0038. Count and Say¶

Table of Contents¶

Problem
Problem Breakdown
Approach 1: Iterative
Approach 2: Recursive
Complexity Comparison
Edge Cases
Common Mistakes
Related Problems
Visual Animation

Problem¶


Leetcode	38. Count and Say
Difficulty	Medium
Tags	`String`

Description¶

The count-and-say sequence is a sequence of digit strings defined by the recursive formula:

countAndSay(1) = "1"

countAndSay(n) is the run-length encoding of countAndSay(n - 1).

Run-length encoding (RLE) is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compress the string "3322251" we replace "33" with "23", replace "222" with "32", replace "5" with "15" and replace "1" with "11". Thus the compressed string becomes "23321511".

Given a positive integer n, return the nth element of the count-and-say sequence.

Examples¶

Example 1:
Input: n = 4
Output: "1211"
Explanation:
  countAndSay(1) = "1"
  countAndSay(2) = RLE of "1" = "11"       (one 1)
  countAndSay(3) = RLE of "11" = "21"      (two 1s)
  countAndSay(4) = RLE of "21" = "1211"    (one 2, one 1)

Example 2:
Input: n = 1
Output: "1"
Explanation: This is the base case.

Constraints¶

1 <= n <= 30

Problem Breakdown¶

1. What is being asked?¶

Generate the nth term of the count-and-say sequence. Each term is produced by "reading aloud" the digits of the previous term — counting consecutive runs of the same digit.

2. What is the input?¶

Parameter	Type	Description
`n`	`int`	The index (1-based) of the desired term in the sequence

Important observations about the input: - n is always at least 1 - The base case is n = 1 which returns "1" - Each term is built from the previous term

3. What is the output?¶

A string — the nth term of the count-and-say sequence
The string only contains digits 1, 2, and 3 (proven by the sequence properties)

4. Constraints analysis¶

Constraint	Significance
`n <= 30`	The 30th term has ~5,808 characters. No performance concerns.
`n >= 1`	Always valid input, base case is trivial.

5. Step-by-step example analysis¶

Building the sequence from n=1 to n=7¶

n=1: "1"
     Read: one 1
n=2: "11"
     Read: two 1s
n=3: "21"
     Read: one 2, one 1
n=4: "1211"
     Read: one 1, one 2, two 1s
n=5: "111221"
     Read: three 1s, two 2s, one 1
n=6: "312211"
     Read: one 3, one 1, two 2s, two 1s
n=7: "13112221"

Detailed RLE process for n=5 -> n=6¶

Input:  "111221"

Scan left to right:
  "111" → three 1s → "31"
  "22"  → two 2s   → "22"
  "1"   → one 1    → "11"

Concatenate: "31" + "22" + "11" = "312211"

Result: countAndSay(6) = "312211"

6. Key Observations¶

Sequential dependency — Each term depends on the previous term. We must build the sequence step by step.
Run-length encoding — The core operation is counting consecutive identical characters.
String grows — The length of each term generally grows, but n <= 30 keeps it manageable.
Only digits 1, 2, 3 — The sequence never produces digits larger than 3 (mathematical property).

7. Pattern identification¶

Pattern	Why it fits	Example
Iterative simulation	Build each term from the previous one	Count and Say (this problem)
Recursion	Natural recursive definition	Base case n=1, recurse on n-1

Chosen pattern: String simulation with RLE Reason: Directly simulate the process described in the problem. Either iterate from 1 to n, or recurse from n down to 1.

Approach 1: Iterative¶

Thought process¶

Start with "1" and build each subsequent term by performing run-length encoding on the current term. Repeat n-1 times to get the nth term.

For RLE: scan the string left to right, count consecutive identical characters, and append "count" + "digit" to the result.

Algorithm (step-by-step)¶

Initialize result = "1"
Repeat n-1 times: a. Initialize an empty string next b. Set i = 0 c. While i < len(result):
- Set digit = result[i], count = 1
- While i + count < len(result) and result[i + count] == digit: increment count
- Append str(count) + digit to next
- Move i += count d. Set result = next
Return result

Pseudocode¶

function countAndSay(n):
    result = "1"
    for step = 2 to n:
        next = ""
        i = 0
        while i < len(result):
            digit = result[i]
            count = 1
            while i + count < len(result) AND result[i + count] == digit:
                count++
            next += str(count) + digit
            i += count
        result = next
    return result

Complexity¶

	Complexity	Explanation
Time	O(n * L)	n iterations, each processing a string of length L (max ~5808 for n=30)
Space	O(L)	Store the current and next strings

Implementation¶

Go¶

// countAndSay — Iterative approach
// Time: O(n * L), Space: O(L)
func countAndSay(n int) string {
    result := "1"

    for step := 2; step <= n; step++ {
        var next strings.Builder
        i := 0

        for i < len(result) {
            digit := result[i]
            count := 1

            // Count consecutive identical digits
            for i+count < len(result) && result[i+count] == digit {
                count++
            }

