Problem Description

You are given three integers x, y, and z, representing the positions of three people on a number line:

• x is the position of Person 1.
• y is the position of Person 2.
• z is the position of Person 3, who does not move.

Both Person 1 and Person 2 move toward Person 3 at the same speed.

Determine which person reaches Person 3 first:

• Return 1 if Person 1 arrives first.
• Return 2 if Person 2 arrives first.
• Return 0 if both arrive at the same time.

Return the result accordingly.

Intuition

To determine which person reaches the stationary Person 3 first, we need to consider the distance each moving person starts from. The distance is calculated as the absolute difference between their starting position and Person 3's position (z).

• For Person 1, the distance to travel is |x - z|.
• For Person 2, the distance to travel is |y - z|.

The person with the smaller distance will reach Person 3 first. Therefore, we compare these distances:

• If |x - z| equals |y - z|, it means both persons arrive simultaneously, so we return 0.
• If |x - z| is less than |y - z|, Person 1 reaches first, so return 1.
• If |y - z| is less than |x - z|, Person 2 reaches first, so return 2.

This approach involves simple arithmetic calculations ensuring efficient comparison of distances.

Learn more about Math patterns.

Solution Approach

The solution leverages basic mathematical computation to determine who reaches Person 3 first. The steps involved are as follows:

1. Calculate Distances:

   • Let a be the distance from Person 1 to Person 3, computed as a = abs(x - z).
   • Let b be the distance from Person 2 to Person 3, computed as b = abs(y - z).

2. Comparison of Distances:

   • If a == b, it indicates that both Person 1 and Person 2 will reach Person 3 simultaneously. Thus, return 0.
   • If a < b, Person 1 is closer to Person 3 than Person 2 is, so return 1.
   • Otherwise, if b < a, Person 2 reaches Person 3 first, so return 2.

This approach efficiently resolves the problem with constant time complexity, as it only involves basic arithmetic operations and comparisons. The solution is implemented in the findClosest method in the reference code:

class Solution:
    def findClosest(self, x: int, y: int, z: int) -> int:
        a = abs(x - z)
        b = abs(y - z)
        return 0 if a == b else (1 if a < b else 2)

This code effectively computes the required distances and returns the desired output based on their comparison.

Example Walkthrough

Let's walk through an example to understand how the solution approach is applied. Suppose we have three integers representing positions on a number line:

• x = 5 (Position of Person 1)
• y = 2 (Position of Person 2)
• z = 4 (Position of Person 3, stationary)

Using the solution approach, let's determine which person reaches Person 3 first:

1. Calculate Distances:

   • Distance a from Person 1 to Person 3:
     [ a = \lvert x - z \rvert = \lvert 5 - 4 \rvert = 1 ]
   • Distance b from Person 2 to Person 3:
     [ b = \lvert y - z \rvert = \lvert 2 - 4 \rvert = 2 ]

2. Comparison of Distances:

   • Compare a and b:
     • Since a < b (1 < 2), Person 1 is closer to Person 3.
   • Thus, Person 1 reaches Person 3 first, and we return 1.

This process shows that Person 1 will arrive at Person 3's position before Person 2, given the calculated distances.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(1) because the computation involves a constant number of arithmetic operations (calculating absolute differences and comparing them), which does not depend on the size of the input.

The space complexity is also O(1) as the code uses a fixed amount of extra space for variables a and b regardless of the input size.

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