Count and Say Problem in DSA C# - Build from Scratch

Count and Say Problem

📖 Scenario: Imagine you are working with a special way to describe numbers called the "Count and Say" sequence. Each number is described by counting the groups of digits in the previous number and saying what you see.For example, starting with "1", the next number is "11" (one 1), then "21" (two 1s), then "1211" (one 2, one 1), and so on.

🎯 Goal: You will build a program that takes a number n and generates the nth term in the Count and Say sequence.

📋 What You'll Learn

Create a string variable current to hold the starting term "1".

Create an integer variable n to hold the term number to generate.

Use a loop to generate terms from 2 up to n.

Inside the loop, build the next term by counting consecutive digits in current.

Print the final term after the loop ends.

💡 Why This Matters

🌍 Real World

The Count and Say sequence is a classic example of string processing and pattern recognition used in coding interviews and algorithm practice.

💼 Career

Understanding how to manipulate strings and use loops effectively is essential for software development, especially in tasks involving data parsing and transformation.

1

Setup the starting term and the term number

Create a string variable called current and set it to "1". Also create an integer variable called n and set it to 5.

DSA C#

# Create a string variable called current and set it to "1"
# Create an integer variable called n and set it to 5
# Your code here

Hint

Start with the first term as "1" and choose the term number 5 to generate.

2

Create a loop to generate terms up to n

Create a for loop with an integer variable i starting from 2 up to and including n.

DSA C#

string current = "1";
int n = 5;

// Create a for loop with i from 2 to n
// Your code here

Hint

Use a for loop to repeat the process from term 2 to term n.

3

Build the next term inside the loop

Inside the for loop, create a string variable nextTerm and set it to empty string "". Use a while loop with an index j to count consecutive digits in current. Append the count and digit to nextTerm. After the while loop, set current to nextTerm.

DSA C#

string current = "1";
int n = 5;

for (int i = 2; i <= n; i++)
{
    // Create nextTerm as empty string
    // Use while loop with index j to count digits
    // Append count and digit to nextTerm
    // Set current to nextTerm
    // Your code here
}

Hint

Use a nested while loop to count how many times the same digit repeats consecutively.

4

Print the final term

After the for loop, write a Console.WriteLine statement to print the value of current.

DSA C#

string current = "1";
int n = 5;

for (int i = 2; i <= n; i++)
{
    string nextTerm = "";
    int j = 0;
    while (j < current.Length)
    {
        char digit = current[j];
        int count = 0;
        while (j < current.Length && current[j] == digit)
        {
            count++;
            j++;
        }
        nextTerm += count.ToString() + digit;
    }
    current = nextTerm;
}

// Print the final current value
// Your code here

Hint

The 5th term in the Count and Say sequence starting from "1" is "111221".

Practice

1. What does the "Count and Say" sequence start with as its first term?

easy

A. "11"

B. "1"

C. "0"

D. "21"

Solution

1. Step 1: Understand the sequence start

   The "Count and Say" sequence always begins with the string "1" as the first term.

2. Step 2: Confirm the initial term

   This initial term is the base for generating subsequent terms by counting and saying digits.

3. Final Answer:

   "1" -> Option B

4. Quick Check:

   First term = "1" [OK]

Hint: Remember: sequence always starts with "1" [OK]

Common Mistakes:

• Thinking it starts with "0"
• Confusing first term with second term
• Assuming it starts with "11" or "21"

2. Which of the following is the correct C# syntax to declare a method named CountAndSay that takes an integer n and returns a string?

easy

A. public string CountAndSay(int n)

B. string CountAndSay(int n) public

C. public CountAndSay(int n) string

D. public void CountAndSay(int n)

Solution

1. Step 1: Identify correct method declaration syntax

   In C#, method declaration starts with access modifier, then return type, method name, and parameters.

2. Step 2: Match syntax to options

   public string CountAndSay(int n) correctly declares a public method returning string with int parameter.

