Question

I want to compare two lists with the same number of elements, and find the number of differences between them. Right now, I have this code (which works):

public static int CountDifferences<T> (this IList<T> list1, IList<T> list2)
{
    if (list1.Count != list2.Count)
        throw new ArgumentException ("Lists must have the same number of elements", "list2");

    int count = 0;
    for (int i = 0; i < list1.Count; i++) {
        if (!EqualityComparer<T>.Default.Equals (list1[i], list2[i]))
            count++;
    }

    return count;
}

This feels messy to me, and it seems like there must be a more elegant way to achieve it. Is there a way, perhaps, to combine the two lists into a single list of tuples, then simple examine each element of the new list to see if both elements are equal?

Answer 1

Since order in the list does count this would be my approach:

public static int CountDifferences<T>(this IList<T> list1, IList<T> list2)
{
    if (list1.Count != list2.Count)
        throw new ArgumentException("Lists must have the same number of elements", "list2");

    int count  = list1.Zip(list2, (a, b) => a.Equals(b) ? 0 : 1).Sum();
    return count;
}

Simply merging the lists using Enumerable.Zip() then summing up the differences, still O(n) but this just enumerates the lists once.

Also this approach would work on any two IEnumerable of the same type since we do not use the list indexer (besides obviously in your count comparison in the guard check).