Calculate the biggest of 3 numbers using Java

The challenge

Task

  • Given three integers a ,b ,creturn the largest number obtained after inserting the following operators and brackets+*()
  • In other words , try every combination of a,b,c with [*+()] , and return the Maximum Obtained

Consider an Example :

With the numbers are 1, 2 and 3 , here are some ways of placing signs and brackets:

  • 1 * (2 + 3) = 5
  • 1 * 2 * 3 = 6
  • 1 + 2 * 3 = 7
  • (1 + 2) * 3 = 9

So the maximum value that you can obtain is 9.

Notes

  • The numbers are always positive.
  • The numbers are in the range (1 ≤ a, b, c ≤ 10).
  • You can use the same operation more than once.
  • It’s not necessary to place all the signs and brackets.
  • Repetition in numbers may occur .
  • You cannot swap the operands. For instance, in the given example you cannot get expression (1 + 3) * 2 = 8.

Input >> Output Examples:

expressionsMatter(1,2,3)  ==>  return 9

Explanation:

After placing signs and brackets, the Maximum value obtained from the expression (1+2) * 3 = 9.

expressionsMatter(1,1,1)  ==>  return 3

Explanation:

After placing signs, the Maximum value obtained from the expression is 1 + 1 + 1 = 3.

expressionsMatter(9,1,1)  ==>  return 18

Explanation:

After placing signs and brackets, the Maximum value obtained from the expression is 9 * (1+1) = 18.

Test cases

import org.junit.Test;
import static org.junit.Assert.assertEquals;
import org.junit.runners.JUnit4;

public class ExpressionsMatter {
   @Test
    public void checkSmallValues() {
        assertEquals(6, Biggest.expressionsMatter(2, 1, 2));
        assertEquals(3, Biggest.expressionsMatter(1, 1, 1));
        assertEquals(4, Biggest.expressionsMatter(2, 1, 1));
        assertEquals(9, Biggest.expressionsMatter(1, 2, 3));
        assertEquals(5, Biggest.expressionsMatter(1, 3, 1));
        assertEquals(8, Biggest.expressionsMatter(2, 2, 2));
    }
  
    @Test
    public void checkIntermediateValues() {
        assertEquals( 20, Biggest.expressionsMatter(5, 1, 3));
        assertEquals(105, Biggest.expressionsMatter(3, 5, 7));
        assertEquals( 35, Biggest.expressionsMatter(5, 6, 1));
        assertEquals(  8, Biggest.expressionsMatter(1, 6, 1));
        assertEquals( 14, Biggest.expressionsMatter(2, 6, 1));
        assertEquals( 48, Biggest.expressionsMatter(6, 7, 1));
    }

    @Test
    public void checkMixedValues() {
        assertEquals( 60, Biggest.expressionsMatter( 2, 10,  3));
        assertEquals( 27, Biggest.expressionsMatter( 1,  8,  3));
        assertEquals(126, Biggest.expressionsMatter( 9,  7,  2));
        assertEquals( 20, Biggest.expressionsMatter( 1,  1, 10));
        assertEquals( 18, Biggest.expressionsMatter( 9,  1,  1));
        assertEquals(300, Biggest.expressionsMatter(10,  5,  6));
        assertEquals( 12, Biggest.expressionsMatter( 1, 10,  1));
    }
}

The solution in Java

Option 1:

public class Biggest {
    public static int expressionsMatter(int a, int b, int c) {
        int out = a+b+c;
      
        if ((a * (b + c))>out) out = a * (b + c);
        if ((a * b * c)>out) out = a * b * c;
        if ((a + b * c)>out) out = a + b * c;
        if ((a + b) * c>out) out = (a + b) * c;
      
        return out;
    }
}

Option 2 (using Math.max):

public class Biggest {
    public static int expressionsMatter(int a, int b, int c) {
      return  Math.max(Math.max(a + b + c, a * b * c),Math.max ((a + b) * c, a * (b + c)));
    }
}

Option 3 (using Vector):

import java.util.*;

public class Biggest {
    public static int expressionsMatter(int a, int b, int c) {
        Vector <Integer> sol = new Vector();

        sol.add(a + b + c);
        sol.add(a * b * c);
        sol.add((a+b)*c);
        sol.add(a*(c+b));

        Collections.sort(sol);
        
        return sol.lastElement();
    }
}

Option 4 (using a stream):

import java.util.stream.IntStream;

public class Biggest {
    public static int expressionsMatter(int a, int b, int c) {
        return IntStream.of(a + b + c, a * b * c, a + b * c, a * b + c, (a + b) * c, a * (b + c)).max().getAsInt();
    }
}
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