LINQ interview questions

In this post we will review a bunch of small and simple LINQ interview questions to get you started. Once you understand these standard query expressions, it should be a breeze for you to answer LINQ questions.

Question: Given an array of numbers, find if ALL numbers are a multiple of a provided number. For example, all of the following numbers - 30, 27, 15, 90, 99, 42, 75 are multiples of 3.

The trick here is to use the Enumerable.All<TSource> method which determines whether all elements of a sequence satisfy a condition.

static void Main(string[] args)
{
    int[] numbers = { 30, 27, 15, 90, 99, 42, 75 };

    bool isMultiple = MultipleTester(numbers, 3);
}

private static bool MultipleTester(int[] numbers, int divisor)
{
    bool isMultiple =
        numbers.All(number => number % divisor == 0);
    
    return isMultiple;
}

Question: Given an array of numbers, find if ANY of the number is divisible by 5. For example, one of the following numbers - 30, 27, 18, 92, 99, 42, 72 is divisible by 5.

Again, the key fact here is to use Enumerable.Any<TSource> method which determines whether any element of a sequence satisfies a condition. The code is very similar to the ALL case.

 

static void Main(string[] args)
{
    int[] numbers = { 30, 27, 18, 92, 99, 42, 72 };
        
    bool isDivisible = IsDivisible(numbers, 3);
}

private static bool IsDivisible(int[] numbers, int divisor)
{
    bool isDivisible =
        numbers.Any(number => number % divisor == 0);
    
    return isDivisible;
}

Another way to present a similar question that utilizes Enumerable.Any<TSource> method is shown below.

Question: Given a GPSManufacturer and GPSDevice data structure as defined below, find all the manufacturers that have at least 1 or more active GPS devices.

class GpsDevice
{
    public string Name;
    public bool IsActive;
}
    
class GpsManufacturer
{
    public string Name;
    public GpsDevice[] Devices;
}
    
static List<GpsManufacturer> gpsManufacturers = 
    new List<GpsManufacturer> 
    {
        new GpsManufacturer 
           {Name = "manf1", 
            Devices = new GpsDevice[] 
            {
                new GpsDevice {Name = "device1", IsActive = true},
                new GpsDevice {Name = "device2", IsActive = false}
            }
        },
        new GpsManufacturer 
           {Name = "manf2", 
            Devices = new GpsDevice[] 
            {
                new GpsDevice {Name = "device1", IsActive = false},
                new GpsDevice {Name = "device2", IsActive = false}
            }
        },
        new GpsManufacturer 
        {Name = "manf3"}
    };

The following function finds out the active GPS manufacturers.

        public static void ActiveGpsManufacturers()
        {
            // Determine which manufacturers have a active device.
            IEnumerable<string> activeManf =
                from manf in gpsManufacturers // for all manufacturers
                where manf.Devices != null && // they have a list of devices
                    // and at least one of the device is active
                      manf.Devices.Any(m => m.IsActive == true)
                select manf.Name; // select them
                
            // just for debugging
            foreach (string name in activeManf)
                Console.WriteLine(name);
        }

As an extension question and using Lambda expressions, consider the following question:

Question: Find the count of all the manufacturers that have at least 1 or more active GPS devices.

 

public static void ActiveGpsManufacturersCount()
{
    // to count the number of active manufacturers
    int activeManfCount =
        gpsManufacturers.Count
            (manf => (manf.Devices != null &&
                      manf.Devices.Any(m => m.IsActive == true)));

    // debugging
    Console.WriteLine("Active Manf count: {0}", activeManfCount);
}

MSDN has extensive documentation on LINQ. I recommend reading about the various functions and then trying out various queries on your own. LINQPad is an excellent tool to learn also.