My first post will hardly be original. In fact it is probably most unoriginal post ever :) I will implement and compare several sorting algorithms in Java. I’m not going into deep details on how each algorithm works but will try to explain big picture and provide nice clean Java code and some illustrations or links to more information. Also I will try to microbenchmark some of those algorithms.

First algorithm – **Insertion sort.**

Its probably simplest (and most inefficient O(n^2)) algorithm of the bunch I will discuss. We start from second element and compare it with the first one. If it is smaller we swap them. Now we have the first one smaller than the second one. Then we take element #3 and compare with the second and first ones. If element #3 is smaller than #2 but bigger than #1 – its new place is between #1 and #2. If element is smaller than both – its new place is as #1. You probably see where it is going. We iterate through the list and find each element its new place. This gif from wiki is very clear illustration of how it works:

And here is JAVA implementation of this algorithm with some comments:

public static int[] insertionSort(int[] data) { // run from second element to last for (int i = 1; i < data.length; i++) { int key = data[i]; int j = i - 1; // take element and run back until smaller is found while (j >= 0 && data[j] > key) { data[j + 1] = data[j]; // move elements to make place for insert j--; } // insert element in position after a smaller element data[j + 1] = key; } return data; }

**Quicksort**

This is probably most used sorting algorithm as in practice it tends to be the fastest one. It is divide and conquer algorithm which selects some element called pivot and moves all elements that are smaller before it and all the greater ones after. Then it takes those 2 sub lists and do the same (recursion). Now we have 4 sub lists and first one contains elements that are smaller then second one which contains elements that are smaller than third one which contains elements smaller than fourth one. We divide and sort until we reach single element and get all the list sorted! Here is the image worth a thousand words:

And JAVA code:

public static int[] quickSort(int[] data) { Sorter.quickSort(data, 0, data.length - 1); return data; } private static void quickSort(int[] data, int sublistFirstIndex, int sublistLastIndex) { if (sublistFirstIndex < sublistLastIndex) { // move smaller elements before pivot and larger after int pivotIndex = partition(data, sublistFirstIndex, sublistLastIndex); // apply recursively to sub lists Sorter.quickSort(data, sublistFirstIndex, pivotIndex - 1); Sorter.quickSort(data, pivotIndex + 1, sublistLastIndex); } } private static int partition(int[] data, int sublistFirstIndex, int sublistLastIndex) { // differently from gif example we take last element of list as pivot int pivotElement = data[sublistLastIndex]; int pivotIndex = sublistFirstIndex - 1; for (int i = sublistFirstIndex; i < sublistLastIndex; i++) { if (data[i] <= pivotElement) { pivotIndex++; ArrayUtils.swap(data, pivotIndex, i); } } ArrayUtils.swap(data, pivotIndex + 1, sublistLastIndex); return pivotIndex + 1; // return index of pivot element }

**Merge sort**

If quicksort was dividing list into smaller and smaller sub lists, merge sort takes opposite approach. It merges sorted sub lists into larger and larger list until it reaches the result of full sorted list. Smallest possible sub list is single element (which can be considered sorted). So we take first an second elements compare them and swap if second one is smaller than first one. Do same with #3 and #4, #5 and #6 etc. Now we have sorted sub lists each containing 2 elements. Do same with those. Sub list with now sorted elements #1 and #2 is merged with sub list with elements #3 and #4 and result is 4 element sorted sub list. And same operation is repeated until we have only single sorted list.

JAVA code:

