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 /*
  * Licensed to the Apache Software Foundation (ASF) under one or more
  * contributor license agreements.  See the NOTICE file distributed with
  * this work for additional information regarding copyright ownership.
  * The ASF licenses this file to You 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.
  */
 
 /*
  * Based on TimSort.java from the Android Open Source Project
  *
  *  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.
  */
 
 package org.apache.spark.util.collection;
 
 import java.util.Comparator;
 
 /**
  * A port of the Android TimSort class, which utilizes a "stable, adaptive, iterative mergesort."
  * See the method comment on sort() for more details.
  *
  * This has been kept in Java with the original style in order to match very closely with the
  * Android source code, and thus be easy to verify correctness. The class is package private. We put
  * a simple Scala wrapper {@link org.apache.spark.util.collection.Sorter}, which is available to
  * package org.apache.spark.
  *
  * The purpose of the port is to generalize the interface to the sort to accept input data formats
  * besides simple arrays where every element is sorted individually. For instance, the AppendOnlyMap
  * uses this to sort an Array with alternating elements of the form [key, value, key, value].
  * This generalization comes with minimal overhead -- see SortDataFormat for more information.
  *
  * We allow key reuse to prevent creating many key objects -- see SortDataFormat.
  *
  * @see org.apache.spark.util.collection.SortDataFormat
  * @see org.apache.spark.util.collection.Sorter
  */
 class TimSort<K, Buffer> {
 
   /**
    * 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.
    */
   private static final int MIN_MERGE = 32;
 
   private final SortDataFormat<K, Buffer> s;
 
   public TimSort(SortDataFormat<K, Buffer> sortDataFormat) {
     this.s = sortDataFormat;
   }
 
   /**
    * 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.
    *
    * @author Josh Bloch
    */
   public void sort(Buffer a, int lo, int hi, Comparator<? super K> c) {
     assert c != null;
 
     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.
      */
     SortState sortState = new SortState(a, c, hi - lo);
     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
       sortState.pushRun(lo, runLen);
       sortState.mergeCollapse();
 
       // Advance to find next run
       lo += runLen;
       nRemaining -= runLen;
     } while (nRemaining != 0);
 
     // Merge all remaining runs to complete sort
     assert lo == hi;
     sortState.mergeForceCollapse();
     assert sortState.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})
    * @param c comparator to used for the sort
    */
   @SuppressWarnings("fallthrough")
   private void binarySort(Buffer a, int lo, int hi, int start, Comparator<? super K> c) {
     assert lo <= start && start <= hi;
     if (start == lo)
       start++;
 
     K key0 = s.newKey();
     K key1 = s.newKey();
 
     Buffer pivotStore = s.allocate(1);
     for ( ; start < hi; start++) {
       s.copyElement(a, start, pivotStore, 0);
       K pivot = s.getKey(pivotStore, 0, key0);
 
       // 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 (c.compare(pivot, s.getKey(a, mid, key1)) < 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 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:  s.copyElement(a, left + 1, a, left + 2);
         case 1:  s.copyElement(a, left, a, left + 1);
           break;
         default: s.copyRange(a, left, a, left + 1, n);
       }
       s.copyElement(pivotStore, 0, a, left);
     }
   }
 
   /**
    * 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 int countRunAndMakeAscending(Buffer a, int lo, int hi, Comparator<? super K> c) {
     assert lo < hi;
     int runHi = lo + 1;
     if (runHi == hi)
       return 1;
 
     K key0 = s.newKey();
     K key1 = s.newKey();
 
     // Find end of run, and reverse range if descending
     if (c.compare(s.getKey(a, runHi++, key0), s.getKey(a, lo, key1)) < 0) { // Descending
       while (runHi < hi && c.compare(s.getKey(a, runHi, key0), s.getKey(a, runHi - 1, key1)) < 0)
         runHi++;
       reverseRange(a, lo, runHi);
     } else {                              // Ascending
       while (runHi < hi && c.compare(s.getKey(a, runHi, key0), s.getKey(a, runHi - 1, key1)) >= 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 void reverseRange(Buffer a, int lo, int hi) {
     hi--;
     while (lo < hi) {
       s.swap(a, lo, hi);
       lo++;
       hi--;
     }
   }
 
   /**
    * 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 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;
   }
 
   private class SortState {
 
     /**
      * The Buffer being sorted.
      */
     private final Buffer a;
 
     /**
      * Length of the sort Buffer.
      */
     private final int aLength;
 
     /**
      * The comparator for this sort.
      */
     private final Comparator<? super K> c;
 
