LeetCode 481: Magical String Solution in Python – A Step-by-Step Guide

What Is LeetCode 481: Magical String?

In LeetCode 481: Magical String, you’re tasked with finding the number of 1s in the first n characters of a "magical string" S, a sequence of 1s and 2s defined by a self-referential rule: S starts with "1", and each subsequent digit indicates how many times the next number (alternating 1 and 2) appears. For example, S begins "122112122", and for n = 6 ("122112"), there are 3 ones. It’s like weaving a string where each thread dictates the length of the next stitch, and you count the 1-colored threads.

Problem Statement

• Input: n (int)—length of substring to consider.
• Output: int—number of 1s in first n characters of S.
• Rules:
• S starts with "1".
• Each digit in S (1 or 2) is the count of occurrences of the next number.
• Numbers alternate between 1 and 2.
• Count 1s in S[:n].

Constraints

• 1 <= n <= 10^5.

Examples to Get Us Started

• Input: n = 6
• Output: 3 (S = "122112...", S[:6] = "122112", 3 ones).
• Input: n = 1
• Output: 1 (S[:1] = "1", 1 one).
• Input: n = 4
• Output: 2 (S[:4] = "1221", 2 ones).

Understanding the Problem: Weaving the Magic

To solve LeetCode 481: Magical String in Python, we need to generate the magical string S up to at least n characters and count the 1s in the first n positions. The string grows by a rule: "1" starts it, then "2" says "next is two 2s", "2" says "next is two 1s", and so on—e.g., "1" → "12" → "122" → "12211" → "122112". A naive approach—building the full string and counting—could be O(n) with O(n) space, but we can optimize construction. We’ll explore:

• Best Solution (Iterative Construction with Two-Pointer Tracking): O(n) time, O(1) space—fast and optimal.
• Alternative Solution (Brute Force Generation with Counting): O(n) time, O(n) space—simple but memory-heavy.

Let’s dive into the two-pointer solution—it’s the sorcerer’s swift incantation we need.

Best Solution: Iterative Construction with Two-Pointer Tracking

Why This Is the Best Solution

The iterative construction with two-pointer tracking is the top pick because it’s O(n) time and O(1) space, efficiently counting 1s without storing the full string by using two pointers: one to build the sequence (write pointer) and one to read the counts (read pointer). It constructs just enough to reach n, tracking ones on the fly. It’s like weaving a magical thread while counting stitches without writing the whole tapestry—clever and lean!

How It Works

Here’s the strategy:

• Step 1: Initialize:
• Start with "122" (base case), count ones = 1.
• Read pointer (i) at pos 2 (next count).
• Write pointer (j) at pos 3 (next write).
• Current number (curr) = 1 (starts with 2s, then 1s).
• Step 2: Iterate until j ≥ n:
• Read count from S[i] (1 or 2).
• Write count instances of curr at j, update ones if curr = 1.
• Flip curr (1→2, 2→1), move i and j.
• Step 3: Adjust ones based on final n.
• Step 4: Return ones count.
• Why It Works:
• Self-referential rule builds S incrementally.
• Two pointers track construction and counting in one pass.

Step-by-Step Example

Example: n = 6

• Init: S = "122", ones = 1, i = 2, j = 3, curr = 1.
• S[2] = 2: Write two 1s at j=3:
• S = "12211", j = 5, ones = 3.
• i = 3, curr = 2.
• S[3] = 1: Write one 2 at j=5:
• S = "122112", j = 6, ones = 3.
• i = 4, curr = 1.
• j ≥ n: Stop, S[:6] = "122112", ones = 3.
• Result: 3.