public static int[] mergeSort(int[] data) { mergeSort(data, 0, data.length - 1); return data; } private static void mergeSort(int[] data, int firstSublistStartIndex, int secondSublistEndIndex) { if (firstSublistStartIndex < secondSublistEndIndex) { int firstSublistLastElementIndex = (firstSublistStartIndex + secondSublistEndIndex) / 2; // recursively reach sub lists that has only one element Sorter.mergeSort(data, firstSublistStartIndex, firstSublistLastElementIndex); Sorter.mergeSort(data, firstSublistLastElementIndex + 1, secondSublistEndIndex); // then start merging them Sorter.merge(data, firstSublistStartIndex, firstSublistLastElementIndex, secondSublistEndIndex); } } private static void merge(int[] data, int firstSublistStartIndex, int firstSublistLastElementIndex, int secondSublistEndIndex) { int firstArrayLenght = firstSublistLastElementIndex - firstSublistStartIndex + 1; int secondArrayLenght = secondSublistEndIndex - firstSublistLastElementIndex; int[] firstArray = new int[firstArrayLenght + 1]; int[] secondArray = new int[secondArrayLenght + 1]; // copy first sub array to separate array for (int i = 0; i < firstArrayLenght; i++) { firstArray[i] = data[firstSublistStartIndex + i]; } // copy second sub array to separate array for (int i = 0; i < secondArrayLenght; i++) { secondArray[i] = data[firstSublistLastElementIndex + 1 + i]; } // optimization. because last element is maximum, we don't need to check // if current index is not out of bounds firstArray[firstArrayLenght] = Integer.MAX_VALUE; secondArray[secondArrayLenght] = Integer.MAX_VALUE; int firstCurrentIndex = 0; int secondCurrentIndex = 0; // merge firstArray and second array to data[p..r] for (int i = firstSublistStartIndex; i <= secondSublistEndIndex; i++) { // compare current values from sub arrays (both values are smallest // in each array) and move smaller one to data[p..r] if (firstArray[firstCurrentIndex] <= secondArray[secondCurrentIndex]) { data[i] = firstArray[firstCurrentIndex]; firstCurrentIndex++; } else { data[i] = secondArray[secondCurrentIndex]; secondCurrentIndex++; } } }

**Heap sort**

What would be naive sort algorithm that comes to mind as first idea to inexperienced programmer? I guess it would be something like run through the list, find smallest element and put it first, then run from second element to the end and again find smallest element and put it to second place etc – until we sort everything. This not very efficient algorithm (O(n^2) is called selection sort. Heap sort is more efficient variation of selection sort. It too selects smallest (largest) element and puts it to beginning (end) of the list. However to select smallest/largest element it uses data structure called heap which make algorithm much more efficient. Heap is binary tree-like data structure whose every node is always smaller(larger) than its children nodes. So in order to select smallest/largest element we just take it from the root node. The algorithm then basically is 2 step process. First heap is built from data. Simple array is enough for that (element #1 becomes root node, #2 and #3 its children, #4 and #5 elements becomes #2’s children etc…). Second step is remove smallest(largest) element from the root node of the heap and place it first(last). Of course after removal heap must be updated so its root is again smallest. Continue until all the list is sorted. Here is gif which hopefully make it more clear:

JAVA code:

public static int[] heapSort(int[] data) { // build heap HeapUtils.buildMaxHeap(data); int heapSize = data.length; for (int i = data.length - 1; i > 0; i--) { // move largest element to the end ArrayUtils.swap(data, 0, i); // rebuild heap with smaller size by 1 (last element already sorted) HeapUtils.maxHeapify(data, 0, --heapSize); } return data; } public static void buildMaxHeap(int[] data) { for (int i = data.length / 2; i >= 0; i--) { maxHeapify(data, i); } } public static void maxHeapify(int[] data, int index) { maxHeapify(data, index, data.length); } // build heap structure "in place" where index is root node and heapSize is number of elements public static void maxHeapify(int[] data, int index, int heapSize) { int leftLeaf = getLeftLeaf(index); int rightLeaf = getRightLeaf(index); int largest = index; if (leftLeaf < heapSize && data[leftLeaf] > data[largest]) { largest = leftLeaf; } if (rightLeaf < heapSize && data[rightLeaf] > data[largest]) { largest = rightLeaf; } if (largest != index) { ArrayUtils.swap(data, index, largest); HeapUtils.maxHeapify(data, largest, heapSize); } }

OK time to do some benchmarks. For that I used JMH micro benchmark framework. I will simply measure sorting of 100 000 random integers. Of course you can clone the project from github (link at the end of the article) and try different variations. How algorithms act with less elements or with more? Is there a problem when array is already sorted (hint see quicksort)? However for now I will simply post results of 100000 random elements sorting. Here we go:

Benchmark | Score | Score error | Units |
---|---|---|---|

Heap sort |
20.789 | 0.657 | ms/op |

Insertion sort |
2381.962 | 38.257 | ms/op |

Arrays.sort |
11.203 | 0.218 | ms/op |

Merge sort |
27.680 | 0.713 | ms/op |

Quicksort |
12.726 | 0.207 | ms/op |

We could have expected somewhat similar results. Of course slowest was insertion sort. Fastest optimized and well implemented JDK sort algorithm. JDK sort algorithm is variation of quicksort so no surprise that custom quicksort implementation was just a little bit slower than JDK version. And then somewhat similar performance of heapsort and mergesort with heapsort being a bit faster.

