Question

1826. Faulty Sensor

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Description

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

Solutions

Solution 1: Traversal

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 $O(1)$.

class Solution:
  def badSensor(self, sensor1: List[int], sensor2: List[int]) -> int:
    i, n = 0, len(sensor1)
    while i < n - 1:
      if sensor1[i] != sensor2[i]:
        break
      i += 1
    while i < n - 1:
      if sensor1[i + 1] != sensor2[i]:
        return 1
      if sensor1[i] != sensor2[i + 1]:
        return 2
      i += 1
    return -1

class Solution {
  public int badSensor(int[] sensor1, int[] sensor2) {
    int i = 0;
    int n = sensor1.length;
    for (; i < n - 1 && sensor1[i] == sensor2[i]; ++i) {
    }
    for (; i < n - 1; ++i) {
      if (sensor1[i + 1] != sensor2[i]) {
        return 1;
      }
      if (sensor1[i] != sensor2[i + 1]) {
        return 2;
      }
    }
    return -1;
  }
}

class Solution {
public:
  int badSensor(vector<int>& sensor1, vector<int>& sensor2) {
    int i = 0;
    int n = sensor1.size();
    for (; i < n - 1 && sensor1[i] == sensor2[i]; ++i) {}
    for (; i < n - 1; ++i) {
      if (sensor1[i + 1] != sensor2[i]) return 1;
      if (sensor1[i] != sensor2[i + 1]) return 2;
    }
    return -1;
  }
};

func badSensor(sensor1 []int, sensor2 []int) int {
  i, n := 0, len(sensor1)
  for ; i < n-1 && sensor1[i] == sensor2[i]; i++ {
  }
  for ; i < n-1; i++ {
    if sensor1[i+1] != sensor2[i] {
      return 1
    }
    if sensor1[i] != sensor2[i+1] {
      return 2
    }
  }
  return -1
}

function badSensor(sensor1: number[], sensor2: number[]): number {
  let i = 0;
  const n = sensor1.length;
  while (i < n - 1) {
    if (sensor1[i] !== sensor2[i]) {
      break;
    }
    ++i;
  }
  while (i < n - 1) {
    if (sensor1[i + 1] !== sensor2[i]) {
      return 1;
    }
    if (sensor1[i] !== sensor2[i + 1]) {
      return 2;
    }
    ++i;
  }
  return -1;
}