Protocol Sensor Architect 5 — Problem Statement & Solution Guide

ArraysEasyRecursive Backtracking

TimeO(n)

SpaceO(1)

Problem Description

You are tasked with designing a recursive backtracking algorithm to process a sequence of integer sensor readings. The goal is to determine the maximum possible sum of a subset of these readings, subject to the constraint that no two selected elements can be adjacent in the original sequence. This problem models a scenario where selecting a sensor for detailed analysis prevents the immediate neighboring sensors from being selected due to signal interference.

Given an array readings of integers, return the maximum sum achievable by selecting a subset of non-adjacent elements. If the array is empty, return 0. You must implement this using a recursive backtracking approach, although the final solution should be optimized to avoid exponential time complexity if possible, or clearly demonstrate the recursive structure with memoization.

The input will be a single array of integers. The output should be a single integer representing the maximum sum. Ensure your solution handles negative numbers correctly, as selecting no elements (sum 0) might be better than selecting a negative element if all elements are negative, but note that the problem typically implies you must select at least one element if the array is non-empty, or you can choose an empty subset. For this specific problem, assume you can choose an empty subset, so the minimum return value is 0.

DSA Pattern Breakdown

"Protocol Sensor Architect 5"

easy

WHY DOES IT MATTER?

The non‑adjacent maximum sum pattern appears in resource allocation, scheduling, and financial portfolio selection where choosing one option precludes its immediate neighbors. Mastering this pattern teaches candidates how to convert combinatorial constraints into linear recurrences.

OPTIMIZATION CHALLENGE

The key insight is recognizing that the optimal solution up to index i depends only on the two previous optimal values, not the entire prefix. This reduces the DP table from O(n) space to two scalar variables, achieving O(1) extra memory.

REAL-WORLD CONNECTION

Consider a distributed sensor network where activating a sensor drains power from its neighboring nodes due to interference. The optimal activation schedule mirrors the non‑adjacent sum problem, ensuring maximal data collection without causing adjacent sensor overload.

During an interview, write the recurrence first, then immediately discuss memoization or rolling variables. This shows you understand both the conceptual DP and the practical space‑optimisation.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem is a classic instance of the "Maximum Sum of Non‑Adjacent Elements" (also known as the House Robber problem). The naive solution enumerates every subset of indices, checking the adjacency constraint, which leads to O(2^n) time and quickly becomes infeasible for n > 30. The optimal paradigm leverages dynamic programming: at each position i we decide whether to take the current reading (adding it to the best solution up to i‑2) or to skip it (carrying forward the best solution up to i‑1). This recurrence, dp[i] = max(dp[i‑1], arr[i] + dp[i‑2]), captures the optimal substructure and overlapping sub‑problems, allowing us to compute the answer in linear time.

Recursive backtracking can be used to illustrate the decision process, but without memoization it degenerates to the exponential brute‑force. By storing intermediate results (either via a DP array or two rolling variables) we eliminate repeated work, turning the exponential tree into a simple linear scan. This shift from exponential to linear time is the hallmark of dynamic programming and is essential for handling large sensor streams that can contain up to 10^5 readings.

Interview Questions on This Problem

Q1How would you modify the solution if the constraint changed to "no three selected elements can be consecutive"?

Introduce a DP state that tracks the number of consecutive picks at the current index. For example, dp[i][0] = max sum ending at i without picking i, dp[i][1] = max sum ending at i with i picked and previous not picked, dp[i][2] = max sum ending at i with i and i‑1 picked. The recurrence ensures we never have three in a row.

Q2Can you solve the problem in O(1) extra space? Explain the technique.

Yes. Since dp[i] only depends on dp[i‑1] and dp[i‑2], we keep two variables, prev1 and prev2, updating them iteratively: cur = max(prev1, arr[i] + prev2); then shift prev2 = prev1, prev1 = cur. This yields O(1) auxiliary space while preserving O(n) time.

Q3Why is a greedy approach (always pick the larger of two adjacent numbers) incorrect for this problem?

Greedy fails because a locally optimal choice can block a larger future sum. For example, array [4, 1, 2, 7, 5] – picking 4 (greedy) prevents picking 7 later, while the optimal solution picks 4 + 7 = 11, which a naive greedy that picks 5 over 7 would miss.

