Count and Say Problem in DSA Java - 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 the count and digit, and repeats until reaching the nth term.

Execution Sample

DSA Java

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 recursively generates the count-and-say sequence for the nth term.

Execution Table

Step	Operation	Current String	Count Consecutive Digits	Build Next String	Resulting String
1	Start with base case	1	N/A	N/A	1
2	Read '1', count digits	1	count=1 of '1'	Append '11'	11
3	Read '11', count digits	11	count=2 of '1'	Append '21'	21
4	Read '21', count digits	21	count=1 of '2', then count=1 of '1'	Append '12' then '11'	1211
5	Read '1211', count digits	1211	count=1 of '1', count=1 of '2', count=2 of '1'	Append '11', '12', '21'	111221
6	Read '111221', count digits	111221	count=3 of '1', count=2 of '2', count=1 of '1'	Append '31', '22', '11'	312211
Exit	Reached nth term	312211	N/A	N/A	312211

💡 Reached the nth term (6), stop recursion and return the string.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
n	6	6	6	6	6	6	6
prev	N/A	1	11	21	1211	111221	312211
count	N/A	1	2	1 then 1	1 then 1 then 2	3 then 2 then 1	N/A
result	N/A	11	21	1211	111221	312211	312211

Key Moments - 3 Insights

Why do we append count and digit only when the next digit is different?

Because we count consecutive identical digits first. When the next digit changes, we know how many times the previous digit appeared, so we append count and digit then (see execution_table rows 3-5).

Why do we append the last counted digits after the loop?

The loop only appends when it sees a digit change. The last group has no next digit to trigger append, so we append it after the loop (see code line after for-loop and execution_table last steps).

How does recursion help in this problem?

Recursion builds the sequence from the base case (n=1) upwards, each call depends on the previous term. This avoids manual iteration and keeps code simple (see execution_sample code and variable 'prev').

Visual Quiz - 3 Questions

Look at the execution_table at Step 4, what is the 'Build Next String' value?

A Append '12' then '11'

B Append '11' then '12'

C Append '21'

D Append '31' then '22' then '11'

💡 Hint

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

At which step does the count of '1's become 3 in the current string?

A Step 5

B Step 4

C Step 6

D Step 3

💡 Hint

Look at 'Count Consecutive Digits' column in execution_table for count=3 of '1'.

If we change the base case to start with '2' instead of '1', what would be the first resulting string after Step 2?

A 11

B 12

C 22

D 21

💡 Hint

The first string says how many times the digit appears, so count=1 of '2' becomes '12', but starting with '2' means the first string is '2', so count=1 of '2' is '12'.

Concept Snapshot

Count and Say Problem:
- Start with '1' as base case.
- For each next term, read previous term.
- Count consecutive digits and say count + digit.
- Repeat until nth term.
- Use recursion or iteration to build sequence.

Full Transcript

The Count and Say problem generates a sequence of strings where each term describes the previous term's digits. Starting with '1', each next term is formed by reading the previous term, counting consecutive digits, and writing the count followed by the digit. This process repeats until the nth term is reached. The code uses recursion to get the previous term, then iterates over it to build the next term. Key points include appending the last counted digits after the loop and understanding how recursion builds the sequence step-by-step.

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