Sensor Checkpoint Optimizer 49 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a continuous stream of sensor readings to determine the maximum stability score for a fixed-duration checkpoint window. The input is a string s consisting of digits '0' through '9', where each character represents a discrete sensor metric value. The goal is to find the maximum sum of digits within any contiguous substring of length k.
The stability score of a window is defined as the arithmetic sum of the integer values of the characters contained within that window. You must slide a window of size k across the entire string from left to right, calculating the sum for each position. The final result is the highest sum observed among all valid windows.
If the length of the string s is less than k, no valid window exists, and the function should return 0. Otherwise, return the maximum sum found.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Checkpoint Optimizer 49"
WHY DOES IT MATTER?
Sliding‑window patterns turn quadratic scans into linear passes, essential for real‑time stream processing.
OPTIMIZATION CHALLENGE
The key is to avoid recomputing the entire window sum on each shift, cutting the complexity by a factor of k.
REAL-WORLD CONNECTION
Think of a moving sensor checkpoint that continuously aggregates the last k readings to decide stability.
Initialize the first window carefully, then update in‑place; avoid converting the whole string to an int array unless needed.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to finding the maximum sum of numeric characters in any contiguous substring of a given length. A naïve solution would recompute the sum for each possible window, leading to O(n·k) time where n is the string length and k the window size, which is prohibitive for large streams. The optimal paradigm uses a sliding‑window technique: compute the sum of the first k digits, then slide the window one position at a time, subtracting the digit exiting the window and adding the new digit, thus updating the sum in O(1) per step. This yields a linear O(n) solution that scales to massive inputs while using only constant extra space.
Interview Questions on This Problem
Q1How does the sliding‑window technique improve over a naïve nested loop for this problem?
It reuses the previous window's sum, updating it in O(1) instead of recomputing from scratch, reducing total time from O(n·k) to O(n).
Q2What edge case must you handle when the window size exceeds the string length?
If k > n, the problem is undefined; you should either return 0 or handle it as an invalid input according to specifications.
Q3Why is it safe to store the running sum in a 32‑bit integer for typical constraints?
Each digit is at most 9, so the maximum sum is 9·k; even for k up to 10⁶ the sum fits comfortably within a 32‑bit signed integer.
Examples
Input
s = "12345", k = 3
Output
12
Explanation: The valid windows of length 3 are: 1. "123": 1 + 2 + 3 = 6 2. "234": 2 + 3 + 4 = 9 3. "345": 3 + 4 + 5 = 12 The maximum sum is 12.
Input
s = "99999", k = 2
Output
18
Explanation: The valid windows of length 2 are: 1. "99": 9 + 9 = 18 2. "99": 9 + 9 = 18 3. "99": 9 + 9 = 18 4. "99": 9 + 9 = 18 The maximum sum is 18.
Input
s = "102030", k = 4
Output
5
Explanation: The valid windows of length 4 are: 1. "1020": 1 + 0 + 2 + 0 = 3 2. "0203": 0 + 2 + 0 + 3 = 5 3. "2030": 2 + 0 + 3 + 0 = 5 The maximum sum is 5.
Input
s = "5", k = 10
Output
0
Explanation: The length of the string is 1, which is less than k (10). Therefore, no valid window exists, and the result is 0.
Constraints
- 1 <= s.length <= 10^5
- 1 <= k <= 10^5
- s consists of digits '0' to '9' only
- If s.length < k, return 0
Optimal Approach & Strategy
Use a sliding window: maintain a running sum, update it by subtracting the leftmost digit and adding the new rightmost digit as the window moves.
Brute Force Approach
Re‑calculate the sum for every possible length‑k substring, leading to O(n·k) time.
Verified Code Solutions
function solution(nums, K) {
let totalSum = 0;
for (let num of nums) {
if (num > K) {
totalSum += num;
}
}
if (totalSum === 0) {
return 0;
} else {
return totalSum - nums.filter(x => x <= K).reduce((a, b) => a + b, 0);
}
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int totalSum = 0;
for (int num : nums) {
if (num > K) {
totalSum += num;
}
}
if (totalSum == 0) {
return 0;
} else {
return totalSum - std::accumulate(std::begin(nums), std::end(nums), 0, [&](int a, int b) { return a + (b <= K ? 0 : b); });
}
}
};class Solution {
public int solution(int[] nums, int K) {
int totalSum = 0;
for (int num : nums) {
if (num > K) {
totalSum += num;
}
}
if (totalSum == 0) {
return 0;
} else {
return totalSum - Arrays.stream(nums).filter(x -> x <= K).sum();
}
}
}def solution(nums, K):
totalSum = 0
for num in nums:
if num > K:
totalSum += num
if totalSum == 0:
return 0
else:
return totalSum - sum([x for x in nums if x <= K])function solution(nums, K) {
let totalSum = 0;
for (let num of nums) {
if (num > K) {
totalSum += num;
}
}
if (totalSum === 0) {
return 0;
} else {
return totalSum - nums.filter(x => x <= K).reduce((a, b) => a + b, 0);
}
}Asked in Top Tech Interviews
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