prev2 = prev1
            prev1 = current
        
        return prev1

/**
 * @param {number[]} metrics
 * @return {number}
 */
var extractMaxSignal = function(metrics) {
    const n = metrics.length;
    if (n === 0) return 0;
    if (n === 1) return metrics[0];
    
    let prev2 = metrics[0];
    let prev1 = Math.max(metrics[0], metrics[1]);
    
    for (let i = 2; i < n; i++) {
        const current = Math.max(prev1, prev2 + metrics[i]);
        prev2 = prev1;
        prev1 = current;
    }
    
    return prev1;
};

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Network Network Extractor 28 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Backtracking

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Network Network Extractor 28 — 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