Problem Description

This problem asks us to work with a special "magical string" that has a unique self-referential property.

The magical string s consists only of the digits '1' and '2' and follows a special rule: when you count the lengths of consecutive groups of the same digit, these counts form the string itself.

Let's understand this with the given example:

• The string starts as: "1221121221221121122..."
• If we group consecutive identical digits: "1 22 11 2 1 22 1 22 11 2 11 22 ..."
• The lengths of these groups are: 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, ...
• Writing these lengths as a string gives us: "122112122122112122..."
• This sequence of lengths is exactly the same as the original string s!

This self-referential property is what makes the string "magical" - the string describes its own structure.

The construction process works as follows:

1. Start with the beginning of the string (we know it starts with "122")
2. Use each digit in the string to determine how many times to repeat the next alternating digit
3. The digits alternate between groups (if one group is all 1s, the next is all 2s, and vice versa)
4. Each digit in position i tells us how many times to repeat the digit in the corresponding group

Given an integer n, you need to find and return the count of 1s in the first n digits of this magical string.

For example, if n = 6, the first 6 digits are "122112", which contains three 1s, so the answer would be 3.

Intuition

The key insight is recognizing that we can build the magical string incrementally by using the string itself as instructions for its own construction.

Think of it like reading a recipe that writes itself as you follow it. Each digit in the string tells us how many times to write the next group of digits, and these groups alternate between 1s and 2s.

We start with what we know for certain: the string begins with "122". Why? Because:

• The first group must be a single 1 (since groups alternate and we need to start somewhere)
• The second group must be 2s, and according to the second position (which is 2), we need two of them
• So we have "1 22" which gives us "122"

Now here's the clever part: we can use a pointer to track which position in the string tells us how to build the next group. We start this pointer at position 2 (index 2) because we've already used positions 0 and 1 to build the first two groups.

As we build each new group:

1. We look at the digit our pointer is pointing to - this tells us how many times to repeat
2. We determine what digit to repeat by alternating (if the last group was 2s, this group is 1s, and vice versa)
3. We append that many digits to our string
4. We move our pointer forward by one position

The beauty of this approach is that we're simultaneously reading from and writing to the same string. The string contains the instructions for its own construction, and we just need to follow those instructions step by step.

We continue this process until we've generated at least n digits, then simply count how many 1s appear in the first n positions.

Solution Approach

Let's walk through the implementation step by step, following the construction process described in the reference approach.

Initial Setup: We start by initializing the string s as a list with [1, 2, 2], representing the first three digits of the magical string. We also set a pointer i = 2 to track which position in the string tells us how to build the next group.

s = [1, 2, 2]
i = 2

Construction Loop: We continue building the string until its length reaches at least n:

while len(s) < n:

Determining the Next Group: For each iteration:

1. First, we identify what the last digit was using pre = s[-1]. This tells us what digit was used in the previous group.
2. Since groups alternate between 1s and 2s, we calculate the current digit as cur = 3 - pre. This clever formula gives us:
   • If pre = 1, then cur = 3 - 1 = 2
   • If pre = 2, then cur = 3 - 2 = 1

Adding the New Group: The digit at position i (i.e., s[i]) tells us how many times to repeat the current digit:

s += [cur] * s[i]

This appends s[i] copies of cur to our string.

Moving the Pointer: After constructing each group, we move the pointer forward:

i += 1

Example Walkthrough: Starting with s = [1, 2, 2] and i = 2:

• Iteration 1: s[2] = 2, pre = 2, cur = 1. Add two 1s → s = [1, 2, 2, 1, 1]
• Iteration 2: s[3] = 1, pre = 1, cur = 2. Add one 2 → s = [1, 2, 2, 1, 1, 2]
• Iteration 3: s[4] = 1, pre = 2, cur = 1. Add one 1 → s = [1, 2, 2, 1, 1, 2, 1]
• Iteration 4: s[5] = 2, pre = 1, cur = 2. Add two 2s → s = [1, 2, 2, 1, 1, 2, 1, 2, 2]

This process continues until we have at least n digits.

Counting the Result: Finally, we take the first n elements and count the number of 1s:

return s[:n].count(1)

The time complexity is O(n) since we generate approximately n digits, and the space complexity is also O(n) for storing the generated string.

Example Walkthrough

Let's trace through the algorithm with n = 10 to find the count of 1s in the first 10 digits of the magical string.