Those are main and most popular sorting methods. There are various variations and hybrid methods as well. For example Timsort – hybrid of merge and insertion sort which is now used in JDK 7 for sorting non primitive types (take a look at source code of Arrays.java). And for primitive types it looks like JDK use some sophisticated quick sort algorithm with double pivots. Why is the difference? I don’t know, but I guess the reason is that merge sort (and Timsort) is stable algorithm meaning that after sorting even if elements were equal they still maintain same sequence as it was in input (for example sorting people by age, both John and Peter is 25 after sorting John still is before Peter like it was before sorting in input). Quicksort does not have this guarantee, but probably is faster in practice and though preferred method for primitive types. So we can see that library developers already is thinking for us and its probably best just to use standard tools and not reinvent the wheel unless you really know what you are doing and really can get better performance. So lets review algorithms and compare them.

Algorithm | Average | Worst case | Best case | Stable | Memory | Best use cases |
---|---|---|---|---|---|---|

Insertion sort |
O(n^2) | O(n^2) | O(n) | Yes | O(1) (in place) | Quite efficient for small lists. Sometimes used in hybrid implementation to sort very small sub lists. It can also sort a list at the same time as it receives it. |

Quicksort |
O(n log(n)) | O(n log(n)) | O(n log(n)) | No | O(log n) | In practice tends to be faster algorithm than others. |

Merge sort |
O(n log(n)) | O(n log(n)) | O(n log(n)) | Yes | O(n) | With RAM based arrays tends to be slower than quicksort, but can be more efficient with slower media. Needs more memory, but is stable. |

Heap sort |
O(n log(n)) | O(n log(n)) | O(n log(n)) | No | O(1) | It’s performed in place and has very close performance to quicksort, so its good choice when memory is the issue. |

OK so those were so called comparison sorting algorithms. And we can’t really do much better than that. Of course there are various improvements as I mentioned. For example quicksort with dual pivots (interesting read here) but still they are no better than O(n logn). But are we sure we can’t do better? And the answer is it depends! Under some circumstances we could. Imagine we know that we need to sort some numbers with range 0 to 99 and each number can appear only once. Yes we could use insertion sort or quicksort for this task. But if you think a bit – a much quicker version would be simply create a new boolean array with 100 elements and then just run through the input and set value in that helper array with key = input element to true. This way we simply need to print all the indexes that has it true in helper array. And that is very quick. That kind of sorting is called distribution sorting. It can be really fast but you need to know range of possible values of elements and of course need additional memory for a helper array. Lets review some distribution sorting algorithms.

**Counting sort**

This is slightly more sophisticated algorithm than the one I described above. The main difference is that we use integers instead of boolean in helper array and counting the number of element occurrences. This allows input to have equal elements because we are “counting” them.

JAVA CODE

public static int[] countingSort(int[] data, int upperBound) { int dataLenght = data.length; int tempArrayLength = upperBound + 1; int[] tempArr = new int[tempArrayLength]; // count how much of each element (index - element, value - count) for (int i = 0; i < dataLenght; i++) { tempArr[data[i]]++; } // count how much of less or equal elements exists for (int i = 1; i < tempArrayLength; i++) { tempArr[i] += tempArr[i - 1]; } int[] sortedData = new int[dataLenght]; // put each element to its place in sorted array for (int i = dataLenght - 1; i >= 0; i--) { int place = tempArr[data[i]] - 1; // element place in result array sortedData[place] = data[i]; tempArr[data[i]]--; // decrease count less or equal elements count } return sortedData; }

**Radix sort**

In radix sort we are sorting by comparing individual digits from the last one to the first one. In essence radix sort is like this: sort elements by last digit. Then sort elements by second to last digit up till we reach first digit and after that all elements are in sorted order.

