Problem:
Score : 300 points

Problem Statement

You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order:

Choose two among A, B and C, then increase both by 1.

Choose one among A, B and C, then increase it by 2.

It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.

Constraints

0 \leq A,B,C \leq 50

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the minimum number of operations required to make A, B and C all equal.

Sample Input 1

2 5 4

Sample Output 1

2

We can make A, B and C all equal by the following operations:

Increase A and C by 1. Now, A, B, C are 3, 5, 5, respectively.

Increase A by 2. Now, A, B, C are 5, 5, 5, respectively.

Sample Input 2

2 6 3

Sample Output 2

5

Sample Input 3

31 41 5

Sample Output 3

23

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;

public class Main {
  public static void main(String[] args) throws IOException {
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    String[] str = in.readLine().split(" ");
    int a = Integer.parseInt(str[0]);
    int b = Integer.parseInt(str[1]);
    int c = Integer.parseInt(str[2]);
    int sum = a+b+c;
    int max = Math.max(Math.max(a,b),c);
    int min = Math.min(Math.min(a,b),c);
    int middle = sum-max-min;
    int count = 0;
    if((3*max)%2 != sum%2) {
      max++;
      count++;
    }
    count+=(3*max-sum)/2;
    
    System.out.println(count);
  }
}