Sensor Checkpoint Optimizer 14 — 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 optimizer value under given operational constraints. The algorithm should find the maximum sum of subarray that does not exceed K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Checkpoint Optimizer 14"
WHY DOES IT MATTER?
Two‑pointer sliding windows turn quadratic subarray problems into linear scans.
OPTIMIZATION CHALLENGE
The key is to maintain a running sum and adjust pointers only when the constraint is violated, cutting redundant recomputation.
REAL-WORLD CONNECTION
It mirrors buffering data streams where you must keep the total payload under a bandwidth cap.
Always track the current sum in a variable; never recompute it from scratch inside the loop.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to finding the longest (or highest‑sum) contiguous segment whose total does not exceed a given limit K. With non‑negative sensor readings, the sum is monotonic as the right pointer expands, allowing a sliding‑window (two‑pointer) technique to shrink the left side whenever the sum would surpass K, guaranteeing that every feasible window is examined in linear time. Naïve enumeration of all O(N^2) subarrays quickly becomes infeasible for large N because each additional element forces a recomputation of sums, leading to timeouts on typical interview constraints. The optimal paradigm leverages the invariant that extending the window only increases the sum, so we can adjust pointers in O(N) while maintaining the current sum, delivering an O(N) time and O(1) extra‑space solution.
Interview Questions on This Problem
Q1Why does the sliding‑window technique work only when the array contains non‑negative numbers?
With non‑negative values, expanding the window never decreases the sum, so any violation of K can be fixed by moving the left pointer forward. Negative numbers could lower the sum after expansion, breaking the monotonicity required for the two‑pointer guarantee.
Q2How would you modify the algorithm if the array could contain negative numbers?
You would need a data structure like a balanced BST or prefix‑sum with binary search to query the largest prefix sum ≤ current‑sum‑K, resulting in O(N log N) time. The simple two‑pointer window no longer suffices because the sum is no longer monotonic.
Q3What is the time‑space trade‑off when using a prefix‑sum + binary search approach versus the sliding window?
Prefix‑sum + binary search uses O(N) extra space for the sorted prefix array but achieves O(N log N) time, whereas sliding window uses O(1) extra space and O(N) time. The latter is preferable when memory is constrained and the input guarantees non‑negative values.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90]
Output
130
Explanation: Step-by-step: Given the input [10, 20, 30, 40, 50, 60, 70, 80, 90], we first initialize two pointers, left and right, to the start of the array. We then calculate the sum of the subarray from left to right until it exceeds K. In this case, the sum of subarray [10, 20, 30, 40] is 100, which is less than or equal to K. We then find the maximum sum of subarray within the window [10, 20, 30, 40] which is 130.
Input
[50, 60, 70, 80, 90]
Output
90
Explanation: Step-by-step: Given the input [50, 60, 70, 80, 90], we first initialize two pointers, left and right, to the start of the array. We then calculate the sum of the subarray from left to right until it exceeds K. In this case, the sum of subarray [50] is 50, which is less than or equal to K. We then find the maximum sum of subarray within the window [50, 60, 70, 80, 90] which is 90.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a sliding window with two pointers and a running sum, adjusting the left pointer whenever the sum exceeds K, achieving O(N) time.
Brute Force Approach
Check every possible subarray, compute its sum, and keep the largest sum ≤ K; this requires O(N^2) time.
Verified Code Solutions
function solution(nums, K) {
let maxSum = -Infinity;
let left = 0;
let currentSum = 0;
for (let right = 0; right < nums.length; right++) {
currentSum += nums[right];
while (currentSum > K) {
currentSum -= nums[left];
left++;
}
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int maxSum = INT_MIN;
int left = 0;
int currentSum = 0;
for (int right = 0; right < nums.size(); right++) {
currentSum += nums[right];
while (currentSum > K) {
currentSum -= nums[left];
left++;
}
maxSum = max(maxSum, currentSum);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums, int K) {
int maxSum = Integer.MIN_VALUE;
int left = 0;
int currentSum = 0;
for (int right = 0; right < nums.length; right++) {
currentSum += nums[right];
while (currentSum > K) {
currentSum -= nums[left];
left++;
}
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}
}def solution(nums, K):
max_sum = float('-inf')
left = 0
current_sum = 0
for right in range(len(nums)):
current_sum += nums[right]
while current_sum > K:
current_sum -= nums[left]
left += 1
max_sum = max(max_sum, current_sum)
return max_sumfunction solution(nums, K) {
let maxSum = -Infinity;
let left = 0;
let currentSum = 0;
for (let right = 0; right < nums.length; right++) {
currentSum += nums[right];
while (currentSum > K) {
currentSum -= nums[left];
left++;
}
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}Asked in Top Tech Interviews
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