/*
 * Copyright (c) 2009, 2013, Oracle and/or its affiliates. All rights reserved.
 * Copyright 2009 Google Inc. All Rights Reserved.
 * ORACLE PROPRIETARY/CONFIDENTIAL. Use is subject to license terms.
 *
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 *
 */

package java.util;

/**
 * This is a near duplicate of {@link TimSort}, modified for use with
 * arrays of objects that implement {@link Comparable}, instead of using
 * explicit comparators.
 *
 * <p>If you are using an optimizing VM, you may find that ComparableTimSort
 * offers no performance benefit over TimSort in conjunction with a
 * comparator that simply returns {@code ((Comparable)first).compareTo(Second)}.
 * If this is the case, you are better off deleting ComparableTimSort to
 * eliminate the code duplication. (See Arrays.java for details.)
 *
 * @author Josh Bloch
 */
class ComparableTimSort {
 /**
 * This is the minimum sized sequence that will be merged. Shorter
 * sequences will be lengthened by calling binarySort. If the entire
 * array is less than this length, no merges will be performed.
 *
 * This constant should be a power of two. It was 64 in Tim Peter's C
 * implementation, but 32 was empirically determined to work better in
 * this implementation. In the unlikely event that you set this constant
 * to be a number that's not a power of two, you'll need to change the
 * {@link #minRunLength} computation.
 *
 * If you decrease this constant, you must change the stackLen
 * computation in the TimSort constructor, or you risk an
 * ArrayOutOfBounds exception. See listsort.txt for a discussion
 * of the minimum stack length required as a function of the length
 * of the array being sorted and the minimum merge sequence length.
 */
 private static final int MIN_MERGE = 32;

 /**
 * The array being sorted.
 */
 private final Object[] a;

 /**
 * When we get into galloping mode, we stay there until both runs win less
 * often than MIN_GALLOP consecutive times.
 */
 private static final int MIN_GALLOP = 7;

 /**
 * This controls when we get *into* galloping mode. It is initialized
 * to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for
 * random data, and lower for highly structured data.
 */
 private int minGallop = MIN_GALLOP;

 /**
 * Maximum initial size of tmp array, which is used for merging. The array
 * can grow to accommodate demand.
 *
 * Unlike Tim's original C version, we do not allocate this much storage
 * when sorting smaller arrays. This change was required for performance.
 */
 private static final int INITIAL_TMP_STORAGE_LENGTH = 256;

 /**
 * Temp storage for merges. A workspace array may optionally be
 * provided in constructor, and if so will be used as long as it
 * is big enough.
 */
 private Object[] tmp;
 private int tmpBase; // base of tmp array slice
 private int tmpLen; // length of tmp array slice

 /**
 * 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 work a workspace array (slice)
 * @param workBase origin of usable space in work array
 * @param workLen usable size of work array
 */
 private ComparableTimSort(Object[] a, Object[] work, int workBase, int workLen) {
 this.a = a;

 // Allocate temp storage (which may be increased later if necessary)
 int len = a.length;
 int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ?
 len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;
 if (work == null || workLen < tlen || workBase + tlen > work.length) {
 tmp = new Object[tlen];
 tmpBase = 0;
 tmpLen = tlen;
 }
 else {
 tmp = work;
 tmpBase = workBase;
 tmpLen = workLen;
 }

 /*
 * 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];
 }

 /*
 * The next method (package private and static) constitutes the
 * entire API of this class.
 */

 /**
 * Sorts the given range, using the given workspace array slice
 * for temp storage when possible. This method is designed to be
 * invoked from public methods (in class Arrays) after performing
 * any necessary array bounds checks and expanding parameters into
 * the required forms.
 *
 * @param a the array to be sorted
 * @param lo the index of the first element, inclusive, to be sorted
 * @param hi the index of the last element, exclusive, to be sorted
 * @param work a workspace array (slice)
 * @param workBase origin of usable space in work array
 * @param workLen usable size of work array
 * @since 1.8
 */
 static void sort(Object[] a, int lo, int hi, Object[] work, int workBase, int workLen) {
 assert a != null && lo >= 0 && lo <= hi && hi <= a.length;

 int nRemaining = hi - lo;
 if (nRemaining < 2)
 return; // Arrays of size 0 and 1 are always sorted

 // If array is small, do a "mini-TimSort" with no merges
 if (nRemaining < MIN_MERGE) {
 int initRunLen = countRunAndMakeAscending(a, lo, hi);
 binarySort(a, lo, hi, lo + initRunLen);
 return;
 }

