LeetCode 1826: Faulty Sensor ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2026-04-05

Submissions: Wrong Answer ×2 → Accepted

Approach
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A faulty sensor drops one value, shifts the rest left by one, and fills the last slot with a random value (guaranteed not to equal the dropped value). At most one sensor is faulty.

Find the first i with sensor1[i] != sensor2[i]:

If sensor2 is faulty, it dropped sensor1[i], so sensor2[i:-1] should equal sensor1[i+1:].
If sensor1 is faulty, likewise sensor1[i:-1] should equal sensor2[i+1:].

Only one explanation fitting identifies the sensor; if both or neither fit, it cannot be determined, so return -1. No mismatch at all also returns -1.

My Attempts
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Attempt 1 compared only the next element: sensor1[i + 1] == sensor2[i] meant sensor2 was faulty.
Attempt 2 compared the whole remaining tail but still did not handle both explanations fitting.

Both fail on sensor1 = [3, 3, 2, 3], sensor2 = [3, 2, 3, 2]. Sensor2 could have dropped the second 3 (with 2 appended), and sensor1 could have dropped sensor2’s first 2 (with 3 appended); both fit, so the answer is -1, but both versions returned 2.

Solution
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class Solution:
    def badSensor(self, sensor1: List[int], sensor2: List[int]) -> int:
        n = len(sensor1)
        for i in range(n - 1):
            if sensor1[i] != sensor2[i]:
                sensor2_bad = sensor1[i + 1:] == sensor2[i:-1]
                sensor1_bad = sensor2[i + 1:] == sensor1[i:-1]
                if sensor1_bad == sensor2_bad:
                    return -1
                return 1 if sensor1_bad else 2
        return -1

The loop stops at n - 2: if only the last position differs, it is just the random fill value, and the faulty sensor cannot be identified.

Complexity
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Time O(n), space O(n) (slices copy).

Mistakes to Avoid
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Concluding from the next element alone.
Not handling the case where both explanations fit.

Interview Explanation
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I find the first mismatch i and test each hypothesis: if sensor2 is faulty, its data from i should equal sensor1’s data from i + 1, ignoring the last slot, and symmetrically for sensor1. Exactly one fitting identifies the sensor; if both or neither fit, I return -1. Time O(n).