Count and Say Problem in DSA C++

Mental Model

We describe a number by counting consecutive digits and saying them out loud as a new string.

Analogy: Imagine reading a row of colored beads and saying how many beads of each color you see in order, then writing that down as a new row.

1 -> 11 -> 21 -> 1211 -> 111221 -> null

Dry Run Walkthrough

Input: n = 5 (generate the 5th term in the count and say sequence)

Goal: Find the 5th term by repeatedly describing the previous term

Step 1: Start with '1' as the first term

1

Why: The sequence always starts with '1'

Step 2: Describe '1': one 1 -> '11'

11

Why: Count the single '1' and say it

Step 3: Describe '11': two 1s -> '21'

21

Why: Count two '1's and say '2' then '1'

Step 4: Describe '21': one 2, one 1 -> '1211'

1211

Why: Count one '2' then one '1'

Step 5: Describe '1211': one 1, one 2, two 1s -> '111221'

111221

Why: Count each group of digits in order

Result:

111221

Annotated Code

DSA C++

#include <iostream>
#include <string>

using namespace std;

string countAndSay(int n) {
    if (n == 1) return "1"; // base case
    string prev = countAndSay(n - 1); // get previous term
    string result = "";
    int count = 1;
    for (int i = 1; i <= (int)prev.size(); i++) {
        if (i < (int)prev.size() && prev[i] == prev[i - 1]) {
            count++; // count repeated digits
        } else {
            result += to_string(count) + prev[i - 1]; // say count and digit
            count = 1; // reset count
        }
    }
    return result;
}

int main() {
    int n = 5;
    cout << countAndSay(n) << endl;
    return 0;
}

if (n == 1) return "1"; // base case

return the first term '1' to start the sequence

string prev = countAndSay(n - 1); // get previous term

recursively get the previous term to describe

if (i < (int)prev.size() && prev[i] == prev[i - 1]) { count++; }

count consecutive same digits

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

append count and digit to result, then reset count

Output

111221

Complexity Analysis

Time: O(m * n) because for each term up to n, we process the string of length m which grows roughly exponentially

Space: O(m) for storing the current term string, where m is the length of the nth term

vs Alternative: Iterative approach can avoid recursion stack but time complexity remains similar; naive manual counting is the same concept

Edge Cases

n = 1 (smallest input)

Returns '1' immediately as base case

DSA C++

if (n == 1) return "1"; // base case

n is large (e.g., 30)

String length grows large, may cause performance issues but algorithm still works

DSA C++

string prev = countAndSay(n - 1); // get previous term

When to Use This Pattern

When you see a problem asking to generate a sequence by describing previous terms, think of the Count and Say pattern because it builds each term from the last by grouping digits.

Common Mistakes

Mistake: Not resetting the count after appending to result
Fix: Reset count to 1 after appending count and digit to result

Mistake: Off-by-one error in loop causing out-of-bounds access
Fix: Loop up to i <= prev.size() and check bounds carefully before accessing prev[i]

Summary

Generates the nth term of the count and say sequence by describing digits of the previous term.

Use when you need to build a sequence where each term describes the previous one by counting digits.

The key insight is to group consecutive digits and say their count followed by the digit to form the next term.

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