     /**
      * 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 Buffer tmp; // Actual runtime type will be Object[], regardless of T
 
     /**
      * Length of the temp storage.
      */
     private int tmpLength = 0;
 
     /**
      * 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
      * @param c the comparator to determine the order of the sort
      */
     private SortState(Buffer a, Comparator<? super K> c, int len) {
       this.aLength = len;
       this.a = a;
       this.c = c;
 
       // Allocate temp storage (which may be increased later if necessary)
       tmpLength = len < 2 * INITIAL_TMP_STORAGE_LENGTH ? len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;
       tmp = s.allocate(tmpLength);
 
       /*
        * 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.
        * The maximum value of 49 allows for an array up to length
        * Integer.MAX_VALUE-4, if array is filled by the worst case stack size
        * increasing scenario. More explanations are given in section 4 of:
        * http://envisage-project.eu/wp-content/uploads/2015/02/sorting.pdf
        */
       int stackLen = (len <    120  ?  5 :
                       len <   1542  ? 10 :
                       len < 119151  ? 24 : 49);
       runBase = new int[stackLen];
       runLen = new int[stackLen];
     }
 
     /**
      * 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.
      *
      * Thanks to Stijn de Gouw, Jurriaan Rot, Frank S. de Boer,
      * Richard Bubel and Reiner Hahnle, this is fixed with respect to
      * the analysis in "On the Worst-Case Complexity of TimSort" by
      * Nicolas Auger, Vincent Jug, Cyril Nicaud, and Carine Pivoteau.
      */
     private void mergeCollapse() {
       while (stackSize > 1) {
         int n = stackSize - 2;
         if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1] ||
             n > 1 && runLen[n-2] <= runLen[n] + runLen[n-1]) {
           if (runLen[n - 1] < runLen[n + 1])
             n--;
         } else if (n < 0 || runLen[n] > runLen[n + 1]) {
           break; // Invariant is established
         }
         mergeAt(n);
       }
     }
 
     /**
      * 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
      */
     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--;
 
       K key0 = s.newKey();
 
       /*
        * 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(s.getKey(a, base2, key0), a, base1, len1, 0, c);
       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(s.getKey(a, base1 + len1 - 1, key0), a, base2, len2, len2 - 1, c);
       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.
      * @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 int gallopLeft(K key, Buffer a, int base, int len, int hint, Comparator<? super K> c) {
       assert len > 0 && hint >= 0 && hint < len;
       int lastOfs = 0;
       int ofs = 1;
       K key0 = s.newKey();
 
       if (c.compare(key, s.getKey(a, base + hint, key0)) > 0) {
         // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
         int maxOfs = len - hint;
         while (ofs < maxOfs && c.compare(key, s.getKey(a, base + hint + ofs, key0)) > 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, s.getKey(a, base + hint - ofs, key0)) <= 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 (c.compare(key, s.getKey(a, base + m, key0)) > 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;
     }
 
     /**
      * 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 int gallopRight(K key, Buffer a, int base, int len, int hint, Comparator<? super K> c) {
       assert len > 0 && hint >= 0 && hint < len;
 
       int ofs = 1;
       int lastOfs = 0;
       K key1 = s.newKey();
 
       if (c.compare(key, s.getKey(a, base + hint, key1)) < 0) {
         // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
         int maxOfs = hint + 1;
         while (ofs < maxOfs && c.compare(key, s.getKey(a, base + hint - ofs, key1)) < 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, s.getKey(a, base + hint + ofs, key1)) >= 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;
       }
       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, s.getKey(a, base + m, key1)) < 0)
           ofs = m;          // key < a[b + m]
         else
           lastOfs = m + 1;  // a[b + m] <= key
       }
       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.)
      *
      * @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) {
       assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
 
       // Copy first run into temp array
       Buffer a = this.a; // For performance
       Buffer tmp = ensureCapacity(len1);
       s.copyRange(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
       s.copyElement(a, cursor2++, a, dest++);
       if (--len2 == 0) {
         s.copyRange(tmp, cursor1, a, dest, len1);
         return;
       }
       if (len1 == 1) {
         s.copyRange(a, cursor2, a, dest, len2);
         s.copyElement(tmp, cursor1, a, dest + len2); // Last elt of run 1 to end of merge
         return;
       }
 
       K key0 = s.newKey();
       K key1 = s.newKey();
 
