Count and Say Problem in DSA C#

Mental Model

You describe a number by counting how many times each digit appears in order, then say that count and digit out loud as a new number.

Analogy: Imagine reading a list of colored beads aloud by saying how many beads of each color you see in a row, then writing down that description as a new list.

1 -> 11 -> 21 -> 1211 -> 111221 -> null
↑ start with '1' and build next by describing current

Dry Run Walkthrough

Input: n = 5 (generate the 5th term starting from '1')

Goal: Find the 5th number in the count and say sequence

Step 1: Start with '1' as the first term

1

Why: The sequence always starts with '1'

Step 2: Read '1': one '1' -> next term is '11'

11

Why: Describe the first term by counting digits

Step 3: Read '11': two '1's -> next term is '21'

21

Why: Count consecutive digits and say them

Step 4: Read '21': one '2', one '1' -> next term is '1211'

1211

Why: Describe each group of digits in order

Step 5: Read '1211': one '1', one '2', two '1's -> next term is '111221'

111221

Why: Count each group and say count then digit

Result:

111221

Annotated Code

DSA C#

using System;

public class CountAndSay {
    public static string CountAndSaySequence(int n) {
        if (n <= 0) return "";
        string result = "1";
        for (int i = 1; i < n; i++) {
            result = NextTerm(result);
        }
        return result;
    }

    private static string NextTerm(string term) {
        string next = "";
        int count = 1;
        for (int i = 1; i < term.Length; i++) {
            if (term[i] == term[i - 1]) {
                count++;
            } else {
                next += count.ToString() + term[i - 1];
                count = 1;
            }
        }
        next += count.ToString() + term[term.Length - 1];
        return next;
    }

    public static void Main() {
        int n = 5;
        string output = CountAndSaySequence(n);
        Console.WriteLine(output);
    }
}

for (int i = 1; i < n; i++) {

repeat building next term until reaching nth term

result = NextTerm(result);

generate next term by describing current term

if (term[i] == term[i - 1]) { count++; } else { next += count.ToString() + term[i - 1]; count = 1; }

count consecutive same digits and append count and digit when sequence changes

next += count.ToString() + term[term.Length - 1];

append last counted group after loop ends

Output

111221

Complexity Analysis

Time: O(m * n) because for each of the n terms we process the string of length m which grows roughly exponentially

Space: O(m) for storing the current term string which grows with each iteration

vs Alternative: A naive approach might try to generate all terms from scratch each time, but this iterative approach builds each term from the previous efficiently

Edge Cases

n = 0 or negative

Returns empty string as no sequence term exists

DSA C#

if (n <= 0) return "";

n = 1

Returns '1' as the first term

DSA C#

string result = "1";

When to Use This Pattern

When asked to generate a sequence where each term describes the previous term's digit counts, use the Count and Say pattern to build terms iteratively.

Common Mistakes

Mistake: Not appending the last counted group after finishing the loop
Fix: Add the last count and digit after the loop ends to complete the term

Mistake: Starting count at 0 instead of 1 for the first digit
Fix: Initialize count to 1 before looping through the string

Summary

Generates the nth term of the count and say sequence by describing digits of the previous term.

Use when you need to build a sequence where each term narrates the count of digits in the previous term.

The key insight is to group consecutive digits and say their count followed by the digit to form the next term.

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