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3516. Find Closest Person

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.

 

Example 1:

Input: x = 2, y = 7, z = 4

Output: 1

Explanation:

  • Person 1 is at position 2 and can reach Person 3 (at position 4) in 2 steps.
  • Person 2 is at position 7 and can reach Person 3 in 3 steps.

Since Person 1 reaches Person 3 first, the output is 1.

Example 2:

Input: x = 2, y = 5, z = 6

Output: 2

Explanation:

  • Person 1 is at position 2 and can reach Person 3 (at position 6) in 4 steps.
  • Person 2 is at position 5 and can reach Person 3 in 1 step.

Since Person 2 reaches Person 3 first, the output is 2.

Example 3:

Input: x = 1, y = 5, z = 3

Output: 0

Explanation:

  • Person 1 is at position 1 and can reach Person 3 (at position 3) in 2 steps.
  • Person 2 is at position 5 and can reach Person 3 in 2 steps.

Since both Person 1 and Person 2 reach Person 3 at the same time, the output is 0.

 

Constraints:

  • 1 <= x, y, z <= 100

Solutions

Solution 1: Mathematics

We calculate the distance \(a\) between the 1st person and the 3rd person, and the distance \(b\) between the 2nd person and the 3rd person.

  • If \(a = b\), it means both people arrive at the same time, return \(0\);
  • If \(a \lt b\), it means the 1st person will arrive first, return \(1\);
  • Otherwise, it means the 2nd person will arrive first, return \(2\).

The time complexity is \(O(1)\), and the space complexity is \(O(1)\).

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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)

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class Solution {
    public int findClosest(int x, int y, int z) {
        int a = Math.abs(x - z);
        int b = Math.abs(y - z);
        return a == b ? 0 : (a < b ? 1 : 2);
    }
}

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class Solution {
public:
    int findClosest(int x, int y, int z) {
        int a = abs(x - z);
        int b = abs(y - z);
        return a == b ? 0 : (a < b ? 1 : 2);
    }
};

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func findClosest(x int, y int, z int) int {
    a, b := abs(x-z), abs(y-z)
    if a == b {
        return 0
    }
    if a < b {
        return 1
    }
    return 2
}

func abs(x int) int {
    if x < 0 {
        return -x
    }
    return x
}

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function findClosest(x: number, y: number, z: number): number {
    const a = Math.abs(x - z);
    const b = Math.abs(y - z);
    return a === b ? 0 : a < b ? 1 : 2;
}

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