Count and Say Problem in DSA C# - Time & Space Complexity

Time Complexity: Count and Say Problem

O(2^n)

Understanding Time Complexity

We want to understand how the time needed to generate the "count and say" sequence grows as we ask for longer sequences.

How does the work increase when we move from one sequence number to the next?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public 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();
}
    

This code generates the nth term of the "count and say" sequence by building it from the previous term.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursion calls itself n times, each time looping through the previous string.
• How many times: The recursion depth is n, and each level loops through a string whose length grows with n.

How Execution Grows With Input

Each new sequence is built by reading the previous sequence, which roughly doubles in length each time.

Input Size (n)	Approx. Operations
1	1 (base case)
5	About 16 operations (string length grows roughly doubling each step)
10	About 512 operations (string length grows exponentially)

Pattern observation: The work grows exponentially because each term depends on the full previous term, which grows roughly twice as long each time.

Final Time Complexity

Time Complexity: O(2^n)

This means the time needed roughly doubles with each increase in n, making it grow very fast as n gets bigger.

Common Mistake

[X] Wrong: "Each step only takes linear time, so overall time is O(n^2)."

[OK] Correct: The string length grows exponentially, so each step processes a much longer string, making total time exponential, not just quadratic.

Interview Connect

Understanding how recursive string building can cause exponential growth helps you explain performance clearly and shows you can analyze complex recursive patterns.

Self-Check

"What if we stored all previous results to avoid repeated work? How would that change the time complexity?"

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