Question

/**
* Write a method named outOfOrder that takes as a parameter an array of double and returns a value of type int.
* This method will test the array for being out of order, meaning that the array violates the conditions:
* array[0] <= arrayValues[0] <= arrayValues[2] <=...
*
* The method returns -1 if the elemnts are not out of order; otherwise , it returns the index of the first element of the array that is out of order.
* For example, consider the declaration:
* double[] arrayValues = {1.2, 2.1, 3.3, 4.5, 2.5, 7.9, 5.4, 8.7, 9.9, 1.5};
* In the array above, array[3] and arrayValues[4] are the first pair out of order, and arrayValues[4] is the frist element out of order, so the method returns 5.
* If the array were sorted, the method would return -1.
*/
public static void runProblem1()
{
  
double[] unSortedArray = {1.2, 2.1, 3.3, 4.5, 2.5, 7.9, 5.4, 8.7, 9.9, 1.5};
double[] sortedArray = {1.3, 2.9, 3.4, 4.6, 6.8, 7.5, 10.1, 12.6, 45.9, 50.6, 63.7};
  
System.out.println("The method will take in an array and evaluate if it is sorted:\nReturn of -1 means it is sorted, any other value will indicate the index that is out of order");
  
System.out.println("The unsorted Array result: "+ outOfOrder(unSortedArray));
System.out.println("The sorted Array result: "+ outOfOrder(sortedArray));
  
}

Answer 1

Answer:

Explanation:

For loop is used to traverse the array. If any element is greater than its next element, position of next element is returned. Otherwise at the end, -1 is returned.

COde:

public static int outOfOrder(double[] a)
{
for(int i=0; i<a.length-1; i++)
{
if(a[i]>a[i+1])
{
return i+2;
}
}
  
return -1;
}
  

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