Count and Say Problem in DSA C++ - Execution Trace

Concept Flow - Count and Say Problem

Start with '1'

↓

Read current string

↓

Count consecutive digits

↓

Build next string: count + digit

↓

Repeat for n times

↓

Output final string

The process starts with '1', then reads the string, counts consecutive digits, builds the next string by saying counts and digits, and repeats until reaching the nth term.

Execution Sample

DSA C++

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

Generates the nth term of the count-and-say sequence by iteratively reading and building strings.

Execution Table

Step	Operation	Current String	Count	Digit Processed	Next String Built
1	Start	1	-	-	1
2	Read '1'	1	1	1	11
3	Read '11'	11	2	1	21
4	Read '21'	21	1	2	1211
5	Read '1211'	1211	1	1	111221
6	Read '111221'	111221	3	1	312211
7	Read '312211'	312211	1	3	13112221
Exit	Reached n=7	13112221	-	-	Done

💡 Reached the 7th term, sequence generation complete.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	Final
result	1	11	21	1211	111221	312211	13112221	13112221
temp	-	11	21	1211	111221	312211	13112221	-
count	-	1-2	1-2	1-1	1-3	1-2	1-2	-

Key Moments - 3 Insights

Why do we add the last counted group after the inner loop ends?

Because the inner loop only adds groups when a digit changes, the last group remains unadded after the loop. We add it after the loop to include the final counted digits, as shown in step 2 and 3 of the execution_table.

Why do we start counting from 1 for each new digit group?

Each new digit group starts with at least one occurrence of that digit, so count is reset to 1 when a new digit is found, as seen in the 'Count' column resetting in steps 3 and 4.

How does the string 'result' change after each iteration?

The 'result' string updates to the newly built 'temp' string after processing all digit groups, progressively growing as the sequence advances, as tracked in the variable_tracker.

Visual Quiz - 3 Questions

Look at the execution_table at Step 4. What is the 'Next String Built' after reading '21'?

A 1211

B 111221

C 21

D 312211

💡 Hint

Check the 'Next String Built' column at Step 4 in the execution_table.

At which step does the 'result' variable first become '111221' according to variable_tracker?

A After 4

B After 5

C After 3

D After 6

💡 Hint

Look at the 'result' row in variable_tracker and find when it equals '111221'.

If the initial string was '2' instead of '1', how would the 'Next String Built' at Step 2 change?

A It would be '11'

B It would be '21'

C It would be '12'

D It would be '22'

💡 Hint

Think about counting digits in the initial string and how the count-and-say works, referencing Step 2 in execution_table.

Concept Snapshot

Count and Say Problem:
- Start with '1' as the first term.
- For each term, read the previous term.
- Count consecutive digits and say count + digit.
- Repeat until nth term.
- Output the final string.

Full Transcript

The Count and Say problem generates a sequence starting from '1'. Each next term is formed by reading the previous term, counting consecutive digits, and writing down the count followed by the digit. This process repeats until the nth term is reached. The code uses a loop to build each term from the previous one by iterating through the string, counting digits, and constructing the next string. Key points include adding the last counted group after the loop and resetting the count for new digits. The sequence grows as shown in the execution table and variable tracker.

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