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.

string countAndSay(int n) {
    if (n == 1) return "1";
    string prev = countAndSay(n - 1);
    string result = "";
    int count = 1;
    for (int i = 1; i < prev.size(); i++) {
        if (prev[i] == prev[i - 1]) count++;
        else {
            result += to_string(count) + prev[i - 1];
            count = 1;
        }
    }
    result += to_string(count) + prev.back();
    return result;
}
    

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 building the next sequence by scanning the previous string.
• How many times: The for-loop runs once for each character in the previous sequence, which grows with n.

How Execution Grows With Input

Each new sequence is built by reading the previous one, which roughly grows in length by about 1.3 to 1.4 times each time (not exactly doubling).

Input Size (n)	Approx. Operations
1	1 (base case)
5	About 15 (sequence length grows around 1.3^n)
10	About 1000 (sequence length grows exponentially)

Pattern observation: The work grows roughly exponentially because each term depends on the full previous term, which grows in length.

Final Time Complexity

Time Complexity: O(2^n) is a common upper bound, but the actual growth rate is approximately O(1.3^n) due to the sequence's growth rate.

This means the time needed grows exponentially with each increase in n, because the sequence length grows exponentially.

Common Mistake

[X] Wrong: "The time grows linearly with n because we just call the function n times."

[OK] Correct: Each call processes a string that grows exponentially, so the work inside each call increases a lot, not just the number of calls.

Interview Connect

Understanding how recursive calls and string processing combine to affect time helps you explain and optimize similar problems in interviews.

Self-Check

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

Practice

1. What does the "Count and Say" sequence represent in simple terms?

easy

A. A sequence where each term describes the count of digits in the previous term

B. A sequence of random numbers generated by counting

C. A sequence where each term is the sum of previous two terms

D. A sequence of prime numbers only

Solution

1. Step 1: Understand the sequence definition

   The Count and Say sequence generates each term by describing the digits of the previous term.

2. Step 2: Example of the sequence

   Starting with "1", the next term is "11" (one 1), then "21" (two 1s), and so on.

3. Final Answer:

   A sequence where each term describes the count of digits in the previous term -> Option A

4. Quick Check:

   Count and Say = describe previous term counts [OK]

Hint: Remember: each term says counts of digits in previous term [OK]

Common Mistakes:

• Thinking it's a sum sequence
• Confusing with random or prime sequences
• Ignoring the description aspect

2. Which of the following is the correct way to start the Count and Say sequence in C++?

easy

A. int s = 1;

B. vector<int> s = {1};

C. char s = '1';

D. string s = "1";

Solution

1. Step 1: Identify the data type for sequence terms

   Each term is a string representing digits, so string type is appropriate.

2. Step 2: Check initialization

   Initializing with string "1" correctly starts the sequence.

3. Final Answer:

   string s = "1"; -> Option D

4. Quick Check:

   Sequence terms are strings, so string initialization [OK]

Hint: Use string type to hold digit sequences [OK]

Common Mistakes:

• Using int instead of string
• Using char which holds only one character
• Using vector<int> unnecessarily

3. What is the output of the following C++ code snippet for n = 4?

string countAndSay(int n) {
    if (n == 1) return "1";
    string prev = countAndSay(n - 1);
    string result = "";
    int count = 1;
    for (int i = 1; i <= prev.size(); i++) {
        if (i < prev.size() && prev[i] == prev[i - 1]) {
            count++;
        } else {
            result += to_string(count) + prev[i - 1];
            count = 1;
        }
    }
    return result;
}

cout << countAndSay(4);

medium

A. 21

B. 111221

C. 1211

D. 312211

Solution

1. Step 1: Trace the sequence up to n=4

   n=1: "1"; n=2: "11" (one 1); n=3: "21" (two 1s); n=4: "1211" (one 2, one 1)

2. Step 2: Confirm code logic matches sequence

   The code counts consecutive digits and appends count + digit, matching the sequence.

3. Final Answer:

   1211 -> Option C

4. Quick Check:

   Count and Say term 4 = 1211 [OK]

Hint: Write terms stepwise to avoid confusion [OK]

Common Mistakes:

• Confusing term 3 and 4
• Miscounting consecutive digits
• Off-by-one errors in loop

4. Identify the bug in this C++ code snippet for Count and Say:

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.size(); i++) {
        if (i + 1 < prev.size() && prev[i] == prev[i + 1]) {
            count++;
        } else {
            result += to_string(count) + prev[i];
            count = 1;
        }
    }
    return result;
}

medium

A. No bug, code is correct

B. Count variable is not reset properly

C. Loop index should start from 1, not 0

D. The condition in if statement should compare prev[i] and prev[i-1]

Solution

1. Step 1: Analyze loop and conditions

   The loop starts at 0 and compares current and next characters correctly.

2. Step 2: Check count reset and result building

   Count increments on matching digits and resets after appending count and digit to result.

3. Final Answer:

   No bug, code is correct -> Option A

4. Quick Check:

   Loop and count logic correct [OK]

Hint: Check loop indices and comparisons carefully [OK]

Common Mistakes:

• Starting loop at 1 causing index errors
• Incorrect comparison direction
• Forgetting to reset count

5. Given the Count and Say sequence starting with "1", what is the 6th term in the sequence?

hard

A. 1211

B. 312211

C. 1113213211

D. 13112221

Solution

1. Step 1: List terms up to 6

   Term 1: "1"; Term 2: "11"; Term 3: "21"; Term 4: "1211"; Term 5: "111221"; Term 6: "312211"

2. Step 2: Confirm term 6 matches option

   Term 6 describes term 5: three 1s, two 2s, one 1 -> "312211"

3. Final Answer:

   312211 -> Option B

4. Quick Check:

   Term 6 = 312211 by counting term 5 digits [OK]

Hint: Write terms stepwise to avoid confusion [OK]

Common Mistakes:

• Mixing term 5 and 6
• Miscounting digits in term 5
• Skipping terms or misreading digits