public class DataProcessor {

    /** This method should return true if and only if every element in the
      * input array is in ascending order, meaning that none of the values
      * in the array are smaller than the previous element, and false 
      * otherwise.  For example, the array {1, 2, 5, 5, 6, 7, 8, 8, 8} 
      * would be in order.  An empty array is considered in order as well.
      */
    public static boolean inOrder(int[] values) {
     // DELETE THE LINE BELOW ONCE YOU IMPLEMENT THE CALL!
        throw new RuntimeException("not implemented!");
    }
 
    /** This method will return the smallest value found in the array.  For
      * example, for the array {4, 2, 3, 1, 5, 1}, the value 1 will be 
      * returned.  If the array is empty, then the built-in constant
      * Integer.MAX_VALUE may be returned.
      */
    public static int smallest(int[] values) {
     // DELETE THE LINE BELOW ONCE YOU IMPLEMENT THE CALL!
        throw new RuntimeException("not implemented!");
    }
 
    /** This method will return the smallest value found in the array which
      * is larger than the input n.  For example, for the input n=2 and the
      * array {4, 2, 3, 1, 5, 1}, the value 3 will be returned.  If no such
      * value exists, then the method will return the value n.
      */
    public static int smallestGreaterThanN(int n, int[] values) {
     // DELETE THE LINE BELOW ONCE YOU IMPLEMENT THE CALL!
        throw new RuntimeException("not implemented!");
    }
 
    /** Return the total number of distinct values contained in the
      * input array.  For example, the array {2, 2, 5, 1, 7, 4, 7} has
      * 5 distinct values (the numbers 1, 2, 4, 5, and 7).
      * 
      * Hint: calls to smallest() and smallestGreaterThanN() may help
      * you solve this.
      */
    public static int numDistinct(int[] values) {
     // DELETE THE LINE BELOW ONCE YOU IMPLEMENT THE CALL!
        throw new RuntimeException("not implemented!");
    }

    /** Determines the number of times element v occurs within the array
      * of values. For example, in the array {1, 2, 5, 5, 3, 2}, the value
      * v=2 occurs twice.
      */  
    public static int numOccurrences(int v, int[] values) {
     // DELETE THE LINE BELOW ONCE YOU IMPLEMENT THE CALL!
        throw new RuntimeException("not implemented!");
    }

    /** Determines the number of elements from the array of values which
      * have a value within the range [a, b) (that is, between a and b,
      * including a but not including b).  For example, if the input array
      * is {1, 2, 3, 4, 2, 3, 4, 3, 4, 4} with a=2 and b=4, the result would
      * be 5 (because there are two 2's and three 3's).
      */
    public static int numInRange(int a, int b, int[] values) {
     // DELETE THE LINE BELOW ONCE YOU IMPLEMENT THE CALL!
        throw new RuntimeException("not implemented!");
    }
 
    /** Finds the arithmetic mean (the average) of the list of numbers.  For
      * example, if the input is {1, 2, 3, 4}, then the mean is 2.5.
      * 
      * Hint: the input values are integers, but the output most likely
      * won't be.  Do you need to account for that somehow?
      */
    public static double mean(int[] values) {
     // DELETE THE LINE BELOW ONCE YOU IMPLEMENT THE CALL!
        throw new RuntimeException("not implemented!");
    }
 
    /** The input of this method includes two arrays, which you may assume
      * are the same size, and whose entries correspond to one another 
      * (for example, the first array may be a list of ages of several people,
      * while the second array may be a list of the heights of the same people).
      * The method will compute the average of values from the second array, 
      * but only when the values correspond to entries from the first list 
      * within the range [a, b).  For example, if the first array is a list
      * of ages, {19, 15, 20, 25, 30, 5, 32}, the second array is a list 
      * of heights in feet, {5.2, 4.9, 5.9, 6.1, 5.85, 3.2, 5.7}, and the
      * range is a=20, b=30, then it would give the average of people aged
      * 20-29, which in this example would be 6.0.
      */
    public static double meanInRange(int a, int b, int[] list, double[] values) {
     // DELETE THE LINE BELOW ONCE YOU IMPLEMENT THE CALL!
        throw new RuntimeException("not implemented!");
    }
 
    /** When performing data analysis, it is often necessary to first clean
      * obvious outliers from the data set, because these may be the result
      * of a clerical error and could throw off analysis.  As an example,
      * when studying medical data, sometimes you will see "zombies" (people
      * with hospital visits occurring after their recorded date of death) or
      * "vampires" (people who are hundreds of years old, because their date
      * of birth was left blank and defaulted to zero when entered into the 
      * the system).  This method will eliminate entries from the array of
      * input values which are outside of the range [a, b), and return the list
      * of entries which remain. For example, for the input a=0, b=125, and
      * {10, -1, 20, 30, 25, 125, 40, 200}, the array which is returned would
      * be {10, 20, 30, 25, 40}.
      * 
      * Hint 1: you don't need to delete elements from the input array. You
      * can create a whole new array to return instead.
      * Hint 2: it may help to know how big your output array is before you
      * begin.  Would any of the other methods you've written help here?
      */
    public static int[] cleanList(int a, int b, int[] list) {
       // DELETE THE LINE BELOW ONCE YOU IMPLEMENT THE CALL!
        throw new RuntimeException("not implemented!");
    }
 
    /** Return the value of the nearest neighbor to v.  If v is some value,
      * look through the list of values to find the one which is closest
      * in value to v.  For example, if v=1.2 and the input array is
      * {3.2, 1.7, 2.4, 0.9, 4.4}, then the method will return 0.9 because
      * it is the value closest to 1.2 in the list.  If there is a tie (for
      * example, both 0.9 and 1.5 are the same distance to 1.2), then return
      * the one which occurs first in the list.  If the list is empty, then
      * return v.
      */
    public static double nearestNeighbor(double v, double[] values) {
     // DELETE THE LINE BELOW ONCE YOU IMPLEMENT THE CALL!
        throw new RuntimeException("not implemented!");  
    }

    /** HONORS SECTION ONLY 
      * (if you are not in the honors section, you do not need to complete 
      * this part, and it is not worth any points if you do):
      * 
      * Find and return the k elements which are closest to v in value, in
      * order of increasing distance from v.  For example, if v=1.2, k=3,
      * and the input array is {3.2, 1.7, 2.4, 0.9, 4.4}, then the method 
      * will return {0.9, 1.7, 2.4}.  If k is larger than the length of
      * the input array, then the output array should only have as many values
      * as you have in the original array.  As with nearestNeighbor, if two elements
      * are the same distance from v, then the one which appears earlier in the
      * original should be considerd closer in the output array.
      */
    public static double[] kNearestNeighbors(int k, double v, double[] values) {
     // DELETE THE LINE BELOW ONCE YOU IMPLEMENT THE CALL!
        throw new RuntimeException("not implemented!");  
    }
 
}