/*
 * 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 arc.struct;

import java.util.Arrays;
import java.util.Comparator;

/**
 * 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.
 * <p>
 * This implementation was adapted from Tim Peters's list sort for Python, which is described in detail here:
 * <p>
 * http://svn.python.org/projects/python/trunk/Objects/listsort.txt
 * <p>
 * Tim's C code may be found here:
 * <p>
 * http://svn.python.org/projects/python/trunk/Objects/listobject.c
 * <p>
 * The underlying techniques are described in this paper (and may have even earlier origins):
 * <p>
 * "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.
 * <p>
 * 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.
 */
class TimSort<T>{
    /**
     * 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.
     * <p>
     * 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.
     * <p>
     * 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;
    /** 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;
    /**
     * Maximum initial size of tmp array, which is used for merging. The array can grow to accommodate demand.
     * <p>
     * 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;
    /**
     * Asserts have been placed in if-statements for performance. To enable them, set this field to true and enable them in VM with
     * a command line flag. If you modify this class, please do test the asserts!
     */
    private static final boolean DEBUG = false;
    private final int[] runBase;
    private final int[] runLen;
    /** The array being sorted. */
    private T[] a;
    /** The comparator for this sort. */
    private Comparator<? super T> c;
    /**
     * 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;
    /** Temp storage for merges. */
    private T[] tmp; // Actual runtime type will be Object[], regardless of T
    private int tmpCount;
    /**
     * 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:
     * <p>
     * runBase[i] + runLen[i] == runBase[i + 1]
     * <p>
     * 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

    TimSort(){
        tmp = (T[])new Object[INITIAL_TMP_STORAGE_LENGTH];
        runBase = new int[40];
        runLen = new int[40];
    }

    /**
     * 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 TimSort(T[] a, Comparator<? super T> c){
        this.a = a;
        this.c = c;

        // Allocate temp storage (which may be increased later if necessary)
        int len = a.length;
        tmp = (T[])new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ? len >>> 1 : INITIAL_TMP_STORAGE_LENGTH];

        /*
         * 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];
    }

    static <T> void sort(T[] a, Comparator<? super T> c){
        sort(a, 0, a.length, c);
    }

    /*
     * The next two methods (which are package private and static) constitute the entire API of this class. Each of these methods
     * obeys the contract of the public method with the same signature in java.util.Arrays.
     */

    static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c){
        if(c == null){
            Arrays.sort(a, lo, hi);
            return;
        }

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

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

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

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

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

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

        // Merge all remaining runs to complete sort
        assert !DEBUG || lo == hi;
        ts.mergeForceCollapse();
        assert !DEBUG || 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).
     * <p>
     * If the initial part of the specified range is already sorted, this method can take advantage of it: the method assumes that
     * the elements from index {@code lo}, inclusive, to {@code start}, exclusive are already sorted.
     * @param a the array in which a range is to be sorted
     * @param lo the index of the first element in the range to be sorted
     * @param hi the index after the last element in the range to be sorted
     * @param start the index of the first element in the range that is not already known to be sorted (@code lo <= start <= hi}
     * @param c comparator to used for the sort
     */
    @SuppressWarnings("fallthrough")
    private static <T> void binarySort(T[] a, int lo, int hi, int start, Comparator<? super T> c){
        assert !DEBUG || lo <= start && start <= hi;
        if(start == lo) start++;
        for(; start < hi; start++){
            T pivot = a[start];

            // Set left (and right) to the index where a[start] (pivot) belongs
            int left = lo;
            int right = start;
            assert !DEBUG || left <= right;
            /*
             * Invariants: pivot >= all in [lo, left). pivot < all in [right, start).
             */
            while(left < right){
                int mid = (left + right) >>> 1;
                if(c.compare(pivot, a[mid]) < 0)
                    right = mid;
                else
                    left = mid + 1;
            }
            assert !DEBUG || 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).
     * <p>
     * A run is the longest ascending sequence with:
     * <p>
     * a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
     * <p>
     * or the longest descending sequence with:
     * <p>
     * a[lo] > a[lo + 1] > a[lo + 2] > ...
     * <p>
     * For its intended use in a stable mergesort, the strictness of the definition of "descending" is needed so that the call can
     * safely reverse a descending sequence without violating stability.
     * @param a the array in which a run is to be counted and possibly reversed
     * @param lo index of the first element in the run
     * @param hi index after the last element that may be contained in the run. It is required that @code{lo < hi}.
     * @param c the comparator to used for the sort
     * @return the length of the run beginning at the specified position in the specified array
     */
    private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi, Comparator<? super T> c){
        assert !DEBUG || lo < hi;
        int runHi = lo + 1;
        if(runHi == hi) return 1;

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

        return runHi - lo;
    }

    /**
     * Reverse the specified range of the specified array.
     * @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}.
     * <p>
     * Roughly speaking, the computation is:
     * <p>
     * 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.
     * <p>
     * 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 !DEBUG || 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;
    }