            // Append count and digit
            next.WriteString(strconv.Itoa(count))
            next.WriteByte(digit)
            i += count
        }

        result = next.String()
    }

    return result
}

Java¶

class Solution {
    // countAndSay — Iterative approach
    // Time: O(n * L), Space: O(L)
    public String countAndSay(int n) {
        String result = "1";

        for (int step = 2; step <= n; step++) {
            StringBuilder next = new StringBuilder();
            int i = 0;

            while (i < result.length()) {
                char digit = result.charAt(i);
                int count = 1;

                // Count consecutive identical digits
                while (i + count < result.length() && result.charAt(i + count) == digit) {
                    count++;
                }

                // Append count and digit
                next.append(count);
                next.append(digit);
                i += count;
            }

            result = next.toString();
        }

        return result;
    }
}

Python¶

class Solution:
    def countAndSay(self, n: int) -> str:
        """
        Iterative approach
        Time: O(n * L), Space: O(L)
        """
        result = "1"

        for step in range(2, n + 1):
            next_result = []
            i = 0

            while i < len(result):
                digit = result[i]
                count = 1

                # Count consecutive identical digits
                while i + count < len(result) and result[i + count] == digit:
                    count += 1

                # Append count and digit
                next_result.append(str(count))
                next_result.append(digit)
                i += count

            result = "".join(next_result)

        return result

Dry Run¶

Input: n = 5

Step 1: result = "1" (base case)

Step 2: Process "1"
  i=0: digit='1', count=1 → "11"
  result = "11"

Step 3: Process "11"
  i=0: digit='1', count=2 → "21"
  result = "21"

Step 4: Process "21"
  i=0: digit='2', count=1 → "12"
  i=1: digit='1', count=1 → "11"
  result = "1211"

Step 5: Process "1211"
  i=0: digit='1', count=1 → "11"
  i=1: digit='2', count=1 → "12"
  i=2: digit='1', count=2 → "21"
  result = "111221"

Result: "111221"

Approach 2: Recursive¶

Thought process¶

The problem has a natural recursive definition: - Base case: countAndSay(1) = "1" - Recursive case: countAndSay(n) = RLE(countAndSay(n-1))

We can directly translate this definition into code. First recursively compute the (n-1)th term, then perform RLE on it.

Algorithm (step-by-step)¶

Base case: if n == 1, return "1"
Recursively compute prev = countAndSay(n - 1)
Perform run-length encoding on prev: a. Scan left to right, count consecutive identical digits b. Append str(count) + digit for each run
Return the RLE result

Pseudocode¶

function countAndSay(n):
    if n == 1:
        return "1"

    prev = countAndSay(n - 1)

    result = ""
    i = 0
    while i < len(prev):
        digit = prev[i]
        count = 1
        while i + count < len(prev) AND prev[i + count] == digit:
            count++
        result += str(count) + digit
        i += count

    return result

Complexity¶

	Complexity	Explanation
Time	O(n * L)	n recursive calls, each processing a string of length L
Space	O(n * L)	Recursion stack depth n, plus strings at each level

Implementation¶

Go¶

// countAndSay — Recursive approach
// Time: O(n * L), Space: O(n * L)
func countAndSay(n int) string {
    // Base case
    if n == 1 {
        return "1"
    }

    // Recursively get the previous term
    prev := countAndSay(n - 1)

    // Perform RLE on the previous term
    var result strings.Builder
    i := 0

    for i < len(prev) {
        digit := prev[i]
        count := 1

        // Count consecutive identical digits
        for i+count < len(prev) && prev[i+count] == digit {
            count++
        }

        // Append count and digit
        result.WriteString(strconv.Itoa(count))
        result.WriteByte(digit)
        i += count
    }

    return result.String()
}

Java¶

class Solution {
    // countAndSay — Recursive approach
    // Time: O(n * L), Space: O(n * L)
    public String countAndSay(int n) {
        // Base case
        if (n == 1) {
            return "1";
        }

        // Recursively get the previous term
        String prev = countAndSay(n - 1);

        // Perform RLE on the previous term
        StringBuilder result = new StringBuilder();
        int i = 0;

        while (i < prev.length()) {
            char digit = prev.charAt(i);
            int count = 1;

            // Count consecutive identical digits
            while (i + count < prev.length() && prev.charAt(i + count) == digit) {
                count++;
            }

            // Append count and digit
            result.append(count);
            result.append(digit);
            i += count;
        }

        return result.toString();
    }
}

Python¶

class Solution:
    def countAndSay(self, n: int) -> str:
        """
        Recursive approach
        Time: O(n * L), Space: O(n * L)
        """
        # Base case
        if n == 1:
            return "1"

        # Recursively get the previous term
        prev = self.countAndSay(n - 1)

        # Perform RLE on the previous term
        result = []
        i = 0

        while i < len(prev):
            digit = prev[i]
            count = 1

            # Count consecutive identical digits
            while i + count < len(prev) and prev[i + count] == digit:
                count += 1