 /**
 * March over the array once, left to right, finding natural runs,
 * extending short natural runs to minRun elements, and merging runs
 * to maintain stack invariant.
 */
 ComparableTimSort ts = new ComparableTimSort(a, work, workBase, workLen);
 int minRun = minRunLength(nRemaining);
 do {
 // Identify next run
 int runLen = countRunAndMakeAscending(a, lo, hi);

 // If run is short, extend to min(minRun, nRemaining)
 if (runLen < minRun) {
 int force = nRemaining <= minRun ? nRemaining : minRun;
 binarySort(a, lo, lo + force, lo + runLen);
 runLen = force;
 }

 // Push run onto pending-run stack, and maybe merge
 ts.pushRun(lo, runLen);
 ts.mergeCollapse();

 // Advance to find next run
 lo += runLen;
 nRemaining -= runLen;
 } while (nRemaining != 0);

 // Merge all remaining runs to complete sort
 assert lo == hi;
 ts.mergeForceCollapse();
 assert ts.stackSize == 1;
 }

 /**
 * Sorts the specified portion of the specified array using a binary
 * insertion sort. This is the best method for sorting small numbers
 * of elements. It requires O(n log n) compares, but O(n^2) data
 * movement (worst case).
 *
 * If the initial part of the specified range is already sorted,
 * this method can take advantage of it: the method assumes that the
 * elements from index {@code lo}, inclusive, to {@code start},
 * exclusive are already sorted.
 *
 * @param a the array in which a range is to be sorted
 * @param lo the index of the first element in the range to be sorted
 * @param hi the index after the last element in the range to be sorted
 * @param start the index of the first element in the range that is
 * not already known to be sorted ({@code lo <= start <= hi})
 */
 @SuppressWarnings({"fallthrough", "rawtypes", "unchecked"})
 private static void binarySort(Object[] a, int lo, int hi, int start) {
 assert lo <= start && start <= hi;
 if (start == lo)
 start++;
 for ( ; start < hi; start++) {
 Comparable pivot = (Comparable) a[start];

 // Set left (and right) to the index where a[start] (pivot) belongs
 int left = lo;
 int right = start;
 assert left <= right;
 /*
 * Invariants:
 * pivot >= all in [lo, left).
 * pivot < all in [right, start).
 */
 while (left < right) {
 int mid = (left + right) >>> 1;
 if (pivot.compareTo(a[mid]) < 0)
 right = mid;
 else
 left = mid + 1;
 }
 assert left == right;

 /*
 * The invariants still hold: pivot >= all in [lo, left) and
 * pivot < all in [left, start), so pivot belongs at left. Note
 * that if there are elements equal to pivot, left points to the
 * first slot after them -- that's why this sort is stable.
 * Slide elements over to make room for pivot.
 */
 int n = start - left; // The number of elements to move
 // Switch is just an optimization for arraycopy in default case
 switch (n) {
 case 2: a[left + 2] = a[left + 1];
 case 1: a[left + 1] = a[left];
 break;
 default: System.arraycopy(a, left, a, left + 1, n);
 }
 a[left] = pivot;
 }
 }

 /**
 * Returns the length of the run beginning at the specified position in
 * the specified array and reverses the run if it is descending (ensuring
 * that the run will always be ascending when the method returns).
 *
 * A run is the longest ascending sequence with:
 *
 * a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
 *
 * or the longest descending sequence with:
 *
 * a[lo] > a[lo + 1] > a[lo + 2] > ...
 *
 * For its intended use in a stable mergesort, the strictness of the
 * definition of "descending" is needed so that the call can safely
 * reverse a descending sequence without violating stability.
 *
 * @param a the array in which a run is to be counted and possibly reversed
 * @param lo index of the first element in the run
 * @param hi index after the last element that may be contained in the run.
 * It is required that {@code lo < hi}.
 * @return the length of the run beginning at the specified position in
 * the specified array
 */
 @SuppressWarnings({"unchecked", "rawtypes"})
 private static int countRunAndMakeAscending(Object[] a, int lo, int hi) {
 assert lo < hi;
 int runHi = lo + 1;
 if (runHi == hi)
 return 1;

 // Find end of run, and reverse range if descending
 if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending
 while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0)
 runHi++;
 reverseRange(a, lo, runHi);
 } else { // Ascending
 while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) >= 0)
 runHi++;
 }

 return runHi - lo;
 }

 /**
 * Reverse the specified range of the specified array.
 *
 * @param a the array in which a range is to be reversed
 * @param lo the index of the first element in the range to be reversed
 * @param hi the index after the last element in the range to be reversed
 */
 private static void reverseRange(Object[] a, int lo, int hi) {
 hi--;
 while (lo < hi) {
 Object t = a[lo];
 a[lo++] = a[hi];
 a[hi--] = t;
 }
 }

