Sensor Packet Aligner 42 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the synchronization of a distributed sensor network. The system generates a sequence of integer readings, where each value represents a specific alignment metric. Your goal is to determine the maximum possible 'alignment score' by selecting a contiguous subsequence of these readings. The alignment score is defined as the sum of the elements in the subsequence, but with a specific constraint: the subsequence must contain at least one element that is a multiple of 42. If no such subsequence exists, the score is 0.
Given an array of integers representing the sensor readings, compute the maximum alignment score. The solution must efficiently handle large datasets by leveraging the properties of contiguous sums and modular arithmetic. You are required to return the highest possible sum of any contiguous subarray that includes at least one multiple of 42. If the array does not contain any multiple of 42, return 0.
This problem requires careful handling of negative values and the boundary conditions where the optimal subarray might start or end at the edges of the input array. The core challenge lies in identifying the optimal window that satisfies the divisibility constraint while maximizing the cumulative sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Packet Aligner 42"
WHY DOES IT MATTER?
The maximum subarray pattern demonstrates how local optimal decisions (choosing to extend or restart a subarray) lead to a globally optimal solution. It is a foundational technique that appears in many interview questions, from financial time‑series analysis to network packet optimization, making it a must‑know for candidates.
OPTIMIZATION CHALLENGE
The key insight is that any negative prefix can only decrease future sums, so it can be discarded immediately. This reduces the problem from quadratic to linear time by avoiding redundant calculations of subarray sums that would never be optimal.
REAL-WORLD CONNECTION
In distributed systems, you often need to identify the longest period of high throughput or the most stable network segment. The maximum subarray algorithm is analogous to finding the contiguous time window where the cumulative performance metric is highest, which is critical for load balancing and fault detection.
When explaining Kadane’s algorithm in an interview, emphasize the "running sum" concept and how resetting it at a negative value is a greedy yet optimal strategy. Show the recurrence diagram quickly to demonstrate the dynamic programming nature.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for the maximum possible sum of any contiguous subsequence of an integer array. A naive approach would examine every possible start and end index, compute the sum of each subarray, and keep track of the maximum. This brute‑force method runs in O(n^2) time and is infeasible for large inputs (e.g., n=10^5). The optimal solution uses a single pass dynamic programming technique known as Kadane’s algorithm. At each position, we maintain the best subarray ending at that index by either extending the previous subarray or starting fresh at the current element. This reduces the problem to a simple recurrence: bestEndingHere = max(arr[i], bestEndingHere + arr[i]), and the overall maximum is the maximum of all bestEndingHere values. The algorithm runs in O(n) time and O(1) space, making it ideal for large datasets.
Kadane’s algorithm is a classic example of the "maximum subarray" pattern, which is a special case of the more general "two pointers" or "sliding window" paradigm. While the two‑pointer technique is often used for subarrays with a bounded sum or length, Kadane’s approach can be viewed as a degenerate sliding window where the window size is implicitly determined by the sign of the running sum. This insight allows the algorithm to discard negative prefixes that would only reduce future sums, effectively shrinking the window on the fly.
Because the problem is a textbook case of dynamic programming over contiguous segments, it is frequently used in technical interviews to assess a candidate’s ability to reason about optimal substructure and to implement efficient linear‑time solutions. Understanding why the greedy choice of discarding negative sums leads to an optimal solution is key to mastering this pattern.
Interview Questions on This Problem
Q1What is the time complexity of Kadane’s algorithm and why is it optimal for the maximum subarray problem?
Kadane’s algorithm runs in O(n) time because it processes each element exactly once, updating the best subarray ending at that element. This is optimal because any algorithm must examine each element at least once to determine whether it contributes to the maximum sum, and no algorithm can do better than linear time for this problem.
Q2How would you modify Kadane’s algorithm to also return the start and end indices of the maximum subarray?
Maintain two additional variables, startCandidate and bestStart. When arr[i] > bestEndingHere + arr[i], set startCandidate = i. When bestEndingHere exceeds the global maximum, update bestStart = startCandidate and bestEnd = i. This tracks the indices of the optimal subarray.
Q3In a distributed sensor network, why might you prefer a streaming implementation of the maximum subarray algorithm over a batch approach?
A streaming implementation processes data as it arrives, requiring only constant memory and no need to store the entire sequence. This is essential in sensor networks where memory is limited and data arrives continuously, allowing real‑time alignment score computation without buffering large amounts of data.
Examples
Input
nums = [10, 20, 42, 30, 5]
Output
107
Explanation: The array contains 42 at index 2. The contiguous subarray [10, 20, 42, 30, 5] sums to 107. Since this subarray includes the multiple of 42, it is a valid candidate. No other subarray containing 42 yields a higher sum. Thus, the maximum alignment score is 107.
Input
nums = [-5, 42, -10, 42, 3]
Output
77
Explanation: There are two multiples of 42 at indices 1 and 3. Consider the subarray [42, -10, 42, 3] which sums to 77. Another valid subarray is [42, -10, 42] summing to 74. The subarray [-5, 42, -10, 42, 3] sums to 72. The maximum sum among all valid subarrays is 77.
Input
nums = [1, 2, 3, 4, 5]
Output
0
Explanation: The array does not contain any element that is a multiple of 42. Therefore, no valid subarray exists that satisfies the constraint. According to the problem statement, the alignment score is 0 in this case.
Input
nums = [42, -100, 42, 10]
Output
94
Explanation: The multiples of 42 are at indices 0 and 2. The subarray [42, -100, 42, 10] sums to -8. However, the subarray [42, 10] is not contiguous with the first 42. Let's look at subarrays containing index 0: [42] sum=42, [42, -100] sum=-58, [42, -100, 42] sum=-16, [42, -100, 42, 10] sum=-8. Subarrays containing index 2: [42] sum=42, [42, 10] sum=52, [-100, 42] sum=-58, [-100, 42, 10] sum=-48, [42, -100, 42] sum=-16, [42, -100, 42, 10] sum=-8. The maximum valid sum is 52 from [42, 10] (indices 2-3). Wait, let's re-evaluate. Is [42] at index 0 valid? Yes, sum 42. Is [42, 10] at indices 2-3 valid? Yes, sum 52. Is there a better one? What about just [42] at index 2? Sum 42. The maximum is 52.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The array may contain negative integers.
- If no multiple of 42 exists in the array, return 0.
Optimal Approach & Strategy
Use Kadane’s algorithm: iterate once, keep a running sum that resets when negative, and track the maximum sum seen. This runs in O(n) time and O(1) space.
Brute Force Approach
Check every possible start and end index, compute the sum of each subarray, and keep the maximum. This takes O(n^2) time and is too slow for large arrays.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
let i = 0;
let j = 0;
while (i < nums.length) {
if (nums[i] > K) {
sum += nums[i];
j++;
}
i++;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
int i = 0;
int j = 0;
while (i < nums.size()) {
if (nums[i] > K) {
sum += nums[i];
j++;
}
i++;
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
int i = 0;
int j = 0;
while (i < nums.length) {
if (nums[i] > K) {
sum += nums[i];
j++;
}
i++;
}
return sum;
}
}def solution(nums, K):
sum = 0
i = 0
j = 0
while i < len(nums):
if nums[i] > K:
sum += nums[i]
j += 1
i += 1
return sumfunction solution(nums, K) {
let sum = 0;
let i = 0;
let j = 0;
while (i < nums.length) {
if (nums[i] > K) {
sum += nums[i];
j++;
}
i++;
}
return sum;
}Asked in Top Tech Interviews
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