Inversions.java


Below is the syntax highlighted version of Inversions.java from §2.2 Mergesort.


/******************************************************************************
 *  Compilation:  javac Inversions.java
 *  Execution:    java Inversions < input.txt
 *  Dependencies: StdIn.java StdOut.java
 *  
 *  Read array of n integers and count number of inversions in n log n time.
 *
 ******************************************************************************/

/**
 *  The {@code Inversions} class provides static methods to count the 
 *  number of <em>inversions</em> in either an array of integers or comparables.
 *  An inversion in an array {@code a[]} is a pair of indicies {@code i} and
 *  {@code j} such that {@code i} < {@code j} and {@code a[i] > a[j]}.
 *  <p>
 *  This implementation uses a generalization of mergesort. The <em>count</em>
 *  operation takes time proportional to <em>n</em> log <em>n</em>,
 *  where <em>n</em> is the number of keys in the array.
 *  <p>
 *  For additional documentation, see <a href="http://algs4.cs.princeton.edu/22mergesort">Section 2.2</a>
 *  of <i>Algorithms, 4th Edition</i> by Robert Sedgewick and Kevin Wayne.
 *
 *  @author Robert Sedgewick
 *  @author Kevin Wayne
 */
public class Inversions {

    // do not instantiate
    private Inversions() { }

    // merge and count
    private static long merge(int[] a, int[] aux, int lo, int mid, int hi) {
        long inversions = 0;

        // copy to aux[]
        for (int k = lo; k <= hi; k++) {
            aux[k] = a[k]; 
        }

        // merge back to a[]
        int i = lo, j = mid+1;
        for (int k = lo; k <= hi; k++) {
            if      (i > mid)           a[k] = aux[j++];
            else if (j > hi)            a[k] = aux[i++];
            else if (aux[j] < aux[i]) { a[k] = aux[j++]; inversions += (mid - i + 1); }
            else                        a[k] = aux[i++];
        }
        return inversions;
    }

    // return the number of inversions in the subarray b[lo..hi]
    // side effect b[lo..hi] is rearranged in ascending order
    private static long count(int[] a, int[] b, int[] aux, int lo, int hi) {
        long inversions = 0;
        if (hi <= lo) return 0;
        int mid = lo + (hi - lo) / 2;
        inversions += count(a, b, aux, lo, mid);  
        inversions += count(a, b, aux, mid+1, hi);
        inversions += merge(b, aux, lo, mid, hi);
        assert inversions == brute(a, lo, hi);
        return inversions;
    }


    /**
     * Returns the number of inversions in the integer array.
     * The argument array is not modified.
     * @param  a the array
     * @return the number of inversions in the array. An inversion is a pair of 
     *         indicies {@code i} and {@code j} such that {@code i < j}
     *         and {@code a[i]} > {@code a[j]}.
     */
    public static long count(int[] a) {
        int[] b   = new int[a.length];
        int[] aux = new int[a.length];
        for (int i = 0; i < a.length; i++)
            b[i] = a[i];
        long inversions = count(a, b, aux, 0, a.length - 1);
        return inversions;
    }



    // merge and count (Comparable version)
    private static <Key extends Comparable<Key>> long merge(Key[] a, Key[] aux, int lo, int mid, int hi) {
        long inversions = 0;

        // copy to aux[]
        for (int k = lo; k <= hi; k++) {
            aux[k] = a[k]; 
        }

        // merge back to a[]
        int i = lo, j = mid+1;
        for (int k = lo; k <= hi; k++) {
            if      (i > mid)                a[k] = aux[j++];
            else if (j > hi)                 a[k] = aux[i++];
            else if (less(aux[j], aux[i])) { a[k] = aux[j++]; inversions += (mid - i + 1); }
            else                             a[k] = aux[i++];
        }
        return inversions;
    }

    // return the number of inversions in the subarray b[lo..hi]
    // side effect b[lo..hi] is rearranged in ascending order
    private static <Key extends Comparable<Key>> long count(Key[] a, Key[] b, Key[] aux, int lo, int hi) {
        long inversions = 0;
        if (hi <= lo) return 0;
        int mid = lo + (hi - lo) / 2;
        inversions += count(a, b, aux, lo, mid);  
        inversions += count(a, b, aux, mid+1, hi);
        inversions += merge(b, aux, lo, mid, hi);
        assert inversions == brute(a, lo, hi);
        return inversions;
    }


    /**
     * Returns the number of inversions in the comparable array.
     * The argument array is not modified.
     * @param  a the array
     * @return the number of inversions in the array. An inversion is a pair of 
     *         indicies {@code i} and {@code j} such that {@code i < j}
     *         and {@code a[i].compareTo(a[j]) > 0}.
     */
    public static <Key extends Comparable<Key>> long count(Key[] a) {
        Key[] b   = a.clone();
        Key[] aux = a.clone();
        long inversions = count(a, b, aux, 0, a.length - 1);
        return inversions;
    }


    // is v < w ?
    private static <Key extends Comparable<Key>> boolean less(Key v, Key w) {
        return (v.compareTo(w) < 0);
    }

    // count number of inversions in a[lo..hi] via brute force (for debugging only)
    private static <Key extends Comparable<Key>> long brute(Key[] a, int lo, int hi) {
        long inversions = 0;
        for (int i = lo; i <= hi; i++)
            for (int j = i + 1; j <= hi; j++)
                if (less(a[j], a[i])) inversions++;
        return inversions;
    }

    // count number of inversions in a[lo..hi] via brute force (for debugging only)
    private static long brute(int[] a, int lo, int hi) {
        long inversions = 0;
        for (int i = lo; i <= hi; i++)
            for (int j = i + 1; j <= hi; j++)
                if (a[j] < a[i]) inversions++;
        return inversions;
    }

    /**
     * Reads in a sequence of integers from standard input and prints
     * the number of inversions.
     *
     * @param args the command-line arguments
     */
    public static void main(String[] args) {
        int[] a = StdIn.readAllInts();
        int n = a.length;
        Integer[] b = new Integer[n];
        for (int i = 0; i < n; i++)
            b[i] = a[i];
        StdOut.println(Inversions.count(a));
        StdOut.println(Inversions.count(b));
    }
}


Copyright © 2000–2016, Robert Sedgewick and Kevin Wayne.
Last updated: Sun Oct 30 07:49:10 EDT 2016.