3. Final Answer:

   public string CountAndSay(int n) -> Option A

4. Quick Check:

   Access modifier + return type + name + parameters = public string CountAndSay(int n) [OK]

Hint: Method syntax: access modifier + return type + name + parameters [OK]

Common Mistakes:

• Placing 'public' after method name
• Using void return type when string is needed
• Incorrect order of return type and method name

3. What is the output of the following C# code snippet?

Console.WriteLine(CountAndSay(4));

string CountAndSay(int n) {
    if (n == 1) return "1";
    string prev = CountAndSay(n - 1);
    StringBuilder result = new StringBuilder();
    int count = 1;
    for (int i = 1; i < prev.Length; i++) {
        if (prev[i] == prev[i - 1]) {
            count++;
        } else {
            result.Append(count).Append(prev[i - 1]);
            count = 1;
        }
    }
    result.Append(count).Append(prev[prev.Length - 1]);
    return result.ToString();
}

medium

A. "312211"

B. "111221"

C. "1211"

D. "21"

Solution

1. Step 1: Trace CountAndSay(4) calls

   CountAndSay(1) = "1"
   CountAndSay(2) = "11" (one 1)
   CountAndSay(3) = "21" (two 1s)
   CountAndSay(4) = "1211" (one 2, one 1)

2. Step 2: Confirm output string for n=4

   The function returns "1211" for input 4 by counting digits in previous term.

3. Final Answer:

   "1211" -> Option C

4. Quick Check:

   Count and Say term 4 = "1211" [OK]

Hint: Build sequence step-by-step from n=1 to n=4 [OK]

Common Mistakes:

• Confusing term 3 and term 4 outputs
• Miscounting repeated digits
• Appending digits in wrong order

4. Identify the bug in this C# method for the Count and Say problem:

string CountAndSay(int n) {
    if (n == 1) return "1";
    string prev = CountAndSay(n - 1);
    string result = "";
    int count = 1;
    for (int i = 0; i < prev.Length; i++) {
        if (i < prev.Length - 1 && prev[i] == prev[i + 1]) {
            count++;
        } else {
            result += count.ToString() + prev[i];
            count = 1;
        }
    }
    return result;
}

medium

A. Using string concatenation instead of StringBuilder causes performance issues

B. The count variable is not reset correctly

C. The loop condition causes index out of range error

D. Loop index should start at 1, not 0

Solution

1. Step 1: Analyze loop and string concatenation

   The loop logic is correct for counting digits, but string concatenation with += inside loop is inefficient.

2. Step 2: Identify performance issue

   Using string concatenation repeatedly creates many string copies; StringBuilder is preferred for better performance.

3. Final Answer:

   Using string concatenation instead of StringBuilder causes performance issues -> Option A

4. Quick Check:

   String concatenation in loop = performance issue [OK]

Hint: Use StringBuilder for string building inside loops [OK]

Common Mistakes:

• Thinking loop index causes error
• Believing count reset is wrong
• Ignoring performance impact of string concatenation

5. Given the Count and Say sequence, what is the 6th term generated by the following C# method?

string CountAndSay(int n) {
    if (n == 1) return "1";
    string prev = CountAndSay(n - 1);
    StringBuilder result = new StringBuilder();
    int count = 1;
    for (int i = 1; i < prev.Length; i++) {
        if (prev[i] == prev[i - 1]) {
            count++;
        } else {
            result.Append(count).Append(prev[i - 1]);
            count = 1;
        }
    }
    result.Append(count).Append(prev[prev.Length - 1]);
    return result.ToString();
}

Console.WriteLine(CountAndSay(6));

hard

A. "21121314"

B. "13112221"

C. "1113213211"

D. "312211"

Solution

1. Step 1: Generate terms up to 6

   Term 1: "1"
   Term 2: "11" (one 1)
   Term 3: "21" (two 1s)
   Term 4: "1211" (one 2, one 1)
   Term 5: "111221" (one 1, one 2, two 1s)
   Term 6: "312211" (three 1s, two 2s, one 1)

2. Step 2: Confirm term 6 output

   The method returns "312211" for input 6, matching the sequence rules.

3. Final Answer:

   "312211" -> Option D

4. Quick Check:

   Count and Say term 6 = "312211" [OK]

Hint: Build sequence stepwise from term 1 to 6 [OK]

Common Mistakes:

• Skipping terms or miscounting digits
• Mixing up digits order in output
• Choosing longer incorrect strings