Count and Say Problem in DSA Java - 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 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) {
        System.out.println(countAndSay(4));
    }
}

A 1211

B 21

C 111221

D 312211

💡 Hint

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

The Count and Say sequence starts with "1" for n=1.

For n=2, it reads "1" as "one 1" → "11".

For n=3, it reads "11" as "two 1s" → "21".

For n=4, it reads "21" as "one 2, then one 1" → "1211".

❓ 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 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) {
        System.out.println(countAndSay(5));
    }
}

A 312211

B 1211

C 111221

D 211211

💡 Hint

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

Sequence steps:

• n=1: "1"
• n=2: "11" (one 1)
• n=3: "21" (two 1s)
• n=4: "1211" (one 2, one 1)
• n=5: "111221" (one 1, one 2, two 1s)

🔧 Debug

advanced

Identify the error in Count and Say implementation

What error will this code produce when calling countAndSay(3)?

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 = 0; 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();
    }
}

A Infinite recursion

B NullPointerException

C No error, outputs "21"

D StringIndexOutOfBoundsException

💡 Hint

Look at the loop index and how characters are accessed.

The loop starts at i=0 and tries to access prev.charAt(i - 1) which is prev.charAt(-1) at first iteration.

This causes StringIndexOutOfBoundsException.

🧠 Conceptual

advanced

Time complexity of Count and Say algorithm

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

A O(2^n)

B O(n * 2^n)

C O(n)

D O(n^2)

💡 Hint

Consider the length growth of the sequence and recursion depth.

The length of the sequence roughly doubles each step, so length at step n is about 2^n.

Each recursive call processes the previous string, so total work is about n times 2^n.

🚀 Application

expert

Count and Say sequence for large n optimization

Which approach is best to efficiently generate the Count and Say sequence for large n (e.g., n=30)?

A Use iterative approach building from n=1 up to n=30

B Use recursion with memoization to store previous results

C Use direct formula to compute nth term without iteration

D Use parallel threads to compute each term independently

💡 Hint

Think about avoiding deep recursion and repeated work.

Recursion for large n causes stack overflow and repeated calls.

There is no direct formula for nth term.

Parallel threads cannot compute terms independently because each depends on the previous.

Iterative approach builds terms one by one efficiently.

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