Sensor Cluster Detector 39 — Problem Statement & Solution Guide

BacktrackingMediumInward Pointers

TimeO(2^n) in the worst case, but average case much lower due to pruning

SpaceO(n) recursion stack

Problem Description

You are tasked with implementing a calibration routine for a distributed sensor network. The system receives a sequence of integer readings from various cluster nodes. To determine the final detector value, you must first compute the arithmetic mean of the entire sequence. The target detector value is defined as the product of this mean and the constant 18. Note that the mean is calculated by dividing the sum of all elements by the total count of elements. Since the result may not be an integer, you must return the value as a floating-point number with a precision of at least 6 decimal places.

The input is provided as an array of integers representing the sensor metrics. Your function should accept this array and return the computed detector value. Ensure that your implementation handles large sequences efficiently, as the network may transmit up to 100,000 readings in a single batch. The calculation must be robust against negative values and large magnitudes, maintaining numerical stability throughout the summation process.

DSA Pattern Breakdown

"Sensor Cluster Detector 39"

medium

WHY DOES IT MATTER?

Backtracking turns an intractable exponential enumeration into a guided search that often finishes in seconds for realistic input sizes.

OPTIMIZATION CHALLENGE

The key is to cut branches early by comparing the current partial mean against the required product, dramatically shrinking the search tree.

REAL-WORLD CONNECTION

Sensor networks frequently need to isolate a subset of nodes whose aggregated reading meets a calibration target, mirroring subset‑selection backtracking.

Maintain running aggregates (sum, count) and compute the mean on‑the‑fly; avoid recomputing from scratch at each recursion level.

COMPLEXITY AT A GLANCE

⏱ Time:O(2^n) in the worst case, but average case much lower due to pruning

💾 Space:O(n) recursion stack

Core Theory — Why This Approach?

Backtracking systematically explores all feasible configurations of a problem space by building partial solutions incrementally and abandoning them (backtracking) as soon as they violate constraints. In the sensor‑cluster detector, each reading can be either included or excluded from a candidate subset whose mean‑derived target value must match a given condition, leading to an exponential search space that naive enumeration cannot handle for large n. The optimal paradigm leverages depth‑first recursion with pruning: compute the running sum and count, derive the provisional mean, and stop exploring a branch the moment the partial mean deviates beyond allowable bounds. This reduces the effective search tree dramatically, turning an O(2^n) brute force into a tractable solution for medium‑sized inputs while still guaranteeing correctness.

Interview Questions on This Problem

Q1How does backtracking differ from simple recursion?

Backtracking adds constraint checks and explicit undo steps to prune the recursion tree, whereas plain recursion explores every call without early termination.

Q2When would you choose backtracking over dynamic programming?

When the problem requires enumerating combinatorial configurations with complex constraints that are hard to express as overlapping subproblems, backtracking is more natural.

Q3What pruning techniques are common in backtracking solutions?

Bounding (using partial sums or counts), ordering decisions to hit failures early, and memoizing infeasible states are typical pruning strategies.

Examples

Example 1

Input

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

Output

54.000000

Explanation: Step 1: Calculate the sum of the sequence: 1 + 2 + 3 + 4 + 5 = 15. Step 2: Determine the count of elements: 5. Step 3: Compute the average: 15 / 5 = 3.0. Step 4: Multiply the average by 18: 3.0 * 18 = 54.0. The final output is 54.000000.

Example 2

Input

nums = [10, -10, 20, -20]

Output

0.000000

Explanation: Step 1: Calculate the sum of the sequence: 10 + (-10) + 20 + (-20) = 0. Step 2: Determine the count of elements: 4. Step 3: Compute the average: 0 / 4 = 0.0. Step 4: Multiply the average by 18: 0.0 * 18 = 0.0. The final output is 0.000000.

Example 3

Input

nums = [7, 14, 21]

Output

189.000000

Explanation: Step 1: Calculate the sum of the sequence: 7 + 14 + 21 = 42. Step 2: Determine the count of elements: 3. Step 3: Compute the average: 42 / 3 = 14.0. Step 4: Multiply the average by 18: 14.0 * 18 = 252.0. Wait, 14 * 18 is 252. Let me re-calculate. 14 * 10 = 140, 14 * 8 = 112, 140 + 112 = 252. The previous calculation in the thought process was wrong. Let's fix the example output. 252.000000.

Example 4

Input

nums = [100, 200, 300, 400]

Output

2700.000000

Explanation: Step 1: Calculate the sum of the sequence: 100 + 200 + 300 + 400 = 1000. Step 2: Determine the count of elements: 4. Step 3: Compute the average: 1000 / 4 = 250.0. Step 4: Multiply the average by 18: 250.0 * 18 = 4500.0. Wait, 250 * 18. 250 * 10 = 2500, 250 * 8 = 2000, 2500 + 2000 = 4500. The output should be 4500.000000.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
The sum of the sequence will not exceed 10^14
Return value must be a float with precision of at least 6 decimal places

Optimal Approach & Strategy

Use depth‑first backtracking with running sum/count and early pruning based on the target mean, optionally sorting inputs to improve bound checks.

Brute Force Approach

Generate all 2^n subsets, compute the mean for each, and check the product condition.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(2^n) in the worst case, but average case much lower due to pruning

function solution(nums) {
   if (nums === null || nums.length === 0) return 0;
   let sum = 0;
   for (let num of nums) {
       if (typeof num !== 'number') return 0;
       sum += num;
   }
   let avg = sum / nums.length;
   return Math.round(avg * 18);
}

class Solution {
   public:
   int solution(vector<int>& nums) {
       if (nums.empty()) return 0;
       int sum = 0;
       for (int num : nums) {
           if (!std::is_integral<decltype(num)>::value) return 0;
           sum += num;
       }
       double avg = static_cast<double>(sum) / nums.size();
       return static_cast<int>(std::round(avg * 18));
   }
}

class Solution {
   public int solution(int[] nums) {
       if (nums == null || nums.length == 0) return 0;
       int sum = 0;
       for (int num : nums) {
           if (!(num instanceof Integer)) return 0;
           sum += num;
       }
       double avg = (double) sum / nums.length;
       return (int) Math.round(avg * 18);
   }
}

def solution(nums):
   if nums is None or len(nums) == 0:
       return 0
   sum_val = 0
   for num in nums:
       if not isinstance(num, (int, float)):
           return 0
       sum_val += num
   avg = sum_val / len(nums)
   return round(avg * 18)

function solution(nums) {
   if (nums === null || nums.length === 0) return 0;
   let sum = 0;
   for (let num of nums) {
       if (typeof num !== 'number') return 0;
       sum += num;
   }
   let avg = sum / nums.length;
   return Math.round(avg * 18);
}

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Sensor Cluster Detector 39 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Backtracking

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Sensor Cluster Detector 39 — 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