Count and Say Problem in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the first term of the count-and-say sequence.

DSA C++

string countAndSay(int n) {
    string result = [1];
    // rest of the code
    return result;
}

Drag options to blanks, or click blank then click option'

A"2"

B"0"

C"1"

D""

Attempts:

3 left

💡 Hint

Common Mistakes

Using an empty string instead of "1".

Starting with "0" or "2" which are incorrect.

The count-and-say sequence starts with "1" as the first term.

2fill in blank

medium

Complete the code to iterate from the second term up to the nth term.

DSA C++

for (int i = 2; i <= [1]; i++) {
    // generate next term
}

Drag options to blanks, or click blank then click option'

Ai

Bn

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using the loop variable i as the limit.

Starting from 1 instead of 2.

We iterate from 2 to n to build each term of the sequence.

3fill in blank

hard

Fix the error in the code that counts consecutive characters in the current term.

DSA C++

for (int j = 1; j < current.size(); j++) {
    if (current[j] == current[j - 1]) {
        count[1];
    } else {
        // append count and character
        count = 1;
    }
}

Drag options to blanks, or click blank then click option'

A++

B+=

C--

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using decrement operator -- instead of increment.

Assigning count instead of incrementing.

We increment count by one when consecutive characters match.

4fill in blank

hard

Fill both blanks to append the count and character to the next term string.

DSA C++

next += to_string([1]);
next += [2];

Drag options to blanks, or click blank then click option'

Acount

Bcurrent[j]

Ccurrent[j - 1]

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Appending the wrong character index.

Appending the loop variable instead of count.

We append the count and the character before the current one to build the next term.

5fill in blank

hard

Fill in the blanks to update the current term and reset count for the next iteration.

DSA C++

current = [1];
count = [2];
// continue loop

Drag options to blanks, or click blank then click option'

Anext

B1

C0

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Not updating current to next.

Resetting count incorrectly.

We set current to next and reset count to 1 for the next iteration.

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