Problem Summary

The count-and-say sequence is defined as follows:

• Term 1 is "1".
• Each subsequent term is generated by reading aloud the digits of the previous term.
• Reading aloud means grouping consecutive identical digits and describing them by:

<number of digits><digit>

For example:

• "1" becomes "11"
• "11" becomes "21"
• "21" becomes "1211"
• "1211" becomes "111221"

Your task:
Given an integer n, return the nth term of this sequence as a string.

Constraints
1 ≤ n ≤ 30

---

Understanding the Sequence

List of first few terms:

1. "1"
2. "11" → one 1
3. "21" → two 1s
4. "1211" → one 2, one 1
5. "111221"→ one 1, one 2, two 1s
6. "312211"

Each term depends only on the previous term.

---

Approach (Simple English)

To generate the nth term:

1. Start with "1" for n = 1.
2. For each step from 2 to n:
   • Read the previous term.
   • Count consecutive identical characters.
   • Build the new string using <count><digit>.

Why This Works

The problem is purely iterative:
each term is generated by processing characters of the previous term in order.

This gives a time complexity roughly proportional to the length of the terms.

---

Algorithm Steps

For each iteration:

1. Initialize an empty StringBuilder.
2. Scan the previous term with a pointer:
   • Count how many times the same digit repeats.
   • Append count and digit.
3. Continue until all characters are processed.

Return the final StringBuilder as a string.

---

Java Code (Iterative Solution)

class Solution {
    public String countAndSay(int n) {

        String result = "1";

        for (int i = 2; i <= n; i++) {
            result = generate(result);
        }

        return result;
    }

    private String generate(String prev) {
        StringBuilder sb = new StringBuilder();

        int count = 1;
        for (int i = 1; i < prev.length(); i++) {
            if (prev.charAt(i) == prev.charAt(i - 1)) {
                count++;
            } else {
                sb.append(count);
                sb.append(prev.charAt(i - 1));
                count = 1;
            }
        }

        // Append the last group
        sb.append(count);
        sb.append(prev.charAt(prev.length() - 1));

        return sb.toString();
    }
}

---

Dry Run

Generate term 5:

Term 1: "1"
Term 2: read "1" → "11"
Term 3: read "11" → "21"
Term 4: read "21" → "1211"
Term 5: read "1211":

• "1" → one 1 → "11"
• "2" → one 2 → "12"
• "11" → two 1s → "21"

Concatenate:
"11" + "12" + "21" = "111221"

So output for n = 5 is "111221".

---

Complexity Analysis

Let Lₙ be the length of the nth term.

Time Complexity:
O(L₁ + L₂ + … + Lₙ)
Which is acceptable because n ≤ 30.

Space Complexity:
O(Lₙ)

---

Edge Cases

• n = 1 → return "1"
• Strings with large repeating groups are handled naturally by the counting logic.
• No tricky corner cases since input is small.