The sorting can be done by having buckets for each number and putting elements into each bucket (like in this explanation) or as in implementation provided here reuse counting sort.

As you can see for this method to work you need to know largest number of digits that elements in your data set might contain (or you will need to determine that dynamically which cost time too). Complexity of algorithm is O(kn) where k is number of digits and n is number of elements. If k is very big this sort might be not as efficient as even comparison sorts.

JAVA code for radix sort:

public static int[] radixSort(int[] data, int numberOfDigits) { int[] sortedData = data; for (int i = 0; i < numberOfDigits; i++) { sortedData = Sorter.countingSortByDigit(sortedData, i, 9); } return sortedData; } private static int[] countingSortByDigit(int[] data, int digitIndex, int upperBound) { int dataLenght = data.length; int tempArrayLength = upperBound + 1; int[] tempArr = new int[tempArrayLength]; // count how much of each element (index - element, value - count) for (int i = 0; i < dataLenght; i++) { int digitFromNumber = MathUtils.getDigitFromNumber(data[i], digitIndex); tempArr[digitFromNumber]++; } // count how much of less or equal elements exists for (int i = 1; i < tempArrayLength; i++) { tempArr[i] += tempArr[i - 1]; } int[] sortedData = new int[dataLenght]; // put each element to its place in sorted array for (int i = dataLenght - 1; i >= 0; i--) { int digitFromNumber = MathUtils.getDigitFromNumber(data[i], digitIndex); int place = tempArr[digitFromNumber] - 1; // element place in result // array sortedData[place] = digitFromNumber; tempArr[digitFromNumber]--; // decrease count less or equal elements // count } return sortedData; }

**Bucket sort**

Bucket sort is method that is taking advantage that small lists are faster to sort than very big ones. So it simply splits input into “buckets” and then uses another sorting algorithm to sort those buckets. It is also possible to use bucket sort recursively too. Elements in each bucket is not overlapping so after each bucket is sorted they can be simply joined into the sorted result list. For example lets say we have 3 buckets and put into first elements 3 and 1, into second elements 5 and 8 and into third 11 9 and 12. After sorting each bucket and joining them into a list we get 1 3 5 8 9 11 12 sorted output. In fact bucket sort could be viewed as generalized version of counting sort. Take a look at explanation here The implementation of bucket sort here is efficient way to sort double numbers with range 0 to 1.

JAVA code:

public static double[] bucketSort(double[] data) { int numberOfBuckets = data.length; double[][] buckets = new double[numberOfBuckets][]; int[] bucketListIndices = new int[numberOfBuckets]; for (int i = 0; i < data.length; i++) { int bucketIndex = (int) ((double) numberOfBuckets * data[i]); if (buckets[bucketIndex] == null) { buckets[bucketIndex] = new double[1]; } else { // increase bucket size by 1 (maybe better redo with linked list) double[] bucket = buckets[bucketIndex]; buckets[bucketIndex] = Arrays.copyOf(bucket, bucket.length + 1); } buckets[bucketIndex][bucketListIndices[i]++] = data[i]; } double[] sortedData = new double[data.length]; int nextIndex = 0; // sort all buckets with insertion sort and merge them into result. for (int bucketIndex = 0; bucketIndex < numberOfBuckets; bucketIndex++) { if (buckets[bucketIndex] != null) { int bucketLength = buckets[bucketIndex].length; System.arraycopy(sortedData, nextIndex, Sorter.insertionSort(buckets[bucketIndex]), 0, bucketLength); nextIndex = nextIndex + bucketLength - 1; } } return sortedData; }

So these are all algorithms I wanted to review here. Keep in mind those are only introductions here and my benchmarks are quite naive. Wikipedia is great starting point if you are interested into deeper understanding of the algorithms and learning all the subtleties and tricks that can improve performance. Hope you enjoyed this article!

You can find all the algorithms and benchmarks in the github project here.

Great block. Thanks. Did you try to plug the code to dmelt (http://jwork.org/dmelt) to run these examples in Python or Groovy?

Thanks. No I didn’t try dmelt (was not aware of it). Kept it simple.