Comparing Arrays LINQ

Question

How can I compare to array in linq and get all elements where there is at least one intersection? Example:

selectes = {1,5,7} 
Bands[0].SongsID {1,9} 
Bands[1].SongsID {5,6}
Bands[2].SongsID {4,6}

I need to select Bands[0] and Bands[1]. I tried this:

var selectes2 = Bands.Where(t => t.SongsID.Intersect(selectes));

Bands class:

public class Band
{
    public int ID                { get; set; }
    public string Name           { get; set; }
    public DateTime YearOfCreate { get; set; }
    public string Country        { get; set; }
    public int[] SongsID         { get; set; }
}

Please change your example to compilable C# code (e.g. what is the type of SongsID?). Consider: What is the return type of Intersect? What type does Where expect the delegate to return? What method will do the conversion? — NetMage
– NetMage, Commented Jun 1, 2018 at 22:41
So you want to select any band that has any songID in your selectes list? — Jonathon Chase
– Jonathon Chase, Commented Jun 1, 2018 at 22:43

Serega · Accepted Answer · 2018-06-01 22:43:27Z

5

var selectes2 = Bands.Where(t => t.SongsID.Intersect(selectes).Any());

answered Jun 1, 2018 at 22:43

Serega

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2 Comments

Mrinal Kamboj Over a year ago

Time complexity is O(N^2), performance would be an issue here

Serega Over a year ago

@MrinalKamboj I know that) Of course we can easy get guaranteed complexity O(N log N) if use SortedSet or even O(N) in average if use HashSet. But I think, author of this question was searching for the easiest and shortest solution)

Jonathon Chase · Accepted Answer · 2018-06-01 22:46:57Z

1

Assuming you mean to select any band that has any song ID that matches your list of ids, you could accomplish that with this:

var matchingBands = Bands.Where(band => band.SongsID.Any(selectes.Contains));

answered Jun 1, 2018 at 22:46

Jonathon Chase

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1 Comment

Mrinal Kamboj Over a year ago

Array Contains has time complexity of O(N), same solution would work better by converting int[] to Hashset<int>. This would bring down solution time complexity to O(N) instead of O(N^2)

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