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Faulty Sensor - Solution & Explanation

EasyPremiumFree on FleetCodeArrayTwo Pointers7 min readAsked at: Meta
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Problem Statement

An experiment is being conducted in a lab. To ensure accuracy, there are two sensors collecting data simultaneously. You are given two arrays sensor1 and sensor2, where sensor1[i] and sensor2[i] are the ith data points collected by the two sensors.

However, this type of sensor has a chance of being defective, which causes exactly one data point to be dropped. After the data is dropped, all the data points to the right of the dropped data are shifted one place to the left, and the last data point is replaced with some random value. It is guaranteed that this random value will not be equal to the dropped value.

  • For example, if the correct data is [1,2,3,4,5] and 3 is dropped, the sensor could return [1,2,4,5,7] (the last position can be any value, not just 7).

We know that there is a defect in at most one of the sensors. Return the sensor number (1 or 2) with the defect. If there is no defect in either sensor or if it is impossible to determine the defective sensor, return -1.

 

Example 1:

Input: sensor1 = [2,3,4,5], sensor2 = [2,1,3,4]
Output: 1
Explanation: Sensor 2 has the correct values.
The second data point from sensor 2 is dropped, and the last value of sensor 1 is replaced by a 5.

Example 2:

Input: sensor1 = [2,2,2,2,2], sensor2 = [2,2,2,2,5]
Output: -1
Explanation: It is impossible to determine which sensor has a defect.
Dropping the last value for either sensor could produce the output for the other sensor.

Example 3:

Input: sensor1 = [2,3,2,2,3,2], sensor2 = [2,3,2,3,2,7]
Output: 2
Explanation: Sensor 1 has the correct values.
The fourth data point from sensor 1 is dropped, and the last value of sensor 1 is replaced by a 7.

 

Constraints:

  • sensor1.length == sensor2.length
  • 1 <= sensor1.length <= 100
  • 1 <= sensor1[i], sensor2[i] <= 100

Approach Overview

Problem Overview: Two sensors record integer readings in arrays of equal length. Exactly one sensor may be faulty and drops a value at some position, causing the remaining values to shift left while the last reading becomes inconsistent. Your task is to determine which sensor is faulty, or return -1 if it cannot be determined.

Approach 1: Brute Force Shift Verification (O(n^2) time, O(1) space)

Start by locating the first index where sensor1[i] != sensor2[i]. From that position, simulate the scenario where each sensor could be faulty. For example, assume sensor1 dropped a value and check whether sensor1[i+1...] matches sensor2[i...n-2]. Then repeat assuming sensor2 dropped a value. Because each check scans the remaining array, this approach may re-traverse large portions of the data, resulting in O(n^2) time in the worst case while using constant extra space.

Approach 2: Single Pass Traversal (O(n) time, O(1) space)

A more efficient strategy uses a single linear scan across the array. Iterate until the first mismatch appears. At that point, compare the remaining segments while skipping one element from either array. If skipping one element in sensor1 aligns the rest of the sequence with sensor2, sensor1 is faulty. If skipping one element in sensor2 aligns the rest with sensor1, sensor2 is faulty. This behaves like a lightweight two pointers comparison where the offset starts after the mismatch. If both possibilities remain valid or neither works, return -1.

The key observation: only the first mismatch matters. Before that point both sensors match perfectly, and after the dropped value every element shifts by exactly one position. A single traversal check confirms which sequence experienced the shift.

Recommended for interviews: The linear traversal approach is the expected solution. It demonstrates that you can detect the first divergence and reason about shifted sequences without unnecessary nested loops. Mentioning the brute force idea first shows clear thinking, but implementing the O(n) traversal with constant space proves strong algorithmic instincts.

Solution

Traverse both arrays, find the first unequal position i. If i \lt n - 1, loop to compare sensor1[i + 1] and sensor2[i], if they are not equal, it indicates that sensor 1 is defective, return 1; otherwise compare sensor1[i] and sensor2[i + 1], if they are not equal, it indicates that sensor 2 is defective, return 2.

If the traversal ends, it means that the defective sensor cannot be determined, return -1.

The time complexity is O(n), where n is the length of the array. The space complexity is O(1).

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Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Brute Force Shift VerificationO(n^2)O(1)Conceptual starting point when reasoning about how a dropped value shifts the sequence
Single Pass Traversal (Shift Check)O(n)O(1)Optimal solution for interviews and production; detect first mismatch and verify the shifted segment

Video Solution

1826. Faulty Sensor (Leetcode Easy) • Programming Live with Larry • 308 views views

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Frequently Asked Questions

Is Faulty Sensor easy or hard?
Faulty Sensor is classified as an Easy problem on LeetCode. The challenge comes from correctly identifying the first mismatch and reasoning about how a dropped value shifts the remainder of the array. Once that observation is made, the implementation is straightforward.
Faulty Sensor Python/Java solution
Both Python and Java implementations follow the same pattern: scan until the first mismatch, then verify alignment by skipping one element in either array. The logic uses simple loops and index comparisons, keeping the runtime O(n) and memory usage O(1).
How to solve Faulty Sensor in O(n)?
Iterate through both arrays until you find the first index where the readings differ. From that point, check whether skipping one value in sensor1 aligns the rest with sensor2 or skipping one value in sensor2 aligns the rest with sensor1. If exactly one alignment works, return that sensor index; otherwise return -1.
What is the best approach for Faulty Sensor?
The optimal approach is a single-pass traversal that finds the first mismatch between the two arrays and then checks which array becomes aligned when one element is skipped. This method runs in O(n) time and O(1) space because it only scans the remaining portion once. Interviewers typically expect this linear check instead of repeated comparisons.
Is Faulty Sensor asked at Google/Amazon/Meta?
Faulty Sensor represents a typical array comparison and edge‑case reasoning problem often used in technical screens. Variants of shifted array detection or mismatch detection appear in interviews at companies like Amazon and Google because they test careful traversal logic and boundary handling.
What data structure is used in Faulty Sensor?
The problem primarily uses arrays and simple index traversal. A two-pointer style comparison is applied after the first mismatch to check which sequence shifted by one element. No additional data structures such as hash maps or stacks are required.
What is the time complexity of Faulty Sensor?
The optimal solution runs in O(n) time where n is the number of readings in the arrays. You scan the arrays until the first mismatch and then compare the remaining elements once more. The algorithm uses O(1) additional space because it only stores a few indices.

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