3830. Find Closest Person¶

Difficulty: Easy

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3830. Find Closest Person

Easy

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

Solution¶

class Solution {
    public int findClosest(int x, int y, int z) {
        int diff1 = Math.abs(x - z), diff2 = Math.abs(y - z);
        if (diff1 == diff2) return 0;
        else if (diff1 < diff2) return 1;
        return 2;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here