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

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

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

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

            if(c.compare(key, a[base + m]) > 0)
                lastOfs = m + 1; // a[base + m] < key
            else
                ofs = m; // key <= a[base + m]
        }
        assert !DEBUG || 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 static <T> int gallopRight(T key, T[] a, int base, int len, int hint, Comparator<? super T> c){
        assert !DEBUG || hint >= 0 && hint < len;

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

            // Make offsets relative to b
            int tmp = lastOfs;
            lastOfs = hint - ofs;
            ofs = hint - tmp;
        }else{ // a[b + hint] <= key
            // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
            int maxOfs = len - hint;
            while(ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0){
                lastOfs = ofs;
                ofs = (ofs << 1) + 1;
                if(ofs <= 0) // int overflow
                    ofs = maxOfs;
            }
            if(ofs > maxOfs) ofs = maxOfs;

            // Make offsets relative to b
            lastOfs += hint;
            ofs += hint;
        }
        assert !DEBUG || -1 <= lastOfs && lastOfs < ofs && ofs <= len;

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

            if(c.compare(key, a[base + m]) < 0)
                ofs = m; // key < a[b + m]
            else
                lastOfs = m + 1; // a[b + m] <= key
        }
        assert !DEBUG || lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
        return ofs;
    }

    /**
     * Checks that fromIndex and toIndex are in range, and throws an appropriate exception if they aren't.
     * @param arrayLen the length of the array
     * @param fromIndex the index of the first element of the range
     * @param toIndex the index after the last element of the range
     * @throws IllegalArgumentException if fromIndex > toIndex
     * @throws ArrayIndexOutOfBoundsException if fromIndex < 0 or toIndex > arrayLen
     */
    private static void rangeCheck(int arrayLen, int fromIndex, int toIndex){
        if(fromIndex > toIndex)
            throw new IllegalArgumentException("fromIndex(" + fromIndex + ") > toIndex(" + toIndex + ")");
        if(fromIndex < 0) throw new ArrayIndexOutOfBoundsException(fromIndex);
        if(toIndex > arrayLen) throw new ArrayIndexOutOfBoundsException(toIndex);
    }

    public void doSort(T[] a, Comparator<T> c, int lo, int hi){
        stackSize = 0;
        rangeCheck(a.length, lo, hi);
        int nRemaining = hi - lo;
        if(nRemaining < 2) return; // Arrays of size 0 and 1 are always sorted

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

        this.a = a;
        this.c = c;
        tmpCount = 0;

        /** 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. */
        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
            pushRun(lo, runLen);
            mergeCollapse();

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

        // Merge all remaining runs to complete sort
        assert !DEBUG || lo == hi;
        mergeForceCollapse();
        assert !DEBUG || stackSize == 1;

        this.a = null;
        this.c = null;
        T[] tmp = this.tmp;
        for(int i = 0, n = tmpCount; i < n; i++)
            tmp[i] = null;
    }

    /**
     * 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:
     * <p>
     * 1. runLen[n - 2] > runLen[n - 1] + runLen[n] 2. runLen[n - 1] > runLen[n]
     * <p>
     * where n is the index of the last run in runLen.
     * <p>
     * This method has been formally verified to be correct after checking the last 4 runs.
     * Checking for 3 runs results in an exception for large arrays.
     * (Source: http://envisage-project.eu/proving-android-java-and-python-sorting-algorithm-is-broken-and-how-to-fix-it/)
     * <p>
     * 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 !DEBUG || stackSize >= 2;
        assert !DEBUG || i >= 0;
        assert !DEBUG || 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 !DEBUG || len1 > 0 && len2 > 0;
        assert !DEBUG || base1 + len1 == base2;

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

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

        /*
         * Find where the last element of run1 goes in run2. Subsequent elements in run2 can be ignored (because they're already in
         * place).
         */
        len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
        assert !DEBUG || 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);
    }

    /**
     * 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.
     * <p>
     * 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 !DEBUG || len1 > 0 && len2 > 0 && base1 + len1 == base2;

        // Copy first run into temp array
        T[] a = this.a; // For performance
        T[] tmp = ensureCapacity(len1);
        System.arraycopy(a, base1, tmp, 0, len1);

        int cursor1 = 0; // Indexes into tmp array
        int cursor2 = base2; // Indexes int a
        int dest = base1; // Indexes int a

        // Move first element of second run and deal with degenerate cases
        a[dest++] = a[cursor2++];
        if(--len2 == 0){
            System.arraycopy(tmp, cursor1, a, dest, len1);
            return;
        }
        if(len1 == 1){
            System.arraycopy(a, cursor2, a, dest, len2);
            a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
            return;
        }

