Sensor Checkpoint Evaluator 30 — Problem Statement & Solution Guide
Problem Description
A distributed sensor network transmits a linear sequence of integer telemetry values. The system architecture requires identifying the central region of the array where signal integrity is most critical. You are provided with an array readings of length n and an integer threshold K. Your task is to determine the sum of all elements in the array that are strictly greater than K.
The evaluation process must be performed by simultaneously traversing the array from both ends toward the center. Initialize two pointers, left at index 0 and right at index n-1. In each iteration, check the values at both pointers. If a value exceeds K, add it to the cumulative sum. Move the left pointer forward and the right pointer backward. Continue this process until the pointers meet or cross. This inward-pointing traversal ensures that every element is evaluated exactly once, maintaining O(n) time complexity and O(1) space complexity.
Return the final computed sum. If no elements exceed the threshold, return 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Checkpoint Evaluator 30"
WHY DOES IT MATTER?
Sliding‑window turns a potentially quadratic scan into linear time, essential for real‑time telemetry processing.
OPTIMIZATION CHALLENGE
The key is to maintain the window sum with O(1) updates instead of recomputing from scratch each step.
REAL-WORLD CONNECTION
Network routers compute moving averages of packet latency using the same constant‑size window technique.
Initialize the first window fully, then reuse that sum; always check bounds before subtracting the trailing element.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The optimal solution leverages the sliding‑window paradigm, which maintains the sum of a fixed‑size contiguous segment while traversing the array once. By adding the incoming element and subtracting the element that exits the window, we avoid recomputing the sum from scratch, reducing the time from O(n·K) to O(n). Naïve approaches recompute each window’s sum independently, leading to quadratic time on large inputs and causing time‑outs. The sliding window is a special case of prefix‑sum optimization where the window size is constant, allowing constant‑time updates and linear overall complexity.
Interview Questions on This Problem
Q1How does a sliding window improve over a naïve nested‑loop solution for fixed‑size subarray sums?
It updates the sum in O(1) by adding the new element and removing the old one, turning O(n·K) into O(n). This eliminates redundant work across overlapping windows.
Q2What edge cases must be considered when K equals the array length or exceeds it?
If K equals n, the answer is the total sum of the array; if K > n, the problem is undefined or should return an error/zero. Guarding against these prevents out‑of‑bounds access.
Q3Can the sliding‑window technique be applied to variable‑size windows, and what changes?
Yes, but you need additional logic to expand or shrink the window based on a condition (e.g., sum ≤ K). The update step then may involve multiple adds/removes per iteration.
Examples
Input
readings = [12, 5, 20, 8, 15], K = 10
Output
47
Explanation: Initialize left=0, right=4, sum=0. Iteration 1: readings[0]=12 > 10, sum=12; readings[4]=15 > 10, sum=27. Move left=1, right=3. Iteration 2: readings[1]=5 <= 10, skip; readings[3]=8 <= 10, skip. Move left=2, right=2. Iteration 3: readings[2]=20 > 10, sum=47. Move left=3, right=1. Pointers cross, stop. Return 47.
Input
readings = [3, 7, 2, 9, 4], K = 5
Output
16
Explanation: Initialize left=0, right=4, sum=0. Iteration 1: readings[0]=3 <= 5, skip; readings[4]=4 <= 5, skip. Move left=1, right=3. Iteration 2: readings[1]=7 > 5, sum=7; readings[3]=9 > 5, sum=16. Move left=2, right=2. Iteration 3: readings[2]=2 <= 5, skip. Move left=3, right=1. Pointers cross, stop. Return 16.
Input
readings = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: Initialize left=0, right=4, sum=0. Iteration 1: readings[0]=1 <= 10, skip; readings[4]=5 <= 10, skip. Move left=1, right=3. Iteration 2: readings[1]=2 <= 10, skip; readings[3]=4 <= 10, skip. Move left=2, right=2. Iteration 3: readings[2]=3 <= 10, skip. Move left=3, right=1. Pointers cross, stop. Return 0.
Input
readings = [100, 200, 300], K = 150
Output
500
Explanation: Initialize left=0, right=2, sum=0. Iteration 1: readings[0]=100 <= 150, skip; readings[2]=300 > 150, sum=300. Move left=1, right=1. Iteration 2: readings[1]=200 > 150, sum=500. Move left=2, right=0. Pointers cross, stop. Return 500.
Constraints
- 1 <= readings.length <= 10^5
- -10^9 <= readings[i] <= 10^9
- -10^9 <= K <= 10^9
- The sum of all elements exceeding K will fit within a 64-bit integer.
Optimal Approach & Strategy
Use a sliding window: compute the first K‑sum, then update it in O(1) while moving the window across the array.
Brute Force Approach
Re‑calculate the sum for each possible K‑length subarray using a nested loop, leading to O(n·K) time.
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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