I have an array:

const list1 = [0, 1, 2];

How do I check if other arrays contain any of the target array elements?

For example:

[2, 3] //returns true;

[2, 3, 4] //returns true;

[3, 4] //returns false;

CodePudding user response：

Using list1.indexWhere(list2.contains) should be fine for small lists, but for large lists, the asymptotic runtime complexity would be O(m * n) where m and n are the sizes of the lists.

A different way to pose the problem of checking if a list contains any element of another list is to check if the set-intersection of two lists is non-empty. The direct way to implement that would be:

var contains = list1.toSet().intersection(list2.toSet()).isNotEmpty;

Since the default Set implementation is a LinkedHashSet, lookups would be O(1), and computing the intersection would be linear with respect to one of the Sets. However, converting each List to a Set would take linear time, making the whole operation take O(m n).

That's asymptotically efficient, but it computes the entire intersection just to determine whether it's empty or not, which is wasteful. You can do a bit better by using .any to stop earlier and noting that .any doesn't benefit from the receiving object being a Set:

var set2 = list2.toSet();
var contains = list1.any(set2.contains);

Note that if you can use Sets in the first place instead of Lists, then the conversion cost would disappear and make the operation O(m).

CodePudding user response：

final contains = list1.indexWhere((e) => list2.contains(e)) > -1

Explanation

indexWhere returns an index of an element where the test function returned true.

contains returns true if the given element is presented in the array.