Home Software Engineering The right way to Resolve Max-Min Arrays in Java

## The problem#

You’re given an array of distinctive parts, and your job is to rearrange the values in order that the primary max worth is adopted by the primary minimal, adopted by second max worth then second min worth, and so on.

For instance:

``````clear up([15,11,10,7,12]) = [15,7,12,10,11]
``````

The primary max is `15` and the primary min is `7`. The second max is `12` and the second min is `10` and so forth.

## The answer in Java code#

Choice 1:

``````import java.util.*;

public static int[] clear up (int[] arr){
Arrays.kind(arr);
int[] solutionArray = new int[arr.length];

for(int i = 0; i < arr.size; i++){
solutionArray[i] = i % 2 == 0 ? arr[arr.length - i/2 - 1] : arr[i/2];
}
return solutionArray;
}
}
``````

Choice 2:

``````import java.util.*;

public static int[] clear up (int[] arr){
Listing<Integer> temp = new ArrayList<Integer>();
Arrays.kind(arr);
for (int i = 0, j = arr.size - 1; i <= j; ++i, --j) {
}
return temp.stream().mapToInt(i -> i).toArray();
}
}
``````

Choice 3:

``````import java.util.stream.IntStream;

public static int[] clear up(int[] arr) {
int[] sorted = IntStream.of(arr).sorted().toArray();
int[] consequence = new int[arr.length];
for (int i = 0, j = arr.size - 1, f = -1; i < arr.size;) {
consequence[i] = sorted[j];
j = (j + arr.size + (f *= -1) * (++i)) % arr.size;
}
return consequence;
}

}
``````

## Take a look at circumstances to validate our answer#

``````import org.junit.Take a look at;
import static org.junit.Assert.assertArrayEquals;
import org.junit.runners.JUnit4;

public class SolutionTest{
@Take a look at
public void basicTests(){