       Comparator<? super K> 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 {
           assert len1 > 1 && len2 > 0;
           if (c.compare(s.getKey(a, cursor2, key0), s.getKey(tmp, cursor1, key1)) < 0) {
             s.copyElement(a, cursor2++, a, dest++);
             count2++;
             count1 = 0;
             if (--len2 == 0)
               break outer;
           } else {
             s.copyElement(tmp, cursor1++, a, dest++);
             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 {
           assert len1 > 1 && len2 > 0;
           count1 = gallopRight(s.getKey(a, cursor2, key0), tmp, cursor1, len1, 0, c);
           if (count1 != 0) {
             s.copyRange(tmp, cursor1, a, dest, count1);
             dest += count1;
             cursor1 += count1;
             len1 -= count1;
             if (len1 <= 1) // len1 == 1 || len1 == 0
               break outer;
           }
           s.copyElement(a, cursor2++, a, dest++);
           if (--len2 == 0)
             break outer;
 
           count2 = gallopLeft(s.getKey(tmp, cursor1, key0), a, cursor2, len2, 0, c);
           if (count2 != 0) {
             s.copyRange(a, cursor2, a, dest, count2);
             dest += count2;
             cursor2 += count2;
             len2 -= count2;
             if (len2 == 0)
               break outer;
           }
           s.copyElement(tmp, cursor1++, a, dest++);
           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) {
         assert len2 > 0;
         s.copyRange(a, cursor2, a, dest, len2);
         s.copyElement(tmp, cursor1, a, dest + len2); //  Last elt of run 1 to end of merge
       } else if (len1 == 0) {
         throw new IllegalArgumentException(
             "Comparison method violates its general contract!");
       } else {
         assert len2 == 0;
         assert len1 > 1;
         s.copyRange(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.)
      *
      * @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) {
       assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
 
       // Copy second run into temp array
       Buffer a = this.a; // For performance
       Buffer tmp = ensureCapacity(len2);
       s.copyRange(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
 
       K key0 = s.newKey();
       K key1 = s.newKey();
 
       // Move last element of first run and deal with degenerate cases
       s.copyElement(a, cursor1--, a, dest--);
       if (--len1 == 0) {
         s.copyRange(tmp, 0, a, dest - (len2 - 1), len2);
         return;
       }
       if (len2 == 1) {
         dest -= len1;
         cursor1 -= len1;
         s.copyRange(a, cursor1 + 1, a, dest + 1, len1);
         s.copyElement(tmp, cursor2, a, dest);
         return;
       }
 
       Comparator<? super K> 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 {
           assert len1 > 0 && len2 > 1;
           if (c.compare(s.getKey(tmp, cursor2, key0), s.getKey(a, cursor1, key1)) < 0) {
             s.copyElement(a, cursor1--, a, dest--);
             count1++;
             count2 = 0;
             if (--len1 == 0)
               break outer;
           } else {
             s.copyElement(tmp, cursor2--, a, dest--);
             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 {
           assert len1 > 0 && len2 > 1;
           count1 = len1 - gallopRight(s.getKey(tmp, cursor2, key0), a, base1, len1, len1 - 1, c);
           if (count1 != 0) {
             dest -= count1;
             cursor1 -= count1;
             len1 -= count1;
             s.copyRange(a, cursor1 + 1, a, dest + 1, count1);
             if (len1 == 0)
               break outer;
           }
           s.copyElement(tmp, cursor2--, a, dest--);
           if (--len2 == 1)
             break outer;
 
           count2 = len2 - gallopLeft(s.getKey(a, cursor1, key0), tmp, 0, len2, len2 - 1, c);
           if (count2 != 0) {
             dest -= count2;
             cursor2 -= count2;
             len2 -= count2;
             s.copyRange(tmp, cursor2 + 1, a, dest + 1, count2);
             if (len2 <= 1)  // len2 == 1 || len2 == 0
               break outer;
           }
           s.copyElement(a, cursor1--, a, dest--);
           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) {
         assert len1 > 0;
         dest -= len1;
         cursor1 -= len1;
         s.copyRange(a, cursor1 + 1, a, dest + 1, len1);
         s.copyElement(tmp, cursor2, a, dest); // Move first elt of run2 to front of merge
       } else if (len2 == 0) {
         throw new IllegalArgumentException(
             "Comparison method violates its general contract!");
       } else {
         assert len1 == 0;
         assert len2 > 0;
         s.copyRange(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.
      *
      * @param minCapacity the minimum required capacity of the tmp array
      * @return tmp, whether or not it grew
      */
     private Buffer ensureCapacity(int minCapacity) {
       if (tmpLength < 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, aLength >>> 1);
 
         tmp = s.allocate(newSize);
         tmpLength = newSize;
       }
       return tmp;
     }
   }
 }

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