Problem Description

You are given two sensor arrays sensor1 and sensor2 that should contain identical data collected from an experiment. However, one of the sensors might be defective.

When a sensor is defective, it drops exactly one data point. After dropping a data point:

• All data points to the right shift one position to the left
• The last position gets filled with a random value (guaranteed to be different from the dropped value)

For example, if the correct data is [1,2,3,4,5] and the value 3 is dropped, the defective sensor might return [1,2,4,5,7] where:

• Values 4 and 5 shifted left
• Position 5 is filled with random value 7

Given the two sensor arrays:

• Return 1 if sensor1 is defective
• Return 2 if sensor2 is defective
• Return -1 if you cannot determine which sensor is defective or if neither is defective

The key insight is that at most one sensor can be defective. By comparing the arrays element by element, you can identify where they first differ. From that point, you can check if one array matches the other when accounting for the shifted pattern that would result from a dropped value.

The solution finds the first position where the arrays differ, then checks if sensor1[i+1] matches sensor2[i] (indicating sensor1 dropped a value at position i) or if sensor2[i+1] matches sensor1[i] (indicating sensor2 dropped a value at position i).

Intuition

When two sensors should have identical data but one might have dropped a value, the arrays will be identical up to the point where the drop occurred. This is our starting observation.

Let's think about what happens when a value is dropped. If we have original data [a, b, c, d, e] and value c is dropped, we get [a, b, d, e, x] where x is some random value. Notice that:

• Elements before the dropped position remain unchanged
• Elements after the dropped position shift left by one
• The last element becomes unpredictable

Now, when comparing two arrays where one might have a dropped value, we'll see matching elements until we reach the drop point. At position i where they first differ, one of two things happened:

1. sensor1 dropped its value at position i, so sensor1[i] now contains what was originally at position i+1
2. sensor2 dropped its value at position i, so sensor2[i] now contains what was originally at position i+1

To determine which sensor is defective, we can check the alignment pattern after position i:

• If sensor1[i+1] == sensor2[i], then sensor1 is shifted (meaning it dropped a value)
• If sensor2[i+1] == sensor1[i], then sensor2 is shifted (meaning it dropped a value)

The beauty of this approach is that we only need to check these alignment patterns starting from where the arrays first differ. If both patterns hold true simultaneously throughout the remaining elements (except possibly the last one), then we cannot determine which sensor is defective, returning -1.

Learn more about Two Pointers patterns.

Solution Approach

The implementation follows a two-phase traversal approach:

Phase 1: Find the first difference

i, n = 0, len(sensor1)
while i < n - 1:
    if sensor1[i] != sensor2[i]:
        break
    i += 1

We iterate through both arrays simultaneously until we find the first position i where sensor1[i] != sensor2[i]. We stop at n-1 because if we reach the last element without finding a difference, the arrays might differ only at the last position (which could be the random replacement value).

Phase 2: Determine which sensor is defective

while i < n - 1:
    if sensor1[i + 1] != sensor2[i]:
        return 1
    if sensor1[i] != sensor2[i + 1]:
        return 2
    i += 1

Starting from the first difference position, we check the alignment patterns:

• If sensor1[i + 1] != sensor2[i], it means sensor1 cannot be the shifted version of the correct data, so sensor1 is defective (return 1)
• If sensor1[i] != sensor2[i + 1], it means sensor2 cannot be the shifted version of the correct data, so sensor2 is defective (return 2)
• If both conditions are false (meaning both alignments match), we continue checking the next position

The loop continues until i < n - 1 because we need to compare sensor1[i+1] and sensor2[i+1], which requires staying within array bounds.

Return -1 for undetermined cases If we complete the second loop without returning, it means:

• Either both sensors show consistent shifted patterns (impossible to determine which is defective)
• Or the arrays only differ at the last position (could be the random value in either sensor)
• Or the arrays are completely identical (no defect)

In all these cases, we return -1 to indicate the defective sensor cannot be determined.

Example Walkthrough

Let's walk through an example where sensor1 = [2,3,5,7] and sensor2 = [2,3,4,5].

Step 1: Find the first difference

• Compare position 0: sensor1[0] = 2, sensor2[0] = 2 → Match, continue
• Compare position 1: sensor1[1] = 3, sensor2[1] = 3 → Match, continue
• Compare position 2: sensor1[2] = 5, sensor2[2] = 4 → First difference found!

We stop at position i = 2 where the arrays first differ.

Step 2: Check alignment patterns Starting from position i = 2:

• Check if sensor1 is defective: Does sensor1[i+1] match sensor2[i]?
  • sensor1[3] = 7 vs sensor2[2] = 4 → No match
• Check if sensor2 is defective: Does sensor2[i+1] match sensor1[i]?
  • sensor2[3] = 5 vs sensor1[2] = 5 → Match!

Since sensor2[3] = sensor1[2], this indicates that sensor2 has a shifted pattern. This means sensor2 dropped a value at position 2.

Verification of the pattern:

• Original correct data was likely: [2, 3, 4, 5]
• sensor1 kept all values: [2, 3, 5, 7] (wait, this doesn't match...)

