Count and Say Problem in DSA Java

Mental Model

We describe a number by counting consecutive digits and saying their counts and values aloud.

Analogy: Imagine reading a row of colored beads and saying how many beads of each color appear in a row before the color changes.

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 digits and say them aloud

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

21

Why: Two 1s means '2' then '1'

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

1211

Why: Say counts and digits in order

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

111221

Why: Count each group of digits consecutively

Result:

111221

Annotated Code

DSA Java

public class CountAndSay {
    public static String countAndSay(int n) {
        if (n == 1) return "1";
        String prev = countAndSay(n - 1);
        StringBuilder result = new StringBuilder();
        int count = 1;
        for (int i = 1; i < prev.length(); i++) {
            if (prev.charAt(i) == prev.charAt(i - 1)) {
                count++;
            } else {
                result.append(count).append(prev.charAt(i - 1));
                count = 1;
            }
        }
        result.append(count).append(prev.charAt(prev.length() - 1));
        return result.toString();
    }

    public static void main(String[] args) {
        int n = 5;
        System.out.println(countAndSay(n));
    }
}

if (n == 1) return "1";

base case: first term is '1'

String prev = countAndSay(n - 1);

recursively get previous term

for (int i = 1; i < prev.length(); i++) {

iterate over previous term characters

if (prev.charAt(i) == prev.charAt(i - 1)) { count++; }

count consecutive same digits

else { result.append(count).append(prev.charAt(i - 1)); count = 1; }

append count and digit when digit changes, reset count

result.append(count).append(prev.charAt(prev.length() - 1));

append last counted group after loop

Output

111221

Complexity Analysis

Time: O(m) where m is the length of the nth term because we process each character once

Space: O(m) for the string builder storing the new term

vs Alternative: Recursive approach is simple but can be slower for large n due to repeated calls; iterative builds terms step-by-step with similar complexity

Edge Cases

n = 1 (smallest input)

Returns '1' immediately as base case

DSA Java

if (n == 1) return "1";

n = 0 or negative (invalid input)

Not handled explicitly, may cause stack overflow or incorrect behavior

DSA Java

No explicit guard; user should input n >= 1

When to Use This Pattern

When asked to generate a sequence where each term describes the previous term's digit counts, use the Count and Say pattern to build terms step-by-step.

Common Mistakes

Mistake: Forgetting to append the last counted group after the loop
Fix: Add a final append after the loop to include the last group

Mistake: Not handling the base case n=1 correctly
Fix: Return '1' immediately when n equals 1

Summary

Generates the nth term of a sequence where each term describes the count of digits in the previous term.

Use when you need to build a sequence by describing groups of digits in the previous term.

The key insight is to count consecutive digits and say their counts and values aloud to form the next term.

Practice

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

easy

A. It reverses the digits of a number.

B. It describes the count of digits seen in the previous term.

C. It sorts numbers in ascending order.

D. It adds two numbers repeatedly.

Solution

1. Step 1: Understand the sequence definition

   The Count and Say sequence describes each term by counting the number of digits appearing consecutively in the previous term.

2. Step 2: Relate to example

   For example, starting with "1", the next term is "11" because there is one '1'. Then "21" because there are two '1's, and so on.

3. Final Answer:

   It describes the count of digits seen in the previous term. -> Option B

4. Quick Check:

   Count and Say = Count digits [OK]

Hint: Remember: count digits, then say them aloud [OK]

Common Mistakes:

• Thinking it sorts digits
• Confusing with reversing digits
• Assuming it adds numbers

2. Which of the following Java method signatures is correct for a function that returns the nth term of the Count and Say sequence?

easy

A. public String countAndSay(int n)

B. public int countAndSay(String n)

C. private void countAndSay(int n)

D. public String countAndSay()

Solution

1. Step 1: Identify correct return type and parameters

   The function should return a String representing the sequence term and take an integer n as input.

2. Step 2: Check method signature options

   public String countAndSay(int n) matches this: public String countAndSay(int n). Others either have wrong return type, parameters, or access modifiers.

3. Final Answer:

   public String countAndSay(int n) -> Option A

4. Quick Check:

   Return String, input int n [OK]

Hint: Return String, input int n for Count and Say method [OK]

Common Mistakes:

• Using void return type
• Wrong parameter type
• Missing input parameter

3. What is the output of the following Java code snippet?

System.out.println(countAndSay(4));

Assuming countAndSay is correctly implemented for the Count and Say sequence.

medium

A. "111221"

B. "21"

C. "1211"

D. "312211"

Solution

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

   Term 1: "1"
   Term 2: "11" (one 1)
   Term 3: "21" (two 1s)
   Term 4: "1211" (one 2, two 1s)

2. Step 2: Confirm output for n=4

   The output for countAndSay(4) is "1211".

3. Final Answer:

   "1211" -> Option C

4. Quick Check:

   Term 4 = "1211" [OK]

Hint: Write terms stepwise to find nth output [OK]

Common Mistakes:

• Confusing term 3 and 4
• Skipping counting digits correctly
• Mixing digits order

4. Identify the error in this Java code snippet for Count and Say:

public 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.length(); i++) {
    if (prev.charAt(i) == prev.charAt(i - 1)) {
      count++;
    } else {
      result += count + prev.charAt(i);
      count = 1;
    }
  }
  return result;
}

medium

A. The method should return an integer, not a String.

B. The code appends count before the correct character in else block.

C. The base case for n == 1 is missing.

D. The loop does not handle the last group of characters after the loop ends.

Solution

1. Step 1: Analyze loop behavior

   The loop processes characters but does not append the last counted group after finishing the loop.

2. Step 2: Identify missing step

   After the loop, the last counted group must be appended to the result to complete the sequence term.

3. Final Answer:

   The loop does not handle the last group of characters after the loop ends. -> Option D

4. Quick Check:

   Missing last group append causes incomplete output [OK]

Hint: Always append last counted group after loop ends [OK]

Common Mistakes:

• Forgetting to append last group
• Appending wrong character
• Incorrect base case

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

hard

A. "312211"

B. "13112221"

C. "1113213211"

D. "21133112"

Solution

1. Step 1: Generate terms up to 6

   Term 1: "1"
   Term 2: "11" (one 1)
   Term 3: "21" (two 1s)
   Term 4: "1211" (one 2, two 1s)
   Term 5: "111221" (three 1s, two 2s, one 1)
   Term 6: "312211" (three 1s, two 2s, one 1)

2. Step 2: Confirm 6th term

   The 6th term is "312211".

3. Final Answer:

   "312211" -> Option A

4. Quick Check:

   Term 6 = "312211" [OK]

Hint: Build terms stepwise to avoid mistakes [OK]

Common Mistakes:

• Skipping terms or mixing order
• Miscounting digits
• Confusing digits in groups