/*
 * Copyright (C) 2008 The Android Open Source Project
 *
 * Licensed under the Apache License, Version 2.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 *      http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */
/*
 * This file originates from https://android.googlesource.com/platform/libcore/+/gingerbread/luni/src/main/java/java/util/TimSort.java
 * and has been modified to sort primitive long arrays instead of object arrays.
 */
//package java.util;

package org.apache.cassandra.utils;

import java.util.Arrays;

A stable, adaptive, iterative mergesort that requires far fewer than
n lg(n) comparisons when running on partially sorted arrays, while
offering performance comparable to a traditional mergesort when run
on random arrays.  Like all proper mergesorts, this sort is stable and
runs O(n log n) time (worst case).  In the worst case, this sort requires
temporary storage space for n/2 object references; in the best case,
it requires only a small constant amount of space.
This implementation was adapted from Tim Peters's list sort for
Python, which is described in detail here:
  http://svn.python.org/projects/python/trunk/Objects/listsort.txt
Tim's C code may be found here:
  http://svn.python.org/projects/python/trunk/Objects/listobject.c
The underlying techniques are described in this paper (and may have
even earlier origins):
 "Optimistic Sorting and Information Theoretic Complexity"
 Peter McIlroy
 SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms),
 pp 467-474, Austin, Texas, 25-27 January 1993.
While the API to this class consists solely of static methods, it is
(privately) instantiable; a TimSort instance holds the state of an ongoing
sort, assuming the input array is large enough to warrant the full-blown
TimSort. Small arrays are sorted in place, using a binary insertion sort.
/**
 * A stable, adaptive, iterative mergesort that requires far fewer than
 * n lg(n) comparisons when running on partially sorted arrays, while
 * offering performance comparable to a traditional mergesort when run
 * on random arrays.  Like all proper mergesorts, this sort is stable and
 * runs O(n log n) time (worst case).  In the worst case, this sort requires
 * temporary storage space for n/2 object references; in the best case,
 * it requires only a small constant amount of space.
 *
 * This implementation was adapted from Tim Peters's list sort for
 * Python, which is described in detail here:
 *
 *   http://svn.python.org/projects/python/trunk/Objects/listsort.txt
 *
 * Tim's C code may be found here:
 *
 *   http://svn.python.org/projects/python/trunk/Objects/listobject.c
 *
 * The underlying techniques are described in this paper (and may have
 * even earlier origins):
 *
 *  "Optimistic Sorting and Information Theoretic Complexity"
 *  Peter McIlroy
 *  SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms),
 *  pp 467-474, Austin, Texas, 25-27 January 1993.
 *
 * While the API to this class consists solely of static methods, it is
 * (privately) instantiable; a TimSort instance holds the state of an ongoing
 * sort, assuming the input array is large enough to warrant the full-blown
 * TimSort. Small arrays are sorted in place, using a binary insertion sort.
 */
public final class LongTimSort {
    
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 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. /**
     * 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.
/**
     * The array being sorted.
     */
    private final long[] a;
    
The comparator for this sort.
/**
     * The comparator for this sort.
     */
    private final LongComparator c;
    
When we get into galloping mode, we stay there until both runs win less
often than MIN_GALLOP consecutive times.
/**
     * 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.
/**
     * 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.
/**
     * 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.
/**
     * Temp storage for merges.
     */
    private long[] 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.
/**
     * 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;
    
Asserts have been placed in if-statements for performace. To enable them,
set this field to true and enable them in VM with a command line flag.
If you modify this class, please do test the asserts!
/**
     * Asserts have been placed in if-statements for performace. To enable them,
     * set this field to true and enable them in VM with a command line flag.
     * If you modify this class, please do test the asserts!
     */
    private static final boolean DEBUG = false;
    
