Sensor Checkpoint Tracker 24 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing sensor and checkpoint metrics, construct an optimal algorithm to evaluate and compute the target tracker value under given operational constraints. The target tracker value is the sum of the elements less than or equal to K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Checkpoint Tracker 24"
WHY DOES IT MATTER?
Backtracking transforms an exponential search into a tractable one by eliminating impossible states early.
OPTIMIZATION CHALLENGE
The key is to reduce the combinatorial explosion from 2ⁿ to a manageable subset via pruning.
REAL-WORLD CONNECTION
It mirrors how autonomous drones discard routes that exceed battery limits before full simulation.
Always maintain a running sum and stop recursion the moment it exceeds K; combine this with sorting for maximal early cuts.
COMPLEXITY AT A GLANCE
O(2ⁿ) worst‑case, much lower average with pruningO(n) recursion stackCore Theory — Why This Approach?
Backtracking systematically explores all possible selections of sensor readings to form candidate subsets, pruning branches that cannot possibly satisfy the constraint sum ≤ K. A naïve exhaustive search enumerates 2ⁿ subsets, leading to exponential time and memory blow‑up for large n, while a well‑designed backtracking algorithm uses ordering, cumulative sum checks, and early termination to discard infeasible paths, dramatically reducing the explored state space.
Interview Questions on This Problem
Q1How does backtracking differ from brute‑force recursion when solving subset‑sum‑like problems?
Backtracking adds pruning conditions that cut off branches as soon as they violate constraints, whereas brute‑force explores every branch regardless of feasibility.
Q2What ordering heuristic improves backtracking performance for the sensor checkpoint problem?
Sorting the elements in descending order lets large values be considered first, causing early sum overflow and faster pruning.
Q3When would you switch from backtracking to a DP solution for this problem?
If K is moderate (e.g., ≤ 10⁵) and n is large, a DP table O(n·K) is more predictable than exponential backtracking.
Examples
Input
[1, 2, 3, 4, 5], K = 3
Output
12
Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and K = 3, we iterate through the array and sum all elements less than or equal to 3, giving output 12.
Input
[10, 20, 30, 40, 50], K = 25
Output
150
Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and K = 25, we iterate through the array and sum all elements less than or equal to 25, giving output 150.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use recursive backtracking with sorted inputs, a running sum, and early exit when the sum exceeds K; optionally memoize states for further speed‑up.
Brute Force Approach
Generate all 2ⁿ subsets and sum each, keeping the maximum sum ≤ K.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num <= K) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int num : nums) {
if (num <= K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num <= K) {
sum += num;
}
}
return sum;
}
}def solution(nums, K):
sum = 0
for num in nums:
if num <= K:
sum += num
return sumfunction solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num <= K) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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