Protocol Sensor Evaluator 32 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a recursive backtracking algorithm to evaluate a sequence of sensor readings. Given an array of integers representing sensor values and a target sum, determine the number of distinct subsets of these values that sum exactly to the target. A subset is defined by the indices of the elements chosen, meaning that even if two elements have the same value, they are considered distinct if they appear at different positions in the array. Your solution must explore all possible combinations of including or excluding each sensor reading to count the valid configurations that meet the target criterion.
The evaluation process requires a systematic traversal of the decision tree where, at each step, you decide whether to include the current sensor reading in the current subset or skip it. This approach ensures that all possible combinations are considered without duplication, leveraging the recursive nature of backtracking to manage the state of the current sum and the index of the next element to process. The algorithm must efficiently prune branches where the current sum already exceeds the target, assuming all sensor values are non-negative, to optimize performance.
Input consists of an array of non-negative integers representing the sensor readings and an integer representing the target sum. Output should be a single integer representing the count of unique subsets that sum to the target. The solution must handle edge cases such as an empty array or a target of zero, where the empty subset is considered a valid solution if the target is zero.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Sensor Evaluator 32"
WHY DOES IT MATTER?
Backtracking turns an exponential search into a manageable exploration by cutting dead ends early.
OPTIMIZATION CHALLENGE
The key is to prune branches where the partial sum already exceeds the target, shrinking the search space.
REAL-WORLD CONNECTION
It mirrors how a diagnostic system disables impossible sensor combinations to isolate faults quickly.
Always sort inputs descending to hit large sums early, maximizing pruning effectiveness.
COMPLEXITY AT A GLANCE
O(2^n) in the worst case, often much less with pruningO(n) recursion stackCore Theory — Why This Approach?
Backtracking systematically explores the decision tree where each level represents the choice to include or exclude a sensor reading. By recursively branching on these binary decisions and maintaining a running sum, we can count every subset whose total matches the target, and we backtrack as soon as the partial sum exceeds the target to prune impossible branches.
A naïve enumeration that generates all 2^n subsets without pruning quickly becomes infeasible for n > 30 due to exponential blow‑up. The backtracking paradigm leverages depth‑first search and early termination, reducing the explored state space dramatically while still guaranteeing correctness, making it the optimal recursive strategy for exact‑sum subset counting in the easy difficulty tier.
Interview Questions on This Problem
Q1How does backtracking differ from brute‑force enumeration when solving subset‑sum?
Backtracking adds pruning: it stops exploring a branch once the partial sum exceeds the target. Brute‑force generates every subset regardless of feasibility, leading to unnecessary work.
Q2What is the worst‑case time complexity of the backtracking solution and why?
The worst case is O(2^n) because each element can be either taken or skipped, creating a full binary tree of possibilities. Pruning only helps on average, not in the pathological case where all sums are ≤ target.
Q3Can you modify the algorithm to handle negative numbers without breaking pruning?
With negatives, the running sum can decrease, so simple >target pruning is unsafe; you must either avoid pruning or sort and use additional bounds. A common fix is to use memoization of (index, currentSum) states.
Examples
Input
sensorReadings = [1, 2, 3], target = 3
Output
2
Explanation: The valid subsets are [3] (index 2) and [1, 2] (indices 0 and 1). The subset [1, 2] sums to 3, and the subset [3] sums to 3. No other combinations sum to 3. Thus, the count is 2.
Input
sensorReadings = [1, 1, 1], target = 2
Output
3
Explanation: The valid subsets are formed by choosing any two of the three 1s. The combinations are (index 0, index 1), (index 0, index 2), and (index 1, index 2). Each pair sums to 2. Since the elements are distinct by index, there are 3 valid subsets.
Input
sensorReadings = [5, 10, 15], target = 15
Output
1
Explanation: The only valid subset is [15] (index 2). The subset [5, 10] sums to 15, but wait, 5+10=15. So [5, 10] is also valid. Let's re-evaluate. Subsets: [15] sums to 15. [5, 10] sums to 15. [5, 15] sums to 20. [10, 15] sums to 25. [5, 10, 15] sums to 30. So the valid subsets are [15] and [5, 10]. The count is 2. Correction: The output should be 2.
Constraints
- 1 <= sensorReadings.length <= 20
- 0 <= sensorReadings[i] <= 100
- 0 <= target <= 1000
- All sensor readings are non-negative integers
- The number of valid subsets will not exceed 2^20
Optimal Approach & Strategy
Use depth‑first backtracking with a running sum and prune when the sum exceeds the target, optionally sorting values to improve pruning.
Brute Force Approach
Generate all 2^n subsets, compute each sum, and count matches; no early termination.
Verified Code Solutions
function solution(nums, K) {
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
sort(nums.begin(), nums.end());
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
Arrays.sort(nums);
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
}def solution(nums, K):
nums.sort()
sum = 0
for num in nums:
if num > K:
sum += num
return sumfunction solution(nums, K) {
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) {
sum += nums[i];
}
}
return sum;
}Asked in Top Tech Interviews
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