Creates a TimSort instance to maintain the state of an ongoing sort.
Params:
a – the array to be sorted
c – the comparator to determine the order of the sort/**
     * Creates a TimSort instance to maintain the state of an ongoing sort.
     *
     * @param a the array to be sorted
     * @param c the comparator to determine the order of the sort
     */
    private LongTimSort(long[] a, LongComparator c) {
        this.a = a;
        this.c = c;
        // Allocate temp storage (which may be increased later if necessary)
        int len = a.length;
        @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
        long[] newArray = new long[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.
     */
    public static void sort(long[] a, LongComparator c) {
        sort(a, 0, a.length, c);
    }
    public static void sort(long[] a, int lo, int hi, LongComparator c) {
        if (c == null) {
            Arrays.sort(a, lo, hi);
            return;
        }
        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, c);
            binarySort(a, lo, hi, lo + initRunLen, c);
            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.
         */
        LongTimSort ts = new LongTimSort(a, c);
        int minRun = minRunLength(nRemaining);
        do {
            // Identify next run
            int runLen = countRunAndMakeAscending(a, lo, hi, c);
            // 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, c);
                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
        if (DEBUG) assert lo == hi;
        ts.mergeForceCollapse();
        if (DEBUG) 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 lo, inclusive, to start, exclusive are already sorted. 
Params:
a – the array in which a range is to be sorted
lo – the index of the first element in the range to be sorted
hi – the index after the last element in the range to be sorted
start – the index of the first element in the range that is
       not already known to be sorted (@code lo <= start <= hi}
c – comparator to used for the sort/**
     * 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}
     * @param c comparator to used for the sort
     */
    @SuppressWarnings("fallthrough")
    private static void binarySort(long[] a, int lo, int hi, int start,
                                       LongComparator c) {
        if (DEBUG) assert lo <= start && start <= hi;
        if (start == lo)
            start++;
        for ( ; start < hi; start++) {
            long pivot = a[start];
            // Set left (and right) to the index where a[start] (pivot) belongs
            int left = lo;
            int right = start;
            if (DEBUG) assert left <= right;
            /*
             * Invariants:
             *   pivot >= all in [lo, left).
             *   pivot <  all in [right, start).
             */
            while (left < right) {
                int mid = (left + right) >>> 1;
                if (c.compare(pivot, a[mid]) < 0)
                    right = mid;
                else
                    left = mid + 1;
            }
            if (DEBUG) 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.
Params:
a – the array in which a run is to be counted and possibly reversed
lo – index of the first element in the run
hi – index after the last element that may be contained in the run.
It is required that @code{lo < hi}.
c – the comparator to used for the sort
Returns:
 the length of the run beginning at the specified position in
         the specified array/**
     * 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}.
     * @param c the comparator to used for the sort
     * @return  the length of the run beginning at the specified position in
     *          the specified array
     */
    private static int countRunAndMakeAscending(long[] a, int lo, int hi,
                                                LongComparator c) {
        if (DEBUG) assert lo < hi;
        int runHi = lo + 1;
        if (runHi == hi)
            return 1;
        // Find end of run, and reverse range if descending
        if (c.compare(a[runHi++], a[lo]) < 0) { // Descending
            while(runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
                runHi++;
            reverseRange(a, lo, runHi);
        } else {                              // Ascending
            while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
                runHi++;
        }
        return runHi - lo;
    }
    
Reverse the specified range of the specified array.
Params:
a – the array in which a range is to be reversed
lo – the index of the first element in the range to be reversed
hi – the index after the last element in the range to be reversed/**
     * 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(long[] a, int lo, int hi) {
        hi--;
        while (lo < hi) {
            long 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 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. 
Params:
n – the length of the array to be sorted
Returns:
the length of the minimum run to be merged/**
     * 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) {
        if (DEBUG) 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.
Params:
runBase – index of the first element in the run
runLen –  the number of elements in the run/**
     * 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.
/**
     * 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.
/**
     * 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.
Params:
i – stack index of the first of the two runs to merge/**
     * 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
     */
    private void mergeAt(int i) {
        if (DEBUG) assert stackSize >= 2;
        if (DEBUG) assert i >= 0;
        if (DEBUG) 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];
        if (DEBUG) assert len1 > 0 && len2 > 0;
        if (DEBUG) 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(a[base2], a, base1, len1, 0, c);
        if (DEBUG) 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(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
        if (DEBUG) 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.
Params:
key – the key whose insertion point to search for
a – the array in which to search
base – the index of the first element in the range
len – the length of the range; must be > 0
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.
c – the comparator used to order the range, and to search
Returns:
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./**
     * 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.
     * @param c the comparator used to order the range, and to search
     * @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(long key, long[] a, int base, int len, int hint,
                                  LongComparator c) {
        if (DEBUG) assert len > 0 && hint >= 0 && hint < len;
        int lastOfs = 0;
        int ofs = 1;
        if (c.compare(key, a[base + hint]) > 0) {
            // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
            int maxOfs = len - hint;
            while (ofs < maxOfs && c.compare(key, 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 && c.compare(key, 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;
        }
        if (DEBUG) 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 (c.compare(key, a[base + m]) > 0)
                lastOfs = m + 1;  // a[base + m] < key
            else
                ofs = m;          // key <= a[base + m]
        }
        if (DEBUG) assert lastOfs == ofs;    // so a[base + ofs - 1] < key <= a[base + ofs]
        return ofs;
    }
    
