Count and Say Problem in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Count and Say for n=4

What is the output of the Count and Say sequence for n=4?

DSA C++

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;
}

cout << countAndSay(4);

A 1211

B 111221

C 21

D 312211

💡 Hint

Trace the sequence from n=1 to n=4 step by step.

The sequence starts as: 1 (n=1), 11 (n=2), 21 (n=3), 1211 (n=4). Each term describes the previous term's digits.

❓ Predict Output

intermediate

Output of Count and Say for n=5

What is the output of the Count and Say sequence for n=5?

DSA C++

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;
}

cout << countAndSay(5);

A 211211

B 312211

C 111221

D 1211

💡 Hint

Build the sequence from n=1 up to n=5.

The sequence terms are: 1, 11, 21, 1211, 111221 for n=1 to 5 respectively.

🔧 Debug

advanced

Identify the error in Count and Say implementation

What error will the following code produce when called with n=3?

DSA C++

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 (prev[i] == prev[i + 1]) {
            count++;
        } else {
            result += to_string(count) + prev[i];
            count = 1;
        }
    }
    return result;
}

cout << countAndSay(3);

A Output: 1211

B Output: 21

C Compilation error

D Runtime error: out_of_range exception

💡 Hint

Check the loop index and array access carefully.

The loop accesses prev[i + 1] without checking bounds, causing out_of_range at last iteration.

🧠 Conceptual

advanced

Time complexity of Count and Say sequence generation

What is the time complexity of generating the nth term in the Count and Say sequence using the recursive approach?

A O(n^2)

B O(2^n)

C O(n)

D O(n!)

💡 Hint

Consider how the length of each term grows and the recursive calls.

Each term roughly doubles in length, and recursion calls n times, leading to exponential time O(2^n).

🚀 Application

expert

Count and Say sequence term length for n=6

What is the length of the 6th term in the Count and Say sequence?

A 6

B 8

C 12

D 10

💡 Hint

Generate terms up to n=6 and count characters.

Terms: 1(1), 11(2), 21(2), 1211(4), 111221(6), 312211(6 chars). The 6th term "312211" has length 6.

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