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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
Code language: PHP (php)

Explanation:

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

expressionsMatter(1,1,1) ==> return 3
Code language: PHP (php)

Explanation:

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

expressionsMatter(9,1,1) ==> return 18
Code language: PHP (php)

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)); } }
Code language: Java (java)

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; } }
Code language: Java (java)

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))); } }
Code language: Java (java)

Option 3 (using Vector):

See also  How to Sum Consecutives in Java
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(); } }
Code language: Java (java)

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(); } }
Code language: Java (java)

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