SelectMany for flattening in C Sharp (C#) - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this SelectMany example?

Consider the following C# code using LINQ's SelectMany method. What will be printed to the console?

C Sharp (C#)

var numbers = new List<List<int>> {
    new List<int> {1, 2},
    new List<int> {3, 4},
    new List<int> {5}
};

var flat = numbers.SelectMany(x => x);
foreach(var num in flat)
{
    Console.Write(num + " ");
}

A 1 3 5

B System.Collections.Generic.IEnumerable`1[System.Int32]

C 1 2 3 4 5

D 1 2 3 4

💡 Hint

SelectMany flattens nested collections into one sequence.

SelectMany takes each inner list and combines all elements into a single sequence. So all numbers from all inner lists are printed in order.

❓ Predict Output

intermediate

What does this SelectMany with index produce?

Look at this C# code using SelectMany with an index parameter. What is the output?

C Sharp (C#)

var data = new List<string[]> {
    new string[] {"a", "b"},
    new string[] {"c"},
    new string[] {"d", "e", "f"}
};

var result = data.SelectMany((arr, i) => arr.Select(s => s + i));
foreach(var s in result)
{
    Console.Write(s + " ");
}

A a0 b0 c1 d2 e2 f2

B a b c d e f

C a1 b1 c2 d3 e4 f5

D a0 b1 c2 d3 e4 f5

💡 Hint

The index i is the position of the outer array in the list.

SelectMany passes the index of each inner array. Each string is concatenated with that index, so first array elements get '0', second '1', third '2'.

🔧 Debug

advanced

Why does this SelectMany code throw an exception?

This code throws an exception at runtime. What is the cause?

C Sharp (C#)

var list = new List<List<int>> {
    new List<int> {1, 2},
    null,
    new List<int> {3}
};

var flat = list.SelectMany(x => x);
foreach(var n in flat)
{
    Console.Write(n + " ");
}

A Output: 1 2 3

B InvalidOperationException because SelectMany expects non-empty lists

C Compile-time error due to null in list

D NullReferenceException because one inner list is null

💡 Hint

Check what happens when SelectMany tries to iterate a null collection.

SelectMany tries to iterate each inner list. When it reaches the null element, it throws NullReferenceException because you cannot iterate null.

📝 Syntax

advanced

Which option correctly uses SelectMany to flatten and filter?

You want to flatten a list of integer arrays and keep only even numbers. Which code snippet does this correctly?

A var result = list.SelectMany(x => x.Filter(n => n % 2 == 0));

B var result = list.SelectMany(x => x.Where(n => n % 2 == 0));

C var result = list.SelectMany(x => x.Select(n => n % 2 == 0));

D var result = list.SelectMany(x => n => n % 2 == 0);

💡 Hint

Use Where inside SelectMany to filter elements.

Option B correctly uses Where to filter even numbers inside each inner collection before flattening. Others have syntax errors or wrong methods.

🚀 Application

expert

How many items are in the flattened list?

Given this nested list, how many integers will be in the flattened list after using SelectMany?

C Sharp (C#)

var nested = new List<List<int>> {
    new List<int> {1, 2, 3},
    new List<int> {},
    new List<int> {4, 5},
    new List<int> {6}
};

var flat = nested.SelectMany(x => x).ToList();
int count = flat.Count;

A 6

B 5

C 7

D 4

💡 Hint

Count all elements in all inner lists combined.

The inner lists have 3 + 0 + 2 + 1 elements = 6 total elements after flattening.

Practice

1. What does the SelectMany method do in C# when used on a list of lists?

easy

A. It sorts each inner list individually.

B. It filters the outer list based on a condition.

C. It flattens the nested lists into a single list containing all elements.

D. It creates a dictionary from the nested lists.

Solution

1. Step 1: Understand the structure of nested lists

   A list of lists means you have many small lists inside one big list.

2. Step 2: Role of SelectMany

   SelectMany takes each small list and combines all their elements into one big flat list.

3. Final Answer:

   It flattens the nested lists into a single list containing all elements. -> Option C

4. Quick Check:

   SelectMany flattens nested lists = B [OK]

Hint: SelectMany flattens nested lists into one list fast [OK]

Common Mistakes:

• Thinking SelectMany sorts or filters instead of flattening
• Confusing SelectMany with Select which keeps nested structure
• Assuming SelectMany creates dictionaries

2. Which of the following is the correct syntax to flatten a list of lists List numbers using SelectMany?

easy

A. var flat = numbers.SelectMany(x => x);

B. var flat = numbers.Select(x => x);

C. var flat = numbers.SelectMany(x => x.ToString());

D. var flat = numbers.SelectMany(x => x.Count());

Solution

1. Step 1: Understand SelectMany syntax

   SelectMany expects a function that returns an inner collection to flatten.