Like gallopLeft, except that if the range contains an element equal to
key, gallopRight returns the index after the rightmost equal element.
Params:
key – the key whose insertion point to search for
a – the array in which to search
base – the index of the first element in the range
len – the length of the range; must be > 0
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.
c – the comparator used to order the range, and to search
Returns:
the int k,  0 <= k <= n such that a[b + k - 1] <= key < a[b + k]/**
     * Like gallopLeft, except that if the range contains an element equal to
     * key, gallopRight returns the index after the rightmost 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.
     * @param c the comparator used to order the range, and to search
     * @return the int k,  0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
     */
    private static int gallopRight(long key, long[] a, int base, int len,
                                       int hint, LongComparator c) {
        if (DEBUG) assert len > 0 && hint >= 0 && hint < len;
        int ofs = 1;
        int lastOfs = 0;
        if (c.compare(key, a[base + hint]) < 0) {
            // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
            int maxOfs = hint + 1;
            while (ofs < maxOfs && c.compare(key, 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 b
            int tmp = lastOfs;
            lastOfs = hint - ofs;
            ofs = hint - tmp;
        } else { // a[b + hint] <= key
            // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
            int maxOfs = len - hint;
            while (ofs < maxOfs && c.compare(key, 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 b
            lastOfs += hint;
            ofs += hint;
        }
        if (DEBUG) assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
        /*
         * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
         * the right of lastOfs but no farther right than ofs.  Do a binary
         * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
         */
        lastOfs++;
        while (lastOfs < ofs) {
            int m = lastOfs + ((ofs - lastOfs) >>> 1);
            if (c.compare(key, a[base + m]) < 0)
                ofs = m;          // key < a[b + m]
            else
                lastOfs = m + 1;  // a[b + m] <= key
        }
        if (DEBUG) assert lastOfs == ofs;    // so a[b + ofs - 1] <= key < a[b + ofs]
        return ofs;
    }
    
