Faulty Sensor

Faulty Sensor - Problem

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.

Input & Output

Example 1 — Sensor1 Missing Element

$ Input: sensor1 = [1,3,4,5], sensor2 = [1,2,3,4]

› Output: 1

💡 Note: Sensor1 is missing element 2. If we remove element 3 from position 1 in sensor1, we get [1,4,5], but this doesn't match sensor2's first 3 elements [1,2,3]. However, sensor1 appears to have dropped element 2, causing all subsequent elements to shift left and the last position to be filled with value 5.

Example 2 — No Defect

$ Input: sensor1 = [1,2,3,4], sensor2 = [1,2,3,4]

› Output: -1

💡 Note: Both sensors have identical readings, so there's no defect in either sensor.

Example 3 — Sensor2 Missing Element

$ Input: sensor1 = [1,2,3,4], sensor2 = [1,3,4,7]

› Output: 2

💡 Note: Sensor2 is missing element 2. The correct sequence should be [1,2,3,4], but sensor2 shows [1,3,4,7] where 2 was dropped and 7 was added at the end.

Constraints

2 ≤ sensor1.length, sensor2.length ≤ 1000
sensor1.length == sensor2.length
1 ≤ sensor1[i], sensor2[i] ≤ 10⁶
At most one sensor is defective

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to find the first position where sensors differ, then test if skipping that position in either sensor creates a valid match. The optimal approach uses two pointers for O(n) time complexity with early termination on first mismatch detection.

Common Approaches

✓ Hash Map

⏱️ Time: N/A Space: N/A

Brute Force - Try All Drop Positions

⏱️ Time: O(n²) Space: O(1)

For each sensor, try dropping each position and see if the remaining elements match the other sensor. This approach systematically checks all possibilities but is inefficient.

Two Pointers - Compare Until Mismatch

⏱️ Time: O(n) Space: O(1)

Use two pointers to compare both sensors simultaneously. When we find the first mismatch, we can determine which sensor is faulty by checking if skipping one element in either sensor allows the rest to match.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Smart Actions

💡 Explanation

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