            # Append count and digit
            result.append(str(count))
            result.append(digit)
            i += count

        return "".join(result)

Dry Run¶

Input: n = 4

countAndSay(4)
  → calls countAndSay(3)
    → calls countAndSay(2)
      → calls countAndSay(1)
      ← returns "1"
    RLE("1"):
      i=0: digit='1', count=1 → "11"
    ← returns "11"
  RLE("11"):
    i=0: digit='1', count=2 → "21"
  ← returns "21"
RLE("21"):
  i=0: digit='2', count=1 → "12"
  i=1: digit='1', count=1 → "11"
← returns "1211"

Call stack unwinding:
  countAndSay(1) = "1"
  countAndSay(2) = RLE("1")   = "11"
  countAndSay(3) = RLE("11")  = "21"
  countAndSay(4) = RLE("21")  = "1211"

Result: "1211"

Complexity Comparison¶

#	Approach	Time	Space	Pros	Cons
1	Iterative	O(n * L)	O(L)	Lower space usage, no stack overhead	Slightly more code
2	Recursive	O(n * L)	O(n * L)	Clean, mirrors problem definition	Extra stack space, risk of stack overflow for large n

Which solution to choose?¶

In an interview: Approach 1 (Iterative) — straightforward, efficient, easy to explain
In production: Approach 1 — better space complexity, no recursion overhead
On Leetcode: Both pass easily since n <= 30
For learning: Approach 2 — helps understand the recursive nature of the problem

Edge Cases¶

#	Case	Input	Expected Output	Reason
1	Base case	`n=1`	`"1"`	First term is defined as "1"
2	Second term	`n=2`	`"11"`	One 1
3	Third term	`n=3`	`"21"`	Two 1s
4	Multiple runs	`n=5`	`"111221"`	Three 1s, two 2s, one 1
5	Maximum input	`n=30`	5808-char string	Must handle long strings

Common Mistakes¶

Mistake 1: Confusing the "count" and "digit" order¶

# WRONG — digit before count
result += digit + str(count)  # "12" instead of "21" for two 1s

# CORRECT — count before digit
result += str(count) + digit  # "21" for two 1s

Reason: Run-length encoding is "how many" followed by "what digit". "21" means "two 1s", not "twelve".

Mistake 2: Not handling consecutive runs correctly¶

# WRONG — only comparing adjacent pairs
for i in range(len(s)):
    if i + 1 < len(s) and s[i] == s[i + 1]:
        count += 1
    else:
        result += str(count) + s[i]
        count = 1

# CORRECT — use a while loop to consume entire run
i = 0
while i < len(s):
    digit = s[i]
    count = 1
    while i + count < len(s) and s[i + count] == digit:
        count += 1
    result += str(count) + digit
    i += count

Reason: The first approach can miscount runs of length 3+ if the logic for resetting count is not carefully handled.

Mistake 3: Off-by-one in iteration count¶

# WRONG — iterating n times instead of n-1
result = "1"
for i in range(n):  # This produces the (n+1)th term!
    result = rle(result)

# CORRECT — iterate n-1 times (we start with the 1st term)
result = "1"
for i in range(n - 1):
    result = rle(result)

Reason: We start with "1" as the 1st term, so we only need n-1 transformations.

#	Problem	Difficulty	Similarity
1	443. String Compression	Medium	Run-length encoding
2	271. Encode and Decode Strings	Medium	String encoding
3	394. Decode String	Medium	String decoding with counts
4	1313. Decompress Run-Length Encoded List	Easy	RLE decompression

Visual Animation¶

Interactive animation: animation.html

In the animation: - Step-by-step sequence building from n=1 to chosen n - Run-length encoding visualization with character grouping - Preset buttons for n=1 through n=7 - Play/Pause/Step/Reset controls with speed slider - Detailed log of each step's RLE process

0038. Count and Say¶

Table of Contents¶

Problem¶

Description¶

Examples¶

Constraints¶

Problem Breakdown¶

1. What is being asked?¶

2. What is the input?¶

3. What is the output?¶

4. Constraints analysis¶

5. Step-by-step example analysis¶

Building the sequence from n=1 to n=7¶

Detailed RLE process for n=5 -> n=6¶

6. Key Observations¶

7. Pattern identification¶

Approach 1: Iterative¶

Thought process¶

Algorithm (step-by-step)¶

Pseudocode¶

Complexity¶

Implementation¶

Go¶

Java¶

Python¶

Dry Run¶

Approach 2: Recursive¶

Thought process¶

Algorithm (step-by-step)¶

Pseudocode¶

Complexity¶

Implementation¶

Go¶

Java¶

Python¶

Dry Run¶

Complexity Comparison¶

Which solution to choose?¶

Edge Cases¶

Common Mistakes¶

Mistake 1: Confusing the "count" and "digit" order¶

Mistake 2: Not handling consecutive runs correctly¶

Mistake 3: Off-by-one in iteration count¶

Related Problems¶

Visual Animation¶