 /**
 * Returns the minimum acceptable run length for an array of the specified
 * length. Natural runs shorter than this will be extended with
 * {@link #binarySort}.
 *
 * Roughly speaking, the computation is:
 *
 * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
 * Else if n is an exact power of 2, return MIN_MERGE/2.
 * Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
 * is close to, but strictly less than, an exact power of 2.
 *
 * For the rationale, see listsort.txt.
 *
 * @param n the length of the array to be sorted
 * @return the length of the minimum run to be merged
 */
 private static int minRunLength(int n) {
 assert n >= 0;
 int r = 0; // Becomes 1 if any 1 bits are shifted off
 while (n >= MIN_MERGE) {
 r |= (n & 1);
 n >>= 1;
 }
 return n + r;
 }

 /**
 * Pushes the specified run onto the pending-run stack.
 *
 * @param runBase index of the first element in the run
 * @param runLen the number of elements in the run
 */
 private void pushRun(int runBase, int runLen) {
 this.runBase[stackSize] = runBase;
 this.runLen[stackSize] = runLen;
 stackSize++;
 }

 /**
 * Examines the stack of runs waiting to be merged and merges adjacent runs
 * until the stack invariants are reestablished:
 *
 * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
 * 2. runLen[i - 2] > runLen[i - 1]
 *
 * This method is called each time a new run is pushed onto the stack,
 * so the invariants are guaranteed to hold for i < stackSize upon
 * entry to the method.
 *
 * 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
 */
 @SuppressWarnings("unchecked")
 private void mergeAt(int i) {
 assert stackSize >= 2;
 assert i >= 0;
 assert i == stackSize - 2 || i == stackSize - 3;

 int base1 = runBase[i];
 int len1 = runLen[i];
 int base2 = runBase[i + 1];
 int len2 = runLen[i + 1];
 assert len1 > 0 && len2 > 0;
 assert base1 + len1 == base2;

 /*
 * Record the length of the combined runs; if i is the 3rd-last
 * run now, also slide over the last run (which isn't involved
 * in this merge). The current run (i+1) goes away in any case.
 */
 runLen[i] = len1 + len2;
 if (i == stackSize - 3) {
 runBase[i + 1] = runBase[i + 2];
 runLen[i + 1] = runLen[i + 2];
 }
 stackSize--;

 /*
 * Find where the first element of run2 goes in run1. Prior elements
 * in run1 can be ignored (because they're already in place).
 */
 int k = gallopRight((Comparable<Object>) a[base2], a, base1, len1, 0);
 assert k >= 0;
 base1 += k;
 len1 -= k;
 if (len1 == 0)
 return;

 /*
 * Find where the last element of run1 goes in run2. Subsequent elements
 * in run2 can be ignored (because they're already in place).
 */
 len2 = gallopLeft((Comparable<Object>) a[base1 + len1 - 1], a,
 base2, len2, len2 - 1);
 assert len2 >= 0;
 if (len2 == 0)
 return;

 // Merge remaining runs, using tmp array with min(len1, len2) elements
 if (len1 <= len2)
 mergeLo(base1, len1, base2, len2);
 else
 mergeHi(base1, len1, base2, len2);
 }

 /**
 * Locates the position at which to insert the specified key into the
 * specified sorted range; if the range contains an element equal to key,
 * returns the index of the leftmost equal element.
 *
 * @param key the key whose insertion point to search for
 * @param a the array in which to search
 * @param base the index of the first element in the range
 * @param len the length of the range; must be > 0
 * @param hint the index at which to begin the search, 0 <= hint < n.
 * The closer hint is to the result, the faster this method will run.
 * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
 * pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
 * In other words, key belongs at index b + k; or in other words,
 * the first k elements of a should precede key, and the last n - k
 * should follow it.
 */
 private static int gallopLeft(Comparable<Object> key, Object[] a,
 int base, int len, int hint) {
 assert len > 0 && hint >= 0 && hint < len;

 int lastOfs = 0;
 int ofs = 1;
 if (key.compareTo(a[base + hint]) > 0) {
 // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
 int maxOfs = len - hint;
 while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) > 0) {
 lastOfs = ofs;
 ofs = (ofs << 1) + 1;
 if (ofs <= 0) // int overflow
 ofs = maxOfs;
 }
 if (ofs > maxOfs)
 ofs = maxOfs;