Initial State:

• s = [1, 2, 2] (the known beginning of the magical string)
• i = 2 (pointer to the instruction digit)
• Current length: 3

Iteration 1:

• We need more digits since len(s) = 3 < 10
• Look at s[i] = s[2] = 2 (this tells us to add 2 digits)
• Previous digit: pre = s[-1] = 2
• Current digit to add: cur = 3 - pre = 3 - 2 = 1
• Add two 1s: s = [1, 2, 2, 1, 1]
• Move pointer: i = 3

Iteration 2:

• Still need more digits since len(s) = 5 < 10
• Look at s[i] = s[3] = 1 (add 1 digit)
• Previous digit: pre = 1
• Current digit: cur = 3 - 1 = 2
• Add one 2: s = [1, 2, 2, 1, 1, 2]
• Move pointer: i = 4

Iteration 3:

• Still need more digits since len(s) = 6 < 10
• Look at s[i] = s[4] = 1 (add 1 digit)
• Previous digit: pre = 2
• Current digit: cur = 3 - 2 = 1
• Add one 1: s = [1, 2, 2, 1, 1, 2, 1]
• Move pointer: i = 5

Iteration 4:

• Still need more digits since len(s) = 7 < 10
• Look at s[i] = s[5] = 2 (add 2 digits)
• Previous digit: pre = 1
• Current digit: cur = 3 - 1 = 2
• Add two 2s: s = [1, 2, 2, 1, 1, 2, 1, 2, 2]
• Move pointer: i = 6

Iteration 5:

• Still need more digits since len(s) = 9 < 10
• Look at s[i] = s[6] = 1 (add 1 digit)
• Previous digit: pre = 2
• Current digit: cur = 3 - 2 = 1
• Add one 1: s = [1, 2, 2, 1, 1, 2, 1, 2, 2, 1]
• Move pointer: i = 7

Final Step:

• Now len(s) = 10, which is enough for n = 10
• Take first 10 digits: [1, 2, 2, 1, 1, 2, 1, 2, 2, 1]
• Count the 1s: positions 0, 3, 4, 6, 9 contain 1s
• Answer: 5

We can verify this is correct by checking the self-referential property:

• Groups: 1 | 22 | 11 | 2 | 1 | 22 | 1
• Group lengths: 1, 2, 2, 1, 1, 2, 1
• This matches the beginning of our string: 1, 2, 2, 1, 1, 2, 1

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm builds the magical string up to length n. The while loop continues until len(s) >= n. In each iteration, we append either 1 or 2 elements to the string (determined by s[i] which is either 1 or 2). The pointer i increments by 1 in each iteration. Since we're building a string of length n and each operation inside the loop takes O(1) time (appending to a list and incrementing a counter), the total time complexity is O(n). The final count(1) operation also takes O(n) time to traverse the first n elements.

Space Complexity: O(n)

The algorithm stores the magical string in list s which grows up to at least n elements (possibly slightly more due to the appending pattern, but still bounded by O(n)). The slicing operation s[:n] creates a temporary list of size n, but this doesn't change the overall space complexity. Other variables (i, pre, cur) use constant space. Therefore, the total space complexity is O(n).

Common Pitfalls

1. Off-by-One Error in Initial String Setup

A common mistake is incorrectly initializing the magical string or the starting index pointer. Some might start with [1, 2] or [1] instead of [1, 2, 2], or set the index pointer to 0 or 1 instead of 2.

Why this happens: The problem description might not clearly indicate where to start, and it's tempting to think the string starts with just [1, 2].

Solution: Always remember that the magical string definitively starts with [1, 2, 2] and the index pointer should start at position 2 (the third element) since the first two positions have already been "used" to generate the initial groups.

2. Incorrect Alternation Logic

Some implementations might use complex if-else statements or modulo operations to alternate between 1 and 2, leading to errors:

# Error-prone approaches:
if previous_element == 1:
    current_element = 2
else:
    current_element = 1

# Or using modulo (incorrect):
current_element = (previous_element % 2) + 1  # This gives wrong results

Solution: Use the simple formula current_element = 3 - previous_element. This elegantly handles the alternation without conditional logic.

3. Building More Elements Than Necessary

The code might generate far more elements than needed if not careful with the loop condition, especially when extend() adds multiple elements at once:

# Inefficient approach that might generate too many elements:
while len(magical_string) <= n:  # Using <= instead of <
    # ... generation logic
    magical_string.extend([current_element] * repeat_count)

Why this matters: While the final slicing [:n] ensures correctness, generating excessive elements wastes time and memory.

Solution: Use while len(magical_string) < n and accept that you might generate a few extra elements (but not excessively). This is optimal since we need to complete full groups.

4. Forgetting Edge Cases

Not handling the case when n = 0 or very small values of n:

# Missing edge case handling:
def magicalString(self, n: int) -> int:
    magical_string = [1, 2, 2]
    # If n <= 3, the loop never executes, but we still slice and count
    # This works but isn't explicit about handling edge cases

Solution: Explicitly check for n == 0 at the beginning and return 0. For other small values, the main logic handles them correctly through slicing.

5. Using String Concatenation Instead of List Operations

Some might try to build the magical string using string concatenation:

# Inefficient approach:
magical_string = "122"
while len(magical_string) < n:
    magical_string += str(current_element) * repeat_count

Why this is problematic: String concatenation in Python creates new string objects each time, leading to O(n²) time complexity instead of O(n).

Solution: Use a list to build the sequence and only convert to string if absolutely necessary. Lists support efficient extend() operations with O(1) amortized append time.