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Magical String

Medium

Amazon

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Topics:

ArraysStringsTwo Pointers

A magical string s consists of only '1' and '2' and obeys the following rules:

The string s is magical because concatenating the number of contiguous occurrences of characters '1' and '2' generates the string s itself.

For example, the first few elements of s is s = "1221121221221121122......". If we group the consecutive 1's and 2's in s, it will be "1 22 11 2 1 22 1 22 11 2 11 22 ......" and the occurrences of 1's or 2's in each group are "1 2 2 1 1 2 1 2 2 1 2 2 ......". You can see that the occurrence sequence is s itself.

Given an integer n, return the number of 1's in the first n numbers in the magical string s.

Example:

Input: n = 6 Output: 3 Explanation: The first 6 elements of the magical string s is "122112" and it contains three 1's, so return 3.

Can you implement a function that calculates the number of 1's in the first n characters of the magical string? How would you optimize your solution?

Magical String

Medium

Amazon

1 view

Topics:

ArraysStringsTwo Pointers

A magical string s consists of only '1' and '2' and obeys the following rules:

The string s is magical because concatenating the number of contiguous occurrences of characters '1' and '2' generates the string s itself.

Given an integer n, return the number of 1's in the first n numbers in the magical string s.

Example:

Input: n = 6 Output: 3 Explanation: The first 6 elements of the magical string s is "122112" and it contains three 1's, so return 3.

Can you implement a function that calculates the number of 1's in the first n characters of the magical string? How would you optimize your solution?

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum length of the magical string we need to generate?
Is the input 'n' always a positive integer?
Are we only concerned with the number of '1's in the generated string, or do we need to actually store and return the entire string?
What should be returned if the input 'n' is 0 or 1?
Can you provide a few examples of the expected output for small values of 'n' to ensure I understand the problem correctly?

Brute Force Solution

Approach

Imagine we're building a special sequence of numbers consisting only of 1s and 2s. The brute force method is like trying to build this sequence by trying out every single combination of 1s and 2s until we find one that matches the weird rules of the sequence.

Here's how the algorithm would work step-by-step:

Start building the sequence from the beginning, one digit at a time.
First, try adding a 1 to the sequence. See if that makes sense according to the rules.
Then, try adding a 2 to the sequence instead. See if that makes sense.
For each of these possibilities, keep going, adding another 1 or 2 at a time and checking if the sequence still follows the rules.
If adding a 1 or 2 breaks the rules, backtrack and try a different number.
Continue this process of trying every possible combination until the sequence reaches the desired length.
Count how many 1s appear in the sequence you just built.

Code Implementation

def magical_string_brute_force(number):
    magical_string = []
    one_count = 0

    def generate_sequence(current_sequence):
        nonlocal one_count

        # If sequence is long enough, count ones
        if len(current_sequence) == number:
            one_count = current_sequence.count(1)
            return True

        # Try adding a 1
        if generate_sequence(current_sequence + [1]):
            return True

        # Try adding a 2
        if generate_sequence(current_sequence + [2]):
            return True

        return False

    # Initiate the process by generating the sequence starting from the beginning
    generate_sequence([])

    return one_count

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force approach constructs the magical string by trying all possible combinations of 1s and 2s. At each position in the sequence (up to length n), we have two choices: either add a 1 or add a 2. This branching creates a binary tree of possibilities. Therefore, the total number of possible sequences explored grows exponentially with the length n. In the worst-case scenario, where the algorithm explores all possible sequences before finding one that satisfies the magical string rules, it examines approximately 2^n sequences. Thus, the time complexity is O(2^n).

Space Complexity

O(N) – The algorithm builds a sequence of 1s and 2s until it reaches a desired length N. In the worst-case scenario, the algorithm explores a significant portion of all possible combinations of 1s and 2s up to length N, requiring the storage of the generated sequence. This sequence is of length N. Therefore, the space required scales linearly with N, resulting in O(N) auxiliary space complexity.

Optimal Solution

Approach

The magical string is built by expanding a core pattern. We don't generate every possible sequence; instead, we use the string itself to guide how it grows. This leads to the creation of the string without needless computation.

Here's how the algorithm would work step-by-step:

Start with a small initial string, like '122'. This is your seed.
Look at the digits in this string, treat them as instructions to add more '1's or '2's to the end.
If you see a '1', add one of the opposite number to the string. If you see a '2', add two of the opposite number to the string.
For example, if your current string is '122', the first digit '1' tells you to add one '1' to the end making it '1221'. Then the two '2's tell you to add two '2's and two '1's to the end, respectively. This would give you '12212211'.
Continue expanding the string in this way, always using the digits from the start of the string as instructions to add more digits at the end. Stop when you have reached the desired length.
Count how many '1's are in the final string and return that count.

Code Implementation

def magical_string(required_length):
    magical_string_array = [1, 2, 2]
    first_index = 2
    second_index = 3
    number_of_ones = 1

    if required_length <= 3:
        for i in range(required_length):
            if magical_string_array[i] == 1:
                number_of_ones += 1
        return number_of_ones -1

    # Expand the string until the required length is met
    while second_index < required_length:
        repetition_count = magical_string_array[first_index]
        number = 3 - magical_string_array[second_index - 1]

        # Use the value from the first pointer to decide how many times to add
        for _ in range(repetition_count):
            magical_string_array.append(number)
            if number == 1:
                number_of_ones += 1
            second_index += 1
            if second_index == required_length:
                break

        first_index += 1

    # Count the number of 1s in the magical string up to required_length
    number_of_ones_in_magical_string = 0
    for i in range(required_length):
        if magical_string_array[i] == 1:
            number_of_ones_in_magical_string += 1

    return number_of_ones_in_magical_string

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the 'magical string' once to generate the sequence up to length n. The expansion process adds digits to the end based on the existing digits. The number of iterations required to reach length n is directly proportional to n. Therefore, the time complexity is O(n).

