How do you solve Magical String on LeetCode?

Magical String (LeetCode #481) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The magical string is generated based on the counts of previous characters, making it recursive in nature.

What is the brute force solution for Magical String?

The brute force approach for Magical String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates the magical string directly by simulating the process of constructing it based on the rules provided. This method is straightforward but inefficient for larger values of n. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Magical String?

Magical String uses the following concepts: Two Pointers, String. The recommended patterns to study are: Two Pointers, Dynamic Programming.

What are the interview tips for Magical String?

Always start with a brute-force solution to understand the problem better before optimizing. Make sure to explain your thought process clearly during the interview; it shows your understanding. Practice dry-running your solution on paper to catch edge cases and ensure correctness.

What are common mistakes when solving Magical String?

Not recognizing the pattern in how the magical string is generated, leading to inefficient solutions. Failing to account for the count of '1's correctly during string construction.

#481

Magical String

Medium

Two PointersStringTwo PointersDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution builds the magical string iteratively while counting '1's, avoiding the need to store the entire string. This approach is efficient and directly addresses the problem requirements.

⚙️

Algorithm

4 steps

1Step 1: Initialize an array to hold the counts of '1's and '2's, starting with the first character '1'.
2Step 2: Use a pointer to track the current position and another to determine how many characters to add based on the last character.
3Step 3: Continue this process until the length of the magical string reaches n, counting '1's as you go.
4Step 4: Return the count of '1's after reaching n.

solution.py19 lines

1# Full working Python code
2
3def magical_string(n):
4    if n == 1:
5        return 1
6    s = [1]
7    count_ones = 1
8    index = 0
9    for i in range(1, n):
10        next_char = 2 if s[-1] == 1 else 1
11        for _ in range(s[index]):
12            if len(s) < n:
13                s.append(next_char)
14                if next_char == 1:
15                    count_ones += 1
16        index += 1
17    return count_ones
18
19print(magical_string(6))  # Output: 3

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once while building it, and the space complexity is O(n) due to storing the characters in an array.

1The magical string is generated based on the counts of previous characters, making it recursive in nature.
2Understanding the pattern of how '1's and '2's are added is crucial for optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.