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17.1: Insertion sort

  • Page ID
    12834
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    We’ll start with insertion sort, mostly because it is simple to describe and implement. It is not very efficient, but it has some redeeming qualities, as we’ll see.

    Rather than explain the algorithm here, I suggest you read the insertion sort Wikipedia page at thinkdast.com/insertsort, which includes pseudocode and animated examples. Come back when you get the general idea.

    Here’s an implementation of insertion sort in Java:

    public class ListSorter<T> {
    
        public void insertionSort(List<T> list, Comparator<T> comparator) {
    
            for (int i=1; i < list.size(); i++) {
                T elt_i = list.get(i);
                int j = i;
                while (j > 0) {
                    T elt_j = list.get(j-1);
                    if (comparator.compare(elt_i, elt_j) >= 0) {
                        break;
                    }
                    list.set(j, elt_j);
                    j--;
                }
                list.set(j, elt_i);
            }
        }
    }
    

    I define a class, ListSorter, as a container for sort algorithms. By using the type parameter, T, we can write methods that work on lists containing any object type.

    insertionSort takes two parameters, a List of any kind and a Comparator that knows how to compare type T objects. It sorts the list “in place”, which means it modifies the existing list and does not have to allocate any new space.

    The following example shows how to call this method with a List of Integer objects:

    List<Integer> list = new ArrayList<Integer>(Arrays.asList(3, 5, 1, 4, 2));
    
    Comparator<Integer> comparator = new Comparator<Integer>() {
        @Override
        public int compare(Integer elt1, Integer elt2) {
            return elt1.compareTo(elt2);
        }
    };
    
    ListSorter<Integer> sorter = new ListSorter<Integer>();
    sorter.insertionSort(list, comparator);
    System.out.println(list);
    

    insertionSort has two nested loops, so you might guess that its run time is quadratic. In this case, that turns out to be correct, but before you jump to that conclusion, you have to check that the number of times each loop runs is proportional to n, the size of the array.

    The outer loop iterates from 1 to list.size(), so it is linear in the size of the list, n. The inner loop iterates from i to 0, so it is also linear in n. Therefore, the total number of times the inner loop runs is quadratic.

    If you are not sure about that, here’s the argument:

    • The first time through, \( i = 1 \) and the inner loop runs at most once.
    • The second time, \( i = 2 \) and the inner loop runs at most twice.
    • The last time, \( i = n - 1 \) and the inner loop runs at most \( n - 1 \) times.

    So the total number of times the inner loop runs is the sum of the series \( 1, 2, \dots , n - 1 \), which is \( n(n-1)/2 \). And the leading term of that expression (the one with the highest exponent) is \( n^2 \).

    In the worst case, insertion sort is quadratic. However:

    If the elements are already sorted, or nearly so, insertion sort is linear. Specifically, if each element is no more than k locations away from where it should be, the inner loop never runs more than k times, and the total run time is \( O(kn) \).

    Because the implementation is simple, the overhead is low; that is, although the run time is \( an^2 \), the coefficient of the leading term, a, is probably small.

    So if we know that the array is nearly sorted, or is not very big, insertion sort might be a good choice. But for large arrays, we can do better. In fact, much better.


    This page titled 17.1: Insertion sort is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Allen B. Downey (Green Tea Press) .

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