Space Complexity

O(N) – The algorithm constructs a string whose length can grow up to N, where N is the input length. This string stores the sequence of '1's and '2's as it's being generated. Therefore, the space required to store this string is proportional to N. No other significant auxiliary data structures are used.

Edge Cases

Case	How to Handle
Input n is zero or negative	Return an empty string as there is no magical string of zero or negative length.
Input n is one	Return '1' as the magical string of length one is simply '1'.
Input n is a very large number	Ensure that the solution's time and space complexity are linear, avoiding exponential growth and potential memory issues.
The generated magical string consists only of '1' for a large portion	The algorithm should still produce valid output, even if the counts mostly generate a specific character.
The generated magical string alternates between '1' and '2' rapidly	The algorithm should handle quickly oscillating patterns without issue.
Integer overflow in any calculations when n is large.	Use appropriate data types (e.g., long) to prevent overflow during the calculation of the magical string indices and counts.
Edge cases where the length calculation produces number slightly bigger than requested.	Make sure to truncate/substring the result to ensure the output magical string matches the requested length 'n'.
Out of bounds array access when generating the count array or result string.	Carefully manage index boundaries when constructing both the count array and the final magical string to avoid errors during access.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum length of the magical string we need to generate?
Is the input 'n' always a positive integer?
Are we only concerned with the number of '1's in the generated string, or do we need to actually store and return the entire string?
What should be returned if the input 'n' is 0 or 1?
Can you provide a few examples of the expected output for small values of 'n' to ensure I understand the problem correctly?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start building the sequence from the beginning, one digit at a time.
First, try adding a 1 to the sequence. See if that makes sense according to the rules.
Then, try adding a 2 to the sequence instead. See if that makes sense.
For each of these possibilities, keep going, adding another 1 or 2 at a time and checking if the sequence still follows the rules.
If adding a 1 or 2 breaks the rules, backtrack and try a different number.
Continue this process of trying every possible combination until the sequence reaches the desired length.
Count how many 1s appear in the sequence you just built.

Code Implementation

def magical_string_brute_force(number):
    magical_string = []
    one_count = 0

    def generate_sequence(current_sequence):
        nonlocal one_count

        # If sequence is long enough, count ones
        if len(current_sequence) == number:
            one_count = current_sequence.count(1)
            return True

        # Try adding a 1
        if generate_sequence(current_sequence + [1]):
            return True

        # Try adding a 2
        if generate_sequence(current_sequence + [2]):
            return True

        return False

    # Initiate the process by generating the sequence starting from the beginning
    generate_sequence([])

    return one_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start with a small initial string, like '122'. This is your seed.
Look at the digits in this string, treat them as instructions to add more '1's or '2's to the end.
If you see a '1', add one of the opposite number to the string. If you see a '2', add two of the opposite number to the string.
For example, if your current string is '122', the first digit '1' tells you to add one '1' to the end making it '1221'. Then the two '2's tell you to add two '2's and two '1's to the end, respectively. This would give you '12212211'.
Continue expanding the string in this way, always using the digits from the start of the string as instructions to add more digits at the end. Stop when you have reached the desired length.
Count how many '1's are in the final string and return that count.

Code Implementation

def magical_string(required_length):
    magical_string_array = [1, 2, 2]
    first_index = 2
    second_index = 3
    number_of_ones = 1

    if required_length <= 3:
        for i in range(required_length):
            if magical_string_array[i] == 1:
                number_of_ones += 1
        return number_of_ones -1

    # Expand the string until the required length is met
    while second_index < required_length:
        repetition_count = magical_string_array[first_index]
        number = 3 - magical_string_array[second_index - 1]

        # Use the value from the first pointer to decide how many times to add
        for _ in range(repetition_count):
            magical_string_array.append(number)
            if number == 1:
                number_of_ones += 1
            second_index += 1
            if second_index == required_length:
                break

        first_index += 1

    # Count the number of 1s in the magical string up to required_length
    number_of_ones_in_magical_string = 0
    for i in range(required_length):
        if magical_string_array[i] == 1:
            number_of_ones_in_magical_string += 1

    return number_of_ones_in_magical_string

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input n is zero or negative	Return an empty string as there is no magical string of zero or negative length.
Input n is one	Return '1' as the magical string of length one is simply '1'.
Input n is a very large number	Ensure that the solution's time and space complexity are linear, avoiding exponential growth and potential memory issues.
The generated magical string consists only of '1' for a large portion	The algorithm should still produce valid output, even if the counts mostly generate a specific character.
The generated magical string alternates between '1' and '2' rapidly	The algorithm should handle quickly oscillating patterns without issue.
Integer overflow in any calculations when n is large.	Use appropriate data types (e.g., long) to prevent overflow during the calculation of the magical string indices and counts.
Edge cases where the length calculation produces number slightly bigger than requested.	Make sure to truncate/substring the result to ensure the output magical string matches the requested length 'n'.
Out of bounds array access when generating the count array or result string.	Carefully manage index boundaries when constructing both the count array and the final magical string to avoid errors during access.