        Comparator<? super T> 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 !DEBUG || len1 > 1 && len2 > 0;
                if(c.compare(a[cursor2], tmp[cursor1]) < 0){
                    a[dest++] = a[cursor2++];
                    count2++;
                    count1 = 0;
                    if(--len2 == 0) break outer;
                }else{
                    a[dest++] = tmp[cursor1++];
                    count1++;
                    count2 = 0;
                    if(--len1 == 1) break outer;
                }
            }while((count1 | count2) < minGallop);

            /*
             * One run is winning so consistently that galloping may be a huge win. So try that, and continue galloping until (if
             * ever) neither run appears to be winning consistently anymore.
             */
            do{
                assert !DEBUG || len1 > 1 && len2 > 0;
                count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
                if(count1 != 0){
                    System.arraycopy(tmp, cursor1, a, dest, count1);
                    dest += count1;
                    cursor1 += count1;
                    len1 -= count1;
                    if(len1 <= 1) // len1 == 1 || len1 == 0
                        break outer;
                }
                a[dest++] = a[cursor2++];
                if(--len2 == 0) break outer;

                count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
                if(count2 != 0){
                    System.arraycopy(a, cursor2, a, dest, count2);
                    dest += count2;
                    cursor2 += count2;
                    len2 -= count2;
                    if(len2 == 0) break outer;
                }
                a[dest++] = tmp[cursor1++];
                if(--len1 == 1) break outer;
                minGallop--;
            }while(count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
            if(minGallop < 0) minGallop = 0;
            minGallop += 2; // Penalize for leaving gallop mode
        } // End of "outer" loop
        this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field

        if(len1 == 1){
            assert !DEBUG || 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 !DEBUG || len2 == 0;
            assert !DEBUG || 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)
     */
    private void mergeHi(int base1, int len1, int base2, int len2){
        assert !DEBUG || len1 > 0 && len2 > 0 && base1 + len1 == base2;

        // Copy second run into temp array
        T[] a = this.a; // For performance
        T[] tmp = ensureCapacity(len2);
        System.arraycopy(a, base2, tmp, 0, len2);

        int cursor1 = base1 + len1 - 1; // Indexes into a
        int cursor2 = len2 - 1; // Indexes into tmp array
        int dest = base2 + len2 - 1; // Indexes into a

        // Move last element of first run and deal with degenerate cases
        a[dest--] = a[cursor1--];
        if(--len1 == 0){
            System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
            return;
        }
        if(len2 == 1){
            dest -= len1;
            cursor1 -= len1;
            System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
            a[dest] = tmp[cursor2];
            return;
        }

        Comparator<? super T> 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 !DEBUG || len1 > 0 && len2 > 1;
                if(c.compare(tmp[cursor2], a[cursor1]) < 0){
                    a[dest--] = a[cursor1--];
                    count1++;
                    count2 = 0;
                    if(--len1 == 0) break outer;
                }else{
                    a[dest--] = tmp[cursor2--];
                    count2++;
                    count1 = 0;
                    if(--len2 == 1) break outer;
                }
            }while((count1 | count2) < minGallop);

            /*
             * One run is winning so consistently that galloping may be a huge win. So try that, and continue galloping until (if
             * ever) neither run appears to be winning consistently anymore.
             */
            do{
                assert !DEBUG || len1 > 0 && len2 > 1;
                count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
                if(count1 != 0){
                    dest -= count1;
                    cursor1 -= count1;
                    len1 -= count1;
                    System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
                    if(len1 == 0) break outer;
                }
                a[dest--] = tmp[cursor2--];
                if(--len2 == 1) break outer;

                count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c);
                if(count2 != 0){
                    dest -= count2;
                    cursor2 -= count2;
                    len2 -= count2;
                    System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
                    if(len2 <= 1) // len2 == 1 || len2 == 0
                        break outer;
                }
                a[dest--] = a[cursor1--];
                if(--len1 == 0) break outer;
                minGallop--;
            }while(count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
            if(minGallop < 0) minGallop = 0;
            minGallop += 2; // Penalize for leaving gallop mode
        } // End of "outer" loop
        this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field

        if(len2 == 1){
            assert !DEBUG || 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 !DEBUG || len1 == 0;
            assert !DEBUG || len2 > 0;
            System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
        }
    }

    /**
     * Ensures that the external array tmp has at least the specified number of elements, increasing its size if necessary. The
     * size increases exponentially to ensure amortized Linear time complexity.
     * @param minCapacity the minimum required capacity of the tmp array
     * @return tmp, whether or not it grew
     */
    private T[] ensureCapacity(int minCapacity){
        tmpCount = Math.max(tmpCount, minCapacity);
        if(tmp.length < minCapacity){
            // Compute smallest power of 2 > minCapacity
            int newSize = minCapacity;
            newSize |= newSize >> 1;
            newSize |= newSize >> 2;
            newSize |= newSize >> 4;
            newSize |= newSize >> 8;
            newSize |= newSize >> 16;
            newSize++;

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

            tmp = (T[])new Object[newSize];
        }
        return tmp;
    }
}
