Sensor Packet Synthesizer 10 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing sensor and packet metrics, construct an optimal algorithm to evaluate and compute the target synthesizer value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Packet Synthesizer 10"
WHY DOES IT MATTER?
The subset‑sum/backtracking pattern is a cornerstone for any problem that asks for all possible ways to satisfy a numeric constraint under combinatorial choices. Mastery of this pattern lets engineers solve scheduling, resource allocation, and configuration generation tasks where exhaustive search is impossible without clever pruning.
OPTIMIZATION CHALLENGE
The key insight is to sort the metrics and prune branches the moment the partial sum exceeds the target. This simple comparison turns a potential O(2^n) enumeration into a dramatically smaller search space, especially when the target is modest relative to the total sum.
REAL-WORLD CONNECTION
Think of a distributed sensor network where each node reports a metric (e.g., bandwidth, latency). The central controller must assemble a set of nodes whose combined capacity exactly matches a required throughput. Enumerating every possible node set is infeasible; backtracking with early cut‑offs mirrors how the controller would discard node groups that already exceed the needed capacity.
During an interview, implement the recursion first without duplicate handling, get it working, then add the sort‑and‑skip‑duplicate guard. This incremental approach shows clear thinking and reduces the chance of bugs.
COMPLEXITY AT A GLANCE
O(2^n) worst‑case, but pruning often reduces it dramaticallyO(n) recursion stack plus O(k) for storing each valid combinationCore Theory — Why This Approach?
Backtracking is a depth‑first search technique that incrementally builds candidates for the solution and abandons a candidate (backtracks) as soon as it determines that this candidate cannot possibly lead to a valid solution. In the Sensor Packet Synthesizer problem we are asked to pick a subset of sensor‑packet metrics that exactly matches a target synthesizer value while respecting constraints such as using each metric at most once and avoiding duplicate subsets. A naive exhaustive enumeration would generate all 2^n subsets, compute their sums, and then filter the ones that meet the target – this quickly becomes infeasible for n > 30 because the combinatorial explosion overwhelms both time and memory.
The optimal paradigm leverages two classic backtracking tricks: sorting the input to enable early pruning (stop exploring a branch when the running sum exceeds the target) and skipping over equal consecutive elements to prevent generating duplicate solutions. By recursively deciding for each index whether to include the current metric, we maintain a running sum and a temporary list. When the sum equals the target we record the current list; when it exceeds the target we backtrack immediately. This reduces the search space dramatically, especially when the target is relatively small compared to the sum of all elements, and guarantees that each unique combination is produced exactly once.
Interview Questions on This Problem
Q1How would you modify the backtracking solution if each sensor metric could be used unlimited times (i.e., you can pick the same element multiple times) while still avoiding duplicate combinations?
Allow the recursive call to stay on the same index after including an element (instead of moving to i+1) so the element can be reused, but still sort the array and skip duplicates only when moving to the next distinct value. This transforms the problem into the classic "Combination Sum" variant.
Q2Explain why sorting the input array is crucial for both pruning and duplicate elimination in this problem.
Sorting places larger values after smaller ones, so when the running sum exceeds the target we can safely break the loop because all subsequent values will only increase the sum further. It also groups equal values together, enabling a simple check (i > start && nums[i] == nums[i‑1]) to skip over duplicates and ensure each unique combination is generated once.
Q3If the target synthesizer value can be as large as 10^9 but the number of metrics n is limited to 20, which algorithmic approach would you choose and why?
A backtracking solution with pruning is still appropriate because the exponential factor depends on n, not the magnitude of the target. With n ≤ 20 the recursion depth is bounded, and pruning based on the sorted order will cut many branches, keeping the runtime acceptable even for a large target.
Examples
Input
nums = [1, 2, 3, 4, 5], K = 3
Output
6
Explanation: Step-by-step: We start with the given sequence of data elements [1, 2, 3, 4, 5]. We need to find the synthesizer value under the given operational constraint K = 3. We can select any combination of elements less than or equal to K to find the optimal synthesizer value. In this case, the optimal combination is [1, 2, 3], which gives us a synthesizer value of 6.
Input
nums = [10, 20, 30, 40, 50], K = 10
Output
100
Explanation: Step-by-step: We start with the given sequence of data elements [10, 20, 30, 40, 50]. We need to find the synthesizer value under the given operational constraint K = 10. We can select any combination of elements less than or equal to K to find the optimal synthesizer value. In this case, the optimal combination is [10, 20, 30, 40], which gives us a synthesizer value of 100.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort the metrics, then use depth‑first backtracking with early pruning when the running sum exceeds the target and skip over duplicate values to avoid repeated combinations.
Brute Force Approach
Generate every possible subset of the metrics (2^n combinations), compute each subset's sum, and collect those that equal the target.
Verified Code Solutions
function solution(nums, K) {
let max = 0;
function backtrack(start, currentSum) {
if (currentSum > K) return;
if (start === nums.length) {
max = Math.max(max, currentSum);
return;
}
backtrack(start + 1, currentSum + nums[start]);
backtrack(start + 1, currentSum);
}
backtrack(0, 0);
return max;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int max = 0;
void backtrack(int start, int currentSum) {
if (currentSum > K) return;
if (start == nums.size()) {
max = max(max, currentSum);
return;
}
backtrack(start + 1, currentSum + nums[start]);
backtrack(start + 1, currentSum);
}
backtrack(0, 0);
return max;
}
};class Solution {
public int solution(int[] nums, int K) {
int max = 0;
public void backtrack(int start, int currentSum) {
if (currentSum > K) return;
if (start == nums.length) {
max = Math.max(max, currentSum);
return;
}
backtrack(start + 1, currentSum + nums[start]);
backtrack(start + 1, currentSum);
}
backtrack(0, 0);
return max;
}
}def solution(nums, K):
max = 0
def backtrack(start, current_sum):
if current_sum > K:
return
if start == len(nums):
max = max(max, current_sum)
return
backtrack(start + 1, current_sum + nums[start])
backtrack(start + 1, current_sum)
backtrack(0, 0)
return maxfunction solution(nums, K) {
let max = 0;
function backtrack(start, currentSum) {
if (currentSum > K) return;
if (start === nums.length) {
max = Math.max(max, currentSum);
return;
}
backtrack(start + 1, currentSum + nums[start]);
backtrack(start + 1, currentSum);
}
backtrack(0, 0);
return max;
}Asked in Top Tech Interviews
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