Question

JAVA

/**
   * posOfLargestElementLtOeT returns the position of the largest element in the
   * array that is less than or equal to the limit parameter if all values are
   * greater than limit, return -1;
   *
   * Precondition: the array is nonempty and all elements are unique. Your
   * solution must go through the array exactly once.
   *
   * <pre>
   * 0 == posOfLargestElementLtOeT(3, new double[] { -7 }) // value:-7 is in pos 0
   * 5 == posOfLargestElementLtOeT(3, new double[] { 11, -4, -7, 7, 8, 1 }), // value:1 is in pos 5
   * -1 == posOfLargestElementLtOeT(-7, new double[] { 1, -4, -5, 7, 8, 11 }), // all elements are > -7
   *
   * The code below is a stub version, you should replace the line of code
   * labeled TODO with code that achieves the above specification
   * </pre>
   */
   public static int posOfLargestElementLtOeT(double limit, double[] list) {
       return -2; // TODO 2: fix this code
   }

Answer 1

Program in java:

//class definition
public class LargestElement {
    //main method definition
    public static void main(String[] args) {
        //calling funtion to find the position of largest number
        System.out.println(posOfLargestElementLtOeT(3, new double[]{-7}));
        System.out.println(posOfLargestElementLtOeT(3, new double[]{11,-4,-7,7,8,1}));
        System.out.println(posOfLargestElementLtOeT(-7, new double[]{1,-4,-5,7,8,11}));
    }
    //declaring the function to calculate largest number
    static int posOfLargestElementLtOeT(int val, double[] array){
        //declare varibale
        int pos=-1;
        double largest=0;
        boolean flag=true;
        //iterate through the array to find the largest number within limit
        for(int i=0;i<array.length;i++){
            //if the value of the element is less than the specified limit
            if(array[i]<val){
                //this condition will be executed only once to store the largest value
                if(flag==true){
                    //storing the first element less than the specified limit as the largest
                    largest=array[i];
                    //marking this position as the largest element position
                    pos=i;
                }
                //checking to find the largest number subsequent from the array less than specified limit
                if(largest<array[i]){
                    //storing the first element less than the specified limit as the largest
                    largest=array[i];
                    //marking this position as the largest element position
                    pos=i;
                }              
            }
        }
        //returning the position of the largest element
        return pos;
    }
}

Output: