/*

 * Copyright 2009 Google Inc.  All Rights Reserved.

 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.

 *

 * This code is free software; you can redistribute it and/or modify it

 * under the terms of the GNU General Public License version 2 only, as

 * published by the Free Software Foundation.  Sun designates this

 * particular file as subject to the "Classpath" exception as provided

 * by Sun in the LICENSE file that accompanied this code.

 *

 * This code is distributed in the hope that it will be useful, but WITHOUT

 * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or

 * FITNESS FOR A PARTICULAR PURPOSE.  See the GNU General Public License

 * version 2 for more details (a copy is included in the LICENSE file that

 * accompanied this code).

 *

 * You should have received a copy of the GNU General Public License version

 * 2 along with this work; if not, write to the Free Software Foundation,

 * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.

 *

 * Please contact Sun Microsystems, Inc., 4150 Network Circle, Santa Clara,

 * CA 95054 USA or visit www.sun.com if you need additional information or

 * have any questions.

 */

package java.util;

/**

 * This is a near duplicate of {@link TimSort}, modified for use with

 * arrays of objects that implement {@link Comparable}, instead of using

 * explicit comparators.

 *

 * <p>If you are using an optimizing VM, you may find that ComparableTimSort

 * offers no performance benefit over TimSort in conjunction with a

 * comparator that simply returns {@code ((Comparable)first).compareTo(Second)}.

 * If this is the case, you are better off deleting ComparableTimSort to

 * eliminate the code duplication.  (See Arrays.java for details.)

 *

 * @author Josh Bloch

 */

class ComparableTimSort {

    /**

     * This is the minimum sized sequence that will be merged.  Shorter

     * sequences will be lengthened by calling binarySort.  If the entire

     * array is less than this length, no merges will be performed.

     *

     * This constant should be a power of two.  It was 64 in Tim Peter's C

     * implementation, but 32 was empirically determined to work better in

     * this implementation.  In the unlikely event that you set this constant

     * to be a number that's not a power of two, you'll need to change the

     * {@link #minRunLength} computation.

     *

     * If you decrease this constant, you must change the stackLen

     * computation in the TimSort constructor, or you risk an

     * ArrayOutOfBounds exception.  See listsort.txt for a discussion

     * of the minimum stack length required as a function of the length

     * of the array being sorted and the minimum merge sequence length.

     */

    private static final int MIN_MERGE = 32;

    /**

     * The array being sorted.

     */

    private final Object[] a;

    /**

     * When we get into galloping mode, we stay there until both runs win less

     * often than MIN_GALLOP consecutive times.

     */

    private static final int  MIN_GALLOP = 7;

    /**

     * This controls when we get *into* galloping mode.  It is initialized

     * to MIN_GALLOP.  The mergeLo and mergeHi methods nudge it higher for

     * random data, and lower for highly structured data.

     */

    private int minGallop = MIN_GALLOP;

    /**

     * Maximum initial size of tmp array, which is used for merging.  The array

     * can grow to accommodate demand.

     *

     * Unlike Tim's original C version, we do not allocate this much storage

     * when sorting smaller arrays.  This change was required for performance.

     */

    private static final int INITIAL_TMP_STORAGE_LENGTH = 256;

    /**

     * Temp storage for merges.

     */

    private Object[] tmp;

    /**

     * A stack of pending runs yet to be merged.  Run i starts at

     * address base[i] and extends for len[i] elements.  It's always

     * true (so long as the indices are in bounds) that:

     *

     *     runBase[i] + runLen[i] == runBase[i + 1]

     *

     * so we could cut the storage for this, but it's a minor amount,

     * and keeping all the info explicit simplifies the code.

     */

    private int stackSize = 0;  // Number of pending runs on stack

    private final int[] runBase;

    private final int[] runLen;

    /**

     * Creates a TimSort instance to maintain the state of an ongoing sort.

     *

     * @param a the array to be sorted

     */

    private ComparableTimSort(Object[] a) {

        this.a = a;

        // Allocate temp storage (which may be increased later if necessary)

        int len = a.length;

        @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})

        Object[] newArray = new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ?

                                       len >>> 1 : INITIAL_TMP_STORAGE_LENGTH];

        tmp = newArray;

        /*

         * Allocate runs-to-be-merged stack (which cannot be expanded).  The

         * stack length requirements are described in listsort.txt.  The C

         * version always uses the same stack length (85), but this was

         * measured to be too expensive when sorting "mid-sized" arrays (e.g.,

         * 100 elements) in Java.  Therefore, we use smaller (but sufficiently

         * large) stack lengths for smaller arrays.  The "magic numbers" in the

         * computation below must be changed if MIN_MERGE is decreased.  See

         * the MIN_MERGE declaration above for more information.

         */

        int stackLen = (len <    120  ?  5 :

                        len <   1542  ? 10 :

                        len < 119151  ? 19 : 40);

        runBase = new int[stackLen];

        runLen = new int[stackLen];

    }

    /*

     * The next two methods (which are package private and static) constitute

     * the entire API of this class.  Each of these methods obeys the contract

     * of the public method with the same signature in java.util.Arrays.

     */

    static void sort(Object[] a) {

          sort(a, 0, a.length);

    }

    static void sort(Object[] a, int lo, int hi) {

        rangeCheck(a.length, lo, hi);

        int nRemaining  = hi - lo;

        if (nRemaining < 2)

            return;  // Arrays of size 0 and 1 are always sorted

        // If array is small, do a "mini-TimSort" with no merges

        if (nRemaining < MIN_MERGE) {

            int initRunLen = countRunAndMakeAscending(a, lo, hi);

            binarySort(a, lo, hi, lo + initRunLen);

            return;

        }

        /**

         * March over the array once, left to right, finding natural runs,

         * extending short natural runs to minRun elements, and merging runs

         * to maintain stack invariant.

         */

        ComparableTimSort ts = new ComparableTimSort(a);

        int minRun = minRunLength(nRemaining);

        do {

            // Identify next run

            int runLen = countRunAndMakeAscending(a, lo, hi);

            // If run is short, extend to min(minRun, nRemaining)

            if (runLen < minRun) {

                int force = nRemaining <= minRun ? nRemaining : minRun;

                binarySort(a, lo, lo + force, lo + runLen);

                runLen = force;

            }

            // Push run onto pending-run stack, and maybe merge

            ts.pushRun(lo, runLen);

            ts.mergeCollapse();

            // Advance to find next run

            lo += runLen;

            nRemaining -= runLen;

        } while (nRemaining != 0);

        // Merge all remaining runs to complete sort

        assert lo == hi;

        ts.mergeForceCollapse();

        assert ts.stackSize == 1;

    }

    /**

     * Sorts the specified portion of the specified array using a binary

     * insertion sort.  This is the best method for sorting small numbers

     * of elements.  It requires O(n log n) compares, but O(n^2) data

     * movement (worst case).

     *

     * If the initial part of the specified range is already sorted,

     * this method can take advantage of it: the method assumes that the

     * elements from index {@code lo}, inclusive, to {@code start},

     * exclusive are already sorted.

     *

     * @param a the array in which a range is to be sorted

     * @param lo the index of the first element in the range to be sorted

     * @param hi the index after the last element in the range to be sorted

     * @param start the index of the first element in the range that is

     *        not already known to be sorted (@code lo <= start <= hi}

     */

    @SuppressWarnings("fallthrough")

    private static void binarySort(Object[] a, int lo, int hi, int start) {

        assert lo <= start && start <= hi;

        if (start == lo)

            start++;

        for ( ; start < hi; start++) {

            @SuppressWarnings("unchecked")

            Comparable<Object> pivot = (Comparable) a[start];

            // Set left (and right) to the index where a[start] (pivot) belongs

            int left = lo;

            int right = start;

            assert left <= right;

            /*

             * Invariants:

             *   pivot >= all in [lo, left).

             *   pivot <  all in [right, start).

             */

            while (left < right) {

                int mid = (left + right) >>> 1;

                if (pivot.compareTo(a[mid]) < 0)

                    right = mid;

                else

                    left = mid + 1;

            }

            assert left == right;

            /*

             * The invariants still hold: pivot >= all in [lo, left) and

             * pivot < all in [left, start), so pivot belongs at left.  Note

             * that if there are elements equal to pivot, left points to the

             * first slot after them -- that's why this sort is stable.

             * Slide elements over to make room to make room for pivot.

             */

            int n = start - left;  // The number of elements to move

            // Switch is just an optimization for arraycopy in default case

            switch(n) {

                case 2:  a[left + 2] = a[left + 1];

                case 1:  a[left + 1] = a[left];

                         break;

                default: System.arraycopy(a, left, a, left + 1, n);

            }

            a[left] = pivot;

        }

    }

    /**

     * Returns the length of the run beginning at the specified position in

     * the specified array and reverses the run if it is descending (ensuring

     * that the run will always be ascending when the method returns).

     *

     * A run is the longest ascending sequence with:

     *

     *    a[lo] <= a[lo + 1] <= a[lo + 2] <= ...

     *

     * or the longest descending sequence with:

     *

     *    a[lo] >  a[lo + 1] >  a[lo + 2] >  ...

     *

     * For its intended use in a stable mergesort, the strictness of the

     * definition of "descending" is needed so that the call can safely

     * reverse a descending sequence without violating stability.

     *

     * @param a the array in which a run is to be counted and possibly reversed

     * @param lo index of the first element in the run

     * @param hi index after the last element that may be contained in the run.

              It is required that @code{lo < hi}.

     * @return  the length of the run beginning at the specified position in

     *          the specified array

     */

    @SuppressWarnings("unchecked")

    private static int countRunAndMakeAscending(Object[] a, int lo, int hi) {

        assert lo < hi;

        int runHi = lo + 1;

        if (runHi == hi)

            return 1;

        // Find end of run, and reverse range if descending

        if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending

            while(runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0)

                runHi++;

            reverseRange(a, lo, runHi);

        } else {                              // Ascending

            while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) >= 0)

                runHi++;

        }

        return runHi - lo;

    }

    /**

     * Reverse the specified range of the specified array.

     *

     * @param a the array in which a range is to be reversed

     * @param lo the index of the first element in the range to be reversed

     * @param hi the index after the last element in the range to be reversed

     */

    private static void reverseRange(Object[] a, int lo, int hi) {

        hi--;

        while (lo < hi) {

            Object t = a[lo];

            a[lo++] = a[hi];

            a[hi--] = t;

        }

    }

    /**

     * Returns the minimum acceptable run length for an array of the specified

     * length. Natural runs shorter than this will be extended with

     * {@link #binarySort}.

     *

     * Roughly speaking, the computation is:

     *

     *  If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).

     *  Else if n is an exact power of 2, return MIN_MERGE/2.

     *  Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k

     *   is close to, but strictly less than, an exact power of 2.

     *

     * For the rationale, see listsort.txt.

     *

     * @param n the length of the array to be sorted

     * @return the length of the minimum run to be merged

     */

    private static int minRunLength(int n) {

        assert n >= 0;

        int r = 0;      // Becomes 1 if any 1 bits are shifted off

        while (n >= MIN_MERGE) {

            r |= (n & 1);

            n >>= 1;

        }

        return n + r;

    }

    /**

     * Pushes the specified run onto the pending-run stack.

     *

     * @param runBase index of the first element in the run

     * @param runLen  the number of elements in the run

     */

    private void pushRun(int runBase, int runLen) {

        this.runBase[stackSize] = runBase;

        this.runLen[stackSize] = runLen;

        stackSize++;

    }

    /**

     * Examines the stack of runs waiting to be merged and merges adjacent runs

     * until the stack invariants are reestablished:

     *

     *     1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]

     *     2. runLen[i - 2] > runLen[i - 1]

     *

     * This method is called each time a new run is pushed onto the stack,

     * so the invariants are guaranteed to hold for i < stackSize upon

     * entry to the method.

     */

    private void mergeCollapse() {

        while (stackSize > 1) {

            int n = stackSize - 2;

            if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {

                if (runLen[n - 1] < runLen[n + 1])

                    n--;

                mergeAt(n);

            } else if (runLen[n] <= runLen[n + 1]) {

                mergeAt(n);

            } else {

                break; // Invariant is established

            }

        }

    }

    /**

     * Merges all runs on the stack until only one remains.  This method is

     * called once, to complete the sort.

     */

    private void mergeForceCollapse() {

        while (stackSize > 1) {

            int n = stackSize - 2;

            if (n > 0 && runLen[n - 1] < runLen[n + 1])

                n--;

            mergeAt(n);

        }

    }

    /**

     * Merges the two runs at stack indices i and i+1.  Run i must be

     * the penultimate or antepenultimate run on the stack.  In other words,

     * i must be equal to stackSize-2 or stackSize-3.

     *

     * @param i stack index of the first of the two runs to merge

     */

    @SuppressWarnings("unchecked")

    private void mergeAt(int i) {

        assert stackSize >= 2;

        assert i >= 0;

        assert i == stackSize - 2 || i == stackSize - 3;

        int base1 = runBase[i];

        int len1 = runLen[i];

        int base2 = runBase[i + 1];

        int len2 = runLen[i + 1];

        assert len1 > 0 && len2 > 0;

        assert base1 + len1 == base2;

        /*

         * Record the length of the combined runs; if i is the 3rd-last

         * run now, also slide over the last run (which isn't involved

         * in this merge).  The current run (i+1) goes away in any case.

         */

        runLen[i] = len1 + len2;

        if (i == stackSize - 3) {

            runBase[i + 1] = runBase[i + 2];

            runLen[i + 1] = runLen[i + 2];

        }

        stackSize--;

        /*

         * Find where the first element of run2 goes in run1. Prior elements

         * in run1 can be ignored (because they're already in place).

         */

        int k = gallopRight((Comparable<Object>) a[base2], a, base1, len1, 0);

        assert k >= 0;

        base1 += k;

        len1 -= k;

        if (len1 == 0)

            return;

        /*

         * Find where the last element of run1 goes in run2. Subsequent elements

         * in run2 can be ignored (because they're already in place).

         */

        len2 = gallopLeft((Comparable<Object>) a[base1 + len1 - 1], a,

                base2, len2, len2 - 1);

        assert len2 >= 0;

        if (len2 == 0)

            return;

        // Merge remaining runs, using tmp array with min(len1, len2) elements

        if (len1 <= len2)

            mergeLo(base1, len1, base2, len2);

        else

            mergeHi(base1, len1, base2, len2);

    }

    /**

     * Locates the position at which to insert the specified key into the

     * specified sorted range; if the range contains an element equal to key,

     * returns the index of the leftmost equal element.

     *

     * @param key the key whose insertion point to search for

     * @param a the array in which to search

     * @param base the index of the first element in the range

     * @param len the length of the range; must be > 0

     * @param hint the index at which to begin the search, 0 <= hint < n.

     *     The closer hint is to the result, the faster this method will run.

     * @return the int k,  0 <= k <= n such that a[b + k - 1] < key <= a[b + k],

     *    pretending that a[b - 1] is minus infinity and a[b + n] is infinity.

     *    In other words, key belongs at index b + k; or in other words,

     *    the first k elements of a should precede key, and the last n - k

     *    should follow it.

     */

    private static int gallopLeft(Comparable<Object> key, Object[] a,

            int base, int len, int hint) {

        assert len > 0 && hint >= 0 && hint < len;

        int lastOfs = 0;

        int ofs = 1;

        if (key.compareTo(a[base + hint]) > 0) {

            // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]

            int maxOfs = len - hint;

            while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) > 0) {

                lastOfs = ofs;

                ofs = (ofs << 1) + 1;

                if (ofs <= 0)   // int overflow

                    ofs = maxOfs;

            }

            if (ofs > maxOfs)

                ofs = maxOfs;

            // Make offsets relative to base

            lastOfs += hint;

            ofs += hint;

        } else { // key <= a[base + hint]

            // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]

            final int maxOfs = hint + 1;

            while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) <= 0) {

                lastOfs = ofs;

                ofs = (ofs << 1) + 1;

                if (ofs <= 0)   // int overflow

                    ofs = maxOfs;

            }

            if (ofs > maxOfs)

                ofs = maxOfs;

            // Make offsets relative to base

            int tmp = lastOfs;

            lastOfs = hint - ofs;

            ofs = hint - tmp;

        }

        assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

        /*

         * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere

         * to the right of lastOfs but no farther right than ofs.  Do a binary

         * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].

         */

        lastOfs++;

        while (lastOfs < ofs) {

            int m = lastOfs + ((ofs - lastOfs) >>> 1);

            if (key.compareTo(a[base + m]) > 0)

                lastOfs = m + 1;  // a[base + m] < key

            else

                ofs = m;          // key <= a[base + m]

        }

        assert lastOfs == ofs;    // so a[base + ofs - 1] < key <= a[base + ofs]

        return ofs;

    }