Let me reconsider. Actually, if sensor2[3] = sensor1[2], it means:

• sensor1 has the correct sequence at positions 2 and beyond
• sensor2 dropped value at position 2, causing everything to shift left
• Original data in sensor1: [2, 3, 5, 7]
• sensor2 dropped the 5 at position 2: [2, 3, ~~5~~, 7] → [2, 3, 7, ?] → [2, 3, 4, 5]

Actually, let me use a clearer example:

Let's walk through where sensor1 = [1,2,4,5,9] and sensor2 = [1,2,3,4,5].

Step 1: Find the first difference

• Position 0: sensor1[0] = 1, sensor2[0] = 1 → Match
• Position 1: sensor1[1] = 2, sensor2[1] = 2 → Match
• Position 2: sensor1[2] = 4, sensor2[2] = 3 → First difference at i = 2

Step 2: Check alignment patterns from position 2

• Check sensor1[i+1] vs sensor2[i]: sensor1[3] = 5 vs sensor2[2] = 3 → No match
• Check sensor2[i+1] vs sensor1[i]: sensor2[3] = 4 vs sensor1[2] = 4 → Match!

Move to position 3:

• Check sensor1[i+1] vs sensor2[i]: sensor1[4] = 9 vs sensor2[3] = 4 → No match
• Check sensor2[i+1] vs sensor1[i]: sensor2[4] = 5 vs sensor1[3] = 5 → Match!

The pattern sensor2[i+1] = sensor1[i] consistently holds, meaning sensor2 is shifted relative to sensor1. This indicates sensor1 dropped a value at position 2.

Understanding why: The original data was [1,2,3,4,5]. Sensor1 dropped the 3, resulting in [1,2,4,5,9] where 9 is the random fill value. Sensor2 correctly shows [1,2,3,4,5].

Result: Return 1 (sensor1 is defective)

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the sensor arrays.

The algorithm consists of two sequential while loops:

• The first loop iterates through the arrays to find the first position where sensor1[i] != sensor2[i]. In the worst case, this takes O(n-1) iterations.
• The second loop continues from position i and iterates through the remaining elements, performing constant-time comparisons at each step. This also takes at most O(n-1-i) iterations.

Since both loops are sequential and not nested, the total time complexity is O(n-1) + O(n-1-i) = O(n).

Space Complexity: O(1)

The algorithm only uses a constant amount of extra space:

• Two integer variables i and n to track the current index and array length
• No additional data structures are created

The space usage remains constant regardless of the input size, resulting in O(1) space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Boundary Handling in the First Loop

A common mistake is using while i < n instead of while i < n - 1 in the first loop. This can cause an index out of bounds error when accessing sensor1[i+1] or sensor2[i+1] in the second loop.

Incorrect:

# This will cause index error if arrays differ at last position
while i < n:
    if sensor1[i] != sensor2[i]:
        break
    i += 1

Correct:

while i < n - 1:
    if sensor1[i] != sensor2[i]:
        break
    i += 1

2. Misunderstanding the Shift Pattern

Many solutions incorrectly assume they need to compare the entire remaining subarrays after finding the first difference. The key insight is that you only need to check element-by-element alignment, not slice comparison.

Incorrect approach:

# Comparing entire subarrays is inefficient and unnecessary
if sensor1[i+1:] == sensor2[i:n-1]:
    return 1
if sensor2[i+1:] == sensor1[i:n-1]:
    return 2

Correct approach:

# Check element-by-element as we iterate
while index < n - 1:
    if sensor1[index + 1] != sensor2[index]:
        return 1
    if sensor1[index] != sensor2[index + 1]:
        return 2
    index += 1

3. Forgetting Edge Cases Where Arrays Are Identical

If both arrays are completely identical, the first loop will reach n-1 without finding any difference. The solution should return -1 in this case since there's no defective sensor.

Missing edge case handling:

# Assuming there's always a difference
i = 0
while sensor1[i] == sensor2[i]:  # This could go out of bounds
    i += 1

Proper handling:

# The loop bounds ensure we don't go out of range
while i < n - 1:
    if sensor1[i] != sensor2[i]:
        break
    i += 1
# If i == n-1, arrays might be identical or differ only at last position

4. Early Return Without Complete Verification

Some implementations return immediately after finding the first matching shifted element, without verifying the entire remaining sequence maintains the shift pattern.

Incorrect:

# Returns too early without verifying the entire pattern
if i < n - 1:
    if sensor1[i + 1] == sensor2[i]:
        return 1  # Wrong! Need to verify entire shifted sequence
    if sensor2[i + 1] == sensor1[i]:
        return 2  # Wrong! Need to verify entire shifted sequence

Correct:

# Continue checking until we find a mismatch in the pattern
while index < n - 1:
    if sensor1[index + 1] != sensor2[index]:
        return 1  # Pattern broken, sensor1 is defective
    if sensor1[index] != sensor2[index + 1]:
        return 2  # Pattern broken, sensor2 is defective
    index += 1