content: /*
   * 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.
   */
  
  package org.apache.carbondata.processing.loading.sort.unsafe.sort;
  
  import java.util.Comparator;
  
  /**
   * A port of the Apache Spark's TimSort and they originally ported from 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 SortDataFormat
   * @see org.apache.spark.util.collection.Sorter
   */
  public 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.
     */
    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);
            s.copyElement(a, left, a, left + 1);
            break;
          }
          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.
         */
        int stackLen = (len <    120  ?  5 :
                        len <   1542  ? 10 :
                        len < 119151  ? 19 : 40);
        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.
       */
      private void mergeCollapse() {
        while (stackSize > 1) {
          int n = stackSize - 2;
          if ((n >= 1 && runLen[n - 1] <= runLen[n] + runLen[n + 1])
              || (n >= 2 && runLen[n - 2] <= runLen[n] + runLen[n - 1])) {
            if (runLen[n - 1] < runLen[n + 1]) {
              n--;
            }
          } else if (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;
      }
    }
  }
  