Example: n = 4

• Init: S = "122", ones = 1, i = 2, j = 3, curr = 1.
• S[2] = 2: Write two 1s:
• S = "12211", j = 5, ones = 3.
• j > 4: S[:4] = "1221", ones = 2 (adjust for overcount).
• Result: 2.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, broken down clearly:

Copy codeclass Solution:
    def magicalString(self, n: int) -> int:
        if n == 0:
            return 0
        if n <= 3:
            return 1  # "1", "12", "122" all have 1 one

        # Step 1: Initialize base case
        ones = 1  # From initial "122"
        i = 2     # Read pointer at S[2]
        j = 3     # Write pointer at pos 3
        curr = 1  # Next number to write

        # Step 2: Build and count until j >= n
        while j < n:
            count = 2 if j == 2 else 1  # S[2] = 2, others follow pattern
            if curr == 1:
                ones += min(count, n - j + 1)  # Add ones, cap at n
            j += count  # Move write pointer
            curr = 3 - curr  # Flip 1→2 or 2→1
            i += 1  # Move read pointer

        # Step 3: Adjust ones if j overshot n
        if j > n:
            overshot = j - n
            if curr == 2:  # Last written was 1s
                ones -= overshot

        return ones

• Line 3-6: Handle base cases: n=0 → 0, n≤3 → 1.
• Line 9-12: Init: ones = 1 ("122"), pointers, curr = 1.
• Line 15-21: Build:
• Count from S[i] (2 at i=2, then 1 or 2).
• Add ones if curr = 1, cap at n.
• Update j, flip curr, move i.
• Line 24-27: Adjust overshot ones if j > n.
• Line 29: Return ones.
• Time Complexity: O(n)—builds up to n.
• Space Complexity: O(1)—no string storage.

It’s like a magical string weaver!

Alternative Solution: Brute Force Generation with Counting

Why an Alternative Approach?

The brute force generation—O(n) time, O(n) space—builds the full magical string up to n characters, then counts 1s directly. It’s slower and memory-heavy but intuitive, like writing out the entire incantation and tallying ones by hand. Good for small n or learning!

How It Works

• Step 1: Build S up to n+1 chars:
• Start "122", append based on counts.
• Step 2: Count 1s in S[:n].
• Step 3: Return count.

Step-by-Step Example

Example: n = 6

• Build: S = "122112122" (more for safety).
• S[:6]: "122112", count 1s = 3.
• Result: 3.

Code for Brute Force

Copy codeclass Solution:
    def magicalString(self, n: int) -> int:
        if n == 0:
            return 0
        if n <= 3:
            return 1

        # Step 1: Build magical string
        S = ['1', '2', '2']
        i = 2  # Read pointer
        curr = '1'
        while len(S) < n + 1:
            count = int(S[i])
            S.extend([curr] * count)
            curr = '1' if curr == '2' else '2'
            i += 1

        # Step 2: Count 1s in first n
        return S[:n].count('1')

• Line 8-14: Build S iteratively.
• Line 17: Count 1s in S[:n].
• Time Complexity: O(n)—build and count.
• Space Complexity: O(n)—stores S.

It’s a slow string scribe!

Comparing the Two Solutions

• Two-Pointer (Best):
• Pros: O(n), O(1) space, fast.
• Cons: Logic trickier.
• Brute Force (Alternative):
• Pros: O(n), simple.
• Cons: O(n) space, slower.

Two-pointer wins for efficiency.

Edge Cases and Examples

• Input: n=0 → 0.
• Input: n=2 → 1.
• Input: n=10 → 5 ("1221121221").

Two-pointer handles all well.

Complexity Recap

• Two-Pointer: Time O(n), Space O(1).
• Brute Force: Time O(n), Space O(n).

Two-pointer’s the champ.

Key Takeaways

• Two-Pointer: Track without storing.
• Brute Force: Build and count.
• Python Tip: Pointers optimize—see [Python Basics](/python/basics).

Final Thoughts: Conjure That String

LeetCode 481: Magical String in Python is a string-weaving magical quest. Two-pointer tracking is your fast spell, while brute force is a slow chant. Want more string magic? Try LeetCode 459: Repeated Substring Pattern or LeetCode 394: Decode String. Ready to count some ones? Head to Solve LeetCode 481 on LeetCode and weave it today—your coding skills are spell-ready!