 // Make offsets relative to base
 lastOfs += hint;
 ofs += hint;
 } else { // key <= a[base + hint]
 // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
 final int maxOfs = hint + 1;
 while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) <= 0) {
 lastOfs = ofs;
 ofs = (ofs << 1) + 1;
 if (ofs <= 0) // int overflow
 ofs = maxOfs;
 }
 if (ofs > maxOfs)
 ofs = maxOfs;

 // Make offsets relative to base
 int tmp = lastOfs;
 lastOfs = hint - ofs;
 ofs = hint - tmp;
 }
 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

 /*
 * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
 * to the right of lastOfs but no farther right than ofs. Do a binary
 * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
 */
 lastOfs++;
 while (lastOfs < ofs) {
 int m = lastOfs + ((ofs - lastOfs) >>> 1);

 if (key.compareTo(a[base + m]) > 0)
 lastOfs = m + 1; // a[base + m] < key
 else
 ofs = m; // key <= a[base + m]
 }
 assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
 return ofs;
 }

 /**
 * 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.
 * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
 */
 private static int gallopRight(Comparable<Object> key, Object[] a,
 int base, int len, int hint) {
 assert len > 0 && hint >= 0 && hint < len;

 int ofs = 1;
 int lastOfs = 0;
 if (key.compareTo(a[base + hint]) < 0) {
 // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
 int maxOfs = hint + 1;
 while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) < 0) {
 lastOfs = ofs;
 ofs = (ofs << 1) + 1;
 if (ofs <= 0) // int overflow
 ofs = maxOfs;
 }
 if (ofs > maxOfs)
 ofs = maxOfs;

 // Make offsets relative to 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 && key.compareTo(a[base + hint + ofs]) >= 0) {
 lastOfs = ofs;
 ofs = (ofs << 1) + 1;
 if (ofs <= 0) // int overflow
 ofs = maxOfs;
 }
 if (ofs > maxOfs)
 ofs = maxOfs;

 // Make offsets relative to 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 (key.compareTo(a[base + m]) < 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)
 */
 @SuppressWarnings({"unchecked", "rawtypes"})
 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
 Object[] a = this.a; // For performance
 Object[] tmp = ensureCapacity(len1);

 int cursor1 = tmpBase; // Indexes into tmp array
 int cursor2 = base2; // Indexes int a
 int dest = base1; // Indexes int a
 System.arraycopy(a, base1, tmp, cursor1, len1);

 // 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;
 }

 int minGallop = this.minGallop; // Use local variable for performance
 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 (((Comparable) a[cursor2]).compareTo(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 {
 assert len1 > 1 && len2 > 0;
 count1 = gallopRight((Comparable) a[cursor2], tmp, cursor1, len1, 0);
 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((Comparable) tmp[cursor1], a, cursor2, len2, 0);
 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) {
 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 {
 assert len2 == 0;
 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.)
 *
 * @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)
 */
 @SuppressWarnings({"unchecked", "rawtypes"})
 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
 Object[] a = this.a; // For performance
 Object[] tmp = ensureCapacity(len2);
 int tmpBase = this.tmpBase;
 System.arraycopy(a, base2, tmp, tmpBase, len2);

 int cursor1 = base1 + len1 - 1; // Indexes into a
 int cursor2 = tmpBase + 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, tmpBase, 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;
 }

 int minGallop = this.minGallop; // Use local variable for performance
 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 (((Comparable) tmp[cursor2]).compareTo(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 {
 assert len1 > 0 && len2 > 1;
 count1 = len1 - gallopRight((Comparable) tmp[cursor2], a, base1, len1, len1 - 1);
 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((Comparable) a[cursor1], tmp, tmpBase, len2, len2 - 1);
 if (count2 != 0) {
 dest -= count2;
 cursor2 -= count2;
 len2 -= count2;
 System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
 if (len2 <= 1)
 break outer; // len2 == 1 || len2 == 0
 }
 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) {
 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 {
 assert len1 == 0;
 assert len2 > 0;
 System.arraycopy(tmp, tmpBase, 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 Object[] ensureCapacity(int minCapacity) {
 if (tmpLen < minCapacity) {
 // Compute smallest power of 2 > minCapacity
 int newSize = -1 >>> Integer.numberOfLeadingZeros(minCapacity);
 newSize++;

 if (newSize < 0) // Not bloody likely!
 newSize = minCapacity;
 else
 newSize = Math.min(newSize, a.length >>> 1);

 @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
 Object[] newArray = new Object[newSize];
 tmp = newArray;
 tmpLen = newSize;
 tmpBase = 0;
 }
 return tmp;
 }

}