Merges two adjacent runs in place, in a stable fashion.  The first
element of the first run must be greater than the first element of the
second run (a[base1] > a[base2]), and the last element of the first run
(a[base1 + len1-1]) must be greater than all elements of the second run.
For performance, this method should be called only when len1 <= len2;
its twin, mergeHi should be called if len1 >= len2.  (Either method
may be called if len1 == len2.)
Params:
base1 – index of first element in first run to be merged
len1 –  length of first run to be merged (must be > 0)
base2 – index of first element in second run to be merged
       (must be aBase + aLen)
len2 –  length of second run to be merged (must be > 0)/**
     * Merges two adjacent runs in place, in a stable fashion.  The first
     * element of the first run must be greater than the first element of the
     * second run (a[base1] > a[base2]), and the last element of the first run
     * (a[base1 + len1-1]) must be greater than all elements of the second run.
     *
     * For performance, this method should be called only when len1 <= len2;
     * its twin, mergeHi should be called if len1 >= len2.  (Either method
     * may be called if len1 == len2.)
     *
     * @param base1 index of first element in first run to be merged
     * @param len1  length of first run to be merged (must be > 0)
     * @param base2 index of first element in second run to be merged
     *        (must be aBase + aLen)
     * @param len2  length of second run to be merged (must be > 0)
     */
    private void mergeLo(int base1, int len1, int base2, int len2) {
        if (DEBUG) assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
        // Copy first run into temp array
        long[] a = this.a; // For performance
        long[] tmp = ensureCapacity(len1);
        System.arraycopy(a, base1, tmp, 0, len1);
        int cursor1 = 0;       // Indexes into tmp array
        int cursor2 = base2;   // Indexes int a
        int dest = base1;      // Indexes int a
        // Move first element of second run and deal with degenerate cases
        a[dest++] = a[cursor2++];
        if (--len2 == 0) {
            System.arraycopy(tmp, cursor1, a, dest, len1);
            return;
        }
        if (len1 == 1) {
            System.arraycopy(a, cursor2, a, dest, len2);
            a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
            return;
        }
        LongComparator c = this.c;  // Use local variable for performance
        int minGallop = this.minGallop;    //  "    "       "     "      "
        outer:
        while (true) {
            int count1 = 0; // Number of times in a row that first run won
            int count2 = 0; // Number of times in a row that second run won
            /*
             * Do the straightforward thing until (if ever) one run starts
             * winning consistently.
             */
            do {
                if (DEBUG) assert len1 > 1 && len2 > 0;
                if (c.compare(a[cursor2], tmp[cursor1]) < 0) {
                    a[dest++] = a[cursor2++];
                    count2++;
                    count1 = 0;
                    if (--len2 == 0)
                        break outer;
                } else {
                    a[dest++] = tmp[cursor1++];
                    count1++;
                    count2 = 0;
                    if (--len1 == 1)
                        break outer;
                }
            } while ((count1 | count2) < minGallop);
            /*
             * One run is winning so consistently that galloping may be a
             * huge win. So try that, and continue galloping until (if ever)
             * neither run appears to be winning consistently anymore.
             */
            do {
                if (DEBUG) assert len1 > 1 && len2 > 0;
                count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
                if (count1 != 0) {
                    System.arraycopy(tmp, cursor1, a, dest, count1);
                    dest += count1;
                    cursor1 += count1;
                    len1 -= count1;
                    if (len1 <= 1) // len1 == 1 || len1 == 0
                        break outer;
                }
                a[dest++] = a[cursor2++];
                if (--len2 == 0)
                    break outer;
                count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
                if (count2 != 0) {
                    System.arraycopy(a, cursor2, a, dest, count2);
                    dest += count2;
                    cursor2 += count2;
                    len2 -= count2;
                    if (len2 == 0)
                        break outer;
                }
                a[dest++] = tmp[cursor1++];
                if (--len1 == 1)
                    break outer;
                minGallop--;
            } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
            if (minGallop < 0)
                minGallop = 0;
            minGallop += 2;  // Penalize for leaving gallop mode
        }  // End of "outer" loop
        this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field
        if (len1 == 1) {
            if (DEBUG) assert len2 > 0;
            System.arraycopy(a, cursor2, a, dest, len2);
            a[dest + len2] = tmp[cursor1]; //  Last elt of run 1 to end of merge
        } else if (len1 == 0) {
            throw new IllegalArgumentException(
                                              "Comparison method violates its general contract!");
        } else {
            if (DEBUG) assert len2 == 0;
            if (DEBUG) assert len1 > 1;
            System.arraycopy(tmp, cursor1, a, dest, len1);
        }
    }
    
Like mergeLo, except that this method should be called only if
len1 >= len2; mergeLo should be called if len1 <= len2.  (Either method
may be called if len1 == len2.)
Params:
base1 – index of first element in first run to be merged
len1 –  length of first run to be merged (must be > 0)
base2 – index of first element in second run to be merged
       (must be aBase + aLen)
len2 –  length of second run to be merged (must be > 0)/**
     * Like mergeLo, except that this method should be called only if
     * len1 >= len2; mergeLo should be called if len1 <= len2.  (Either method
     * may be called if len1 == len2.)
     *
     * @param base1 index of first element in first run to be merged
     * @param len1  length of first run to be merged (must be > 0)
     * @param base2 index of first element in second run to be merged
     *        (must be aBase + aLen)
     * @param len2  length of second run to be merged (must be > 0)
     */
    private void mergeHi(int base1, int len1, int base2, int len2) {
        if (DEBUG) assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
        // Copy second run into temp array
        long[] a = this.a; // For performance
        long[] tmp = ensureCapacity(len2);
        System.arraycopy(a, base2, tmp, 0, len2);
        int cursor1 = base1 + len1 - 1;  // Indexes into a
        int cursor2 = len2 - 1;          // Indexes into tmp array
        int dest = base2 + len2 - 1;     // Indexes into a
        // Move last element of first run and deal with degenerate cases
        a[dest--] = a[cursor1--];
        if (--len1 == 0) {
            System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
            return;
        }
        if (len2 == 1) {
            dest -= len1;
            cursor1 -= len1;
            System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
            a[dest] = tmp[cursor2];
            return;
        }
        LongComparator c = this.c;  // Use local variable for performance
        int minGallop = this.minGallop;    //  "    "       "     "      "
        outer:
        while (true) {
            int count1 = 0; // Number of times in a row that first run won
            int count2 = 0; // Number of times in a row that second run won
            /*
             * Do the straightforward thing until (if ever) one run
             * appears to win consistently.
             */
            do {
                if (DEBUG) assert len1 > 0 && len2 > 1;
                if (c.compare(tmp[cursor2], a[cursor1]) < 0) {
                    a[dest--] = a[cursor1--];
                    count1++;
                    count2 = 0;
                    if (--len1 == 0)
                        break outer;
                } else {
                    a[dest--] = tmp[cursor2--];
                    count2++;
                    count1 = 0;
                    if (--len2 == 1)
                        break outer;
                }
            } while ((count1 | count2) < minGallop);
            /*
             * One run is winning so consistently that galloping may be a
             * huge win. So try that, and continue galloping until (if ever)
             * neither run appears to be winning consistently anymore.
             */
            do {
                if (DEBUG) assert len1 > 0 && len2 > 1;
                count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
                if (count1 != 0) {
                    dest -= count1;
                    cursor1 -= count1;
                    len1 -= count1;
                    System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
                    if (len1 == 0)
                        break outer;
                }
                a[dest--] = tmp[cursor2--];
                if (--len2 == 1)
                    break outer;
                count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c);
                if (count2 != 0) {
                    dest -= count2;
                    cursor2 -= count2;
                    len2 -= count2;
                    System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
                    if (len2 <= 1)  // len2 == 1 || len2 == 0
                        break outer;
                }
                a[dest--] = a[cursor1--];
                if (--len1 == 0)
                    break outer;
                minGallop--;
            } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
            if (minGallop < 0)
                minGallop = 0;
            minGallop += 2;  // Penalize for leaving gallop mode
        }  // End of "outer" loop
        this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field
        if (len2 == 1) {
            if (DEBUG) assert len1 > 0;
            dest -= len1;
            cursor1 -= len1;
            System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
            a[dest] = tmp[cursor2];  // Move first elt of run2 to front of merge
        } else if (len2 == 0) {
            throw new IllegalArgumentException(
                                              "Comparison method violates its general contract!");
        } else {
            if (DEBUG) assert len1 == 0;
            if (DEBUG) assert len2 > 0;
            System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
        }
    }
    