2. Step 2: Check each option

   var flat = numbers.SelectMany(x => x); returns each inner list as is, which is correct. var flat = numbers.Select(x => x); uses Select, not flattening. Options C and D return wrong types (string and int), not collections.

3. Final Answer:

   var flat = numbers.SelectMany(x => x); -> Option A

4. Quick Check:

   SelectMany with identity function flattens = A [OK]

Hint: Use SelectMany with identity function to flatten lists [OK]

Common Mistakes:

• Using Select instead of SelectMany
• Returning non-collections inside SelectMany
• Trying to flatten by counting elements

3. What is the output of the following code?

var data = new List {
  new List<int> {1, 2},
  new List<int> {3, 4},
  new List<int> {5}
};
var result = data.SelectMany(x => x).ToList();
Console.WriteLine(string.Join(",", result));

medium

A. 12345

B. 1,2;3,4;5

C. System.Collections.Generic.List`1[System.Int32]

D. 1,2,3,4,5

Solution

1. Step 1: Understand SelectMany flattening

   SelectMany flattens all inner lists into one list: [1,2,3,4,5].

2. Step 2: Output formatting

   string.Join with "," joins elements with commas, so output is "1,2,3,4,5".

3. Final Answer:

   1,2,3,4,5 -> Option D

4. Quick Check:

   Flattened list joined by commas = D [OK]

Hint: SelectMany flattens, string.Join prints with commas [OK]

Common Mistakes:

• Expecting nested list output instead of flat
• Confusing ToString output of list object
• Missing commas in output

4. Identify the error in this code snippet that tries to flatten a list of lists:

List words = new() {
  new List<string> {"apple", "banana"},
  new List<string> {"cat", "dog"}
};
var flatWords = words.SelectMany(w => w.Length).ToList();

medium

A. The list initialization syntax is incorrect.

B. Using w.Length returns int, not a collection, causing a compile error.

C. Missing parentheses after SelectMany call.

D. ToList() cannot be called after SelectMany.

Solution

1. Step 1: Analyze the lambda inside SelectMany

   The lambda uses w => w.Length, which returns an int, not a collection.

2. Step 2: Understand SelectMany requirements

   SelectMany expects a function returning a collection to flatten, not a single int value.

3. Final Answer:

   Using w.Length returns int, not a collection, causing a compile error. -> Option B

4. Quick Check:

   SelectMany needs collections, not int = C [OK]

Hint: SelectMany lambda must return collections, not single values [OK]

Common Mistakes:

• Returning single values instead of collections in SelectMany
• Confusing Length property with collection
• Assuming ToList() causes error

5. Given a list of students where each student has a list of grades, how would you use SelectMany to get a list of all grades above 80 from all students?

class Student {
  public string Name { get; set; }
  public List<int> Grades { get; set; }
}

List<Student> students = new() {
  new Student { Name = "Anna", Grades = new List<int> { 90, 70 } },
  new Student { Name = "Ben", Grades = new List<int> { 85, 60 } }
};

// Fill in the blank to get all grades > 80
var highGrades = students.SelectMany(__________).ToList();

hard

A. s => s.Grades.Where(g => g > 80)

B. s => s.Grades

C. s => s.Name

D. s => s.Grades.Count

Solution

1. Step 1: Understand the goal

   We want all grades above 80 from all students combined.

2. Step 2: Use SelectMany with filtering inside

   s => s.Grades.Where(g => g > 80) filters grades > 80 inside SelectMany, flattening only those grades.

3. Step 3: Why not others?

   s => s.Grades flattens all grades, including low ones like 70 and 60. Options C and D return wrong types (string and int).

4. Final Answer:

   s => s.Grades.Where(g => g > 80) -> Option A

5. Quick Check:

   SelectMany with filtered inner lists = A [OK]

Hint: Filter inside SelectMany to flatten only needed items [OK]

Common Mistakes:

• Flattening all grades without filtering
• Returning non-collections in SelectMany
• Using wrong property like Name or Count