Count and Say Problem in DSA Java - Time & Space Complexity

Time Complexity: Count and Say Problem

O(2^n)

Understanding Time Complexity

We want to understand how the time needed to generate the "count and say" sequence grows as we ask for more terms.

How does the work increase when we move from one term to the next?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public 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();
}
    

This code generates the nth term of the "count and say" sequence by building it from the previous term.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursion calls itself to get the previous term, then loops through that term's characters.
• How many times: The recursion happens n times, and each loop runs over the length of the previous term.

How Execution Grows With Input

Each term is built by reading the previous term, which grows roughly by a factor of about 1.3 to 1.5 each time.

Input Size (n)	Approx. Total Operations
1	1
5	~10
10	~100

Pattern observation: The length of each term grows roughly exponentially, so the work to build each term also grows quickly.

Final Time Complexity

Time Complexity: O(2^n)

This means the time needed roughly doubles with each new term, because each term depends on the full previous term.

Common Mistake

[X] Wrong: "Each term takes only linear time O(n) because we just loop once."

[OK] Correct: Each term depends on the previous term, which grows in length exponentially, so the total work adds up exponentially, not just linearly.

Interview Connect

Understanding how recursive calls and growing input sizes affect time helps you explain and analyze similar problems clearly and confidently.

Self-Check

"What if we stored all previous terms instead of recomputing them? How would the time complexity change?"

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