Ensures that the external array tmp has at least the specified
number of elements, increasing its size if necessary.  The size
increases exponentially to ensure amortized linear time complexity.
Params:
minCapacity – the minimum required capacity of the tmp array
Returns:
tmp, whether or not it grew/**
     * Ensures that the external array tmp has at least the specified
     * number of elements, increasing its size if necessary.  The size
     * increases exponentially to ensure amortized linear time complexity.
     *
     * @param minCapacity the minimum required capacity of the tmp array
     * @return tmp, whether or not it grew
     */
    private long[] ensureCapacity(int minCapacity) {
        if (tmp.length < minCapacity) {
            // Compute smallest power of 2 > minCapacity
            int newSize = minCapacity;
            newSize |= newSize >> 1;
            newSize |= newSize >> 2;
            newSize |= newSize >> 4;
            newSize |= newSize >> 8;
            newSize |= newSize >> 16;
            newSize++;
            if (newSize < 0) // Not bloody likely!
                newSize = minCapacity;
            else
                newSize = Math.min(newSize, a.length >>> 1);
            @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
            long[] newArray = new long[newSize];
            tmp = newArray;
        }
        return tmp;
    }
    
Checks that fromIndex and toIndex are in range, and throws an
appropriate exception if they aren't.
Params:
arrayLen – the length of the array
fromIndex – the index of the first element of the range
toIndex – the index after the last element of the range
Throws:
IllegalArgumentException – if fromIndex > toIndex
ArrayIndexOutOfBoundsException – if fromIndex < 0
        or toIndex > arrayLen/**
     * Checks that fromIndex and toIndex are in range, and throws an
     * appropriate exception if they aren't.
     *
     * @param arrayLen the length of the array
     * @param fromIndex the index of the first element of the range
     * @param toIndex the index after the last element of the range
     * @throws IllegalArgumentException if fromIndex > toIndex
     * @throws ArrayIndexOutOfBoundsException if fromIndex < 0
     *         or toIndex > arrayLen
     */
    private static void rangeCheck(int arrayLen, int fromIndex, int toIndex) {
        if (fromIndex > toIndex)
            throw new IllegalArgumentException("fromIndex(" + fromIndex +
                                               ") > toIndex(" + toIndex+")");
        if (fromIndex < 0)
            throw new ArrayIndexOutOfBoundsException(fromIndex);
        if (toIndex > arrayLen)
            throw new ArrayIndexOutOfBoundsException(toIndex);
    }
    
    // addition to original file
    
    @FunctionalInterface
    public static interface LongComparator
    {
        int compare(long o1, long o2);
    }
}