Examples

Example 1

Input

readings = [3, 2, 7, 10]

Output

Explanation: Step 1: Consider the first element 3. If we pick 3, we cannot pick 2. We then solve for the subarray [7, 10]. Step 2: For [7, 10], if we pick 7, we cannot pick 10. Sum = 7. If we skip 7, we pick 10. Sum = 10. Max for subarray is 10. Step 3: Total if picking 3 is 3 + 10 = 13. Step 4: Consider skipping 3. We solve for [2, 7, 10]. Step 5: For [2, 7, 10], if we pick 2, we solve for [10] (skipping 7). Max is 10. Total = 2 + 10 = 12. If we skip 2, we solve for [7, 10]. Max is 10. Total = 10. Step 6: Max for [2, 7, 10] is 12. Step 7: Compare picking 3 (13) vs skipping 3 (12). The maximum is 13.

Example 2

Input

readings = [5, 1, 1, 5]

Output

Explanation: Step 1: Pick 5 (index 0). Cannot pick 1 (index 1). Solve for [1, 5] (indices 2, 3). Step 2: For [1, 5], pick 1 (index 2) -> cannot pick 5. Sum = 1. Skip 1 -> pick 5. Sum = 5. Max is 5. Step 3: Total if picking first 5 is 5 + 5 = 10. Step 4: Skip first 5. Solve for [1, 1, 5]. Step 5: For [1, 1, 5], pick 1 (index 1) -> solve for [5]. Sum = 1 + 5 = 6. Skip 1 -> solve for [1, 5]. Max for [1, 5] is 5. Total = 5. Step 6: Max for [1, 1, 5] is 6. Step 7: Compare 10 vs 6. Maximum is 10.

Example 3

Input

readings = [-1, -2, -3]

Output

Explanation: Step 1: All elements are negative. Step 2: The problem allows selecting an empty subset, which yields a sum of 0. Step 3: Selecting any single element results in a negative sum (-1, -2, or -3). Step 4: Since 0 is greater than any negative sum, the optimal choice is to select no elements. Step 5: Return 0.

Example 4

Input

readings = [100]

Output

Explanation: Step 1: The array contains a single element, 100. Step 2: There are no adjacent elements to conflict with. Step 3: The only non-empty subset is [100]. Step 4: The sum is 100. Step 5: Return 100.

Constraints

1 <= readings.length <= 10^5
-10^9 <= readings[i] <= 10^9
The sum of all elements may exceed the range of a 32-bit integer, so use 64-bit integer arithmetic.

Optimal Approach & Strategy

Use dynamic programming with the recurrence dp[i] = max(dp[i‑1], arr[i] + dp[i‑2]) and iterate linearly, optionally compressing the DP array to two variables.

Brute Force Approach

Enumerate every subset of indices, filter those with no adjacent picks, and compute their sums; keep the maximum.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums, target) {
   nums.sort((a, b) => b - a);
   let start = 0;
   let end = nums.length - 1;
   let maxSum = 0;
   while (start <= end) {
       let sum = nums[start] + nums[end];
       if (sum <= target) {
           maxSum = Math.max(maxSum, sum);
           start++;
       } else {
           end--;
       }
   }
   return maxSum;
}

class Solution {
   public:
       int solution(vector<int>& nums, int target) {
           sort(nums.begin(), nums.end(), greater<int>());
           int start = 0;
           int end = nums.size() - 1;
           int maxSum = 0;
           while (start <= end) {
               int sum = nums[start] + nums[end];
               if (sum <= target) {
                   maxSum = max(maxSum, sum);
                   start++;
               } else {
                   end--;
               }
           }
           return maxSum;
       }
   };

class Solution {
   public int solution(int[] nums, int target) {
       Arrays.sort(nums);
       int start = 0;
       int end = nums.length - 1;
       int maxSum = 0;
       while (start <= end) {
           int sum = nums[start] + nums[end];
           if (sum <= target) {
               maxSum = Math.max(maxSum, sum);
               start++;
           } else {
               end--;
           }
       }
       return maxSum;
   }
}

def solution(nums, target):
   nums.sort(reverse=True)
   start = 0
   end = len(nums) - 1
   max_sum = 0
   while start <= end:
       sum = nums[start] + nums[end]
       if sum <= target:
           max_sum = max(max_sum, sum)
           start += 1
       else:
           end -= 1
   return max_sum

function solution(nums, target) {
   nums.sort((a, b) => b - a);
   let start = 0;
   let end = nums.length - 1;
   let maxSum = 0;
   while (start <= end) {
       let sum = nums[start] + nums[end];
       if (sum <= target) {
           maxSum = Math.max(maxSum, sum);
           start++;
       } else {
           end--;
       }
   }
   return maxSum;
}

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Protocol Sensor Architect 5 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Arrays

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Protocol Sensor Architect 5 — 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