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Convert to Base -2

Medium

Asked by:

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Topics:

Bit Manipulation

Given an integer n, return a binary string representing its representation in base -2.

Note that the returned string should not have leading zeros unless the string is "0".

Example 1:

Input: n = 2
Output: "110"
Explantion: (-2)² + (-2)¹ = 2

Example 2:

Input: n = 3
Output: "111"
Explantion: (-2)² + (-2)¹ + (-2)⁰ = 3

Example 3:

Input: n = 4
Output: "100"
Explantion: (-2)² = 4

Constraints:

0 <= n <= 10⁹

Try coding it in LeetCode

Convert to Base -2

Medium

Asked by:

28 views

Topics:

Bit Manipulation

Given an integer n, return a binary string representing its representation in base -2.

Note that the returned string should not have leading zeros unless the string is "0".

Example 1:

Input: n = 2
Output: "110"
Explantion: (-2)² + (-2)¹ = 2

Example 2:

Input: n = 3
Output: "111"
Explantion: (-2)² + (-2)¹ + (-2)⁰ = 3

Example 3:

Input: n = 4
Output: "100"
Explantion: (-2)² = 4

Constraints:

0 <= n <= 10⁹

Try coding it in LeetCode

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 range of the input integer `n`? What is the minimum and maximum possible value for n?
What should the output be if the input is zero?
Is the input guaranteed to be an integer, or do I need to handle other data types like floats or strings?
Are there any leading zeros allowed in the output string? If not, how should the most significant digit be determined?
Are there any specific performance constraints I should be aware of, considering the potential size of the input number?

Brute Force Solution

Approach

To convert a number to base -2, the brute force method essentially tries out all possible combinations of 0s and 1s to see if they add up to the original number when interpreted in base -2. Think of it like guessing and checking every possible binary number until we find one that works when we treat it as a base -2 number.

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

Start by testing the simplest possibilities, like 0, 1, 10, 11, 100, 101, and so on. Each of these represents a number in base -2.
For each of these possible numbers, calculate its value in regular base 10 (our everyday number system), remembering to use the rules for base -2.
Compare the base 10 value you calculated to the original number you are trying to convert. If they are the same, you have found your answer! Stop.
If the base 10 value is not equal to the original number, move on to the next possibility and repeat the calculation and comparison steps.
Continue generating and testing these combinations until you find one that matches the original number. Because we can always add a leading '1100' as many times as we want, we will eventually find an answer (or the original number was 0).

Code Implementation

def convert_to_base_negative_two_brute_force(number):
    maximum_sequence_length = 1
    while True:
        for i in range(2**maximum_sequence_length):
            binary_representation = bin(i)[2:].zfill(maximum_sequence_length)
            
            base_negative_two_value = 0
            for digit_index, digit in enumerate(reversed(binary_representation)):
                if digit == '1':
                    # Calculate the value of the digit in base -2
                    base_negative_two_value += int(digit) * (-2)**digit_index
            
            if base_negative_two_value == number:
                # Return if we find a match.
                return binary_representation

        # Increase the sequence length
        maximum_sequence_length += 1

        if maximum_sequence_length > 20:
            # Avoid infinite loops for very large or impossible to convert numbers
            return "Cannot represent the number in base -2"

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach tests all possible combinations of 0s and 1s. In the worst-case scenario, the algorithm explores all binary strings of length n before finding a match, where n is the length of the base -2 representation. Generating each combination takes constant time, but the number of combinations grows exponentially. Therefore, the time complexity is proportional to 2 raised to the power of n, resulting in O(2^n).

Space Complexity

O(1) – The provided plain English explanation describes a brute-force approach that generates and tests potential base -2 representations. It does not explicitly mention storing a large number of intermediate results or using any data structures that scale with the input number N. The process seems to involve generating a candidate binary string (e.g., '101'), calculating its base-10 value, and comparing it to N. The space used to store the current candidate binary string and its base-10 value is constant, irrespective of N. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

Converting to base -2 requires a different way of thinking about remainders. Instead of dividing and taking the remainder as is, we might need to adjust the quotient and remainder to ensure the remainder is always positive.

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

Begin by repeatedly dividing the number by -2.
Observe the remainder after each division. If the remainder is negative, adjust it by adding 2 to the remainder and adding 1 to the quotient.
Record the adjusted remainders (which will always be either 0 or 1). These become the digits of the base -2 representation.
Repeat the division and remainder adjustment until the quotient becomes 0.
The digits written down are read in reverse order to obtain the base -2 representation.
In the special case that the original number is 0, the result is simply 0.

Code Implementation

def convert_to_base_negative_two(number):
    if number == 0:
        return "0"

    base_negative_two_representation = ""

    while number != 0:
        remainder = number % -2

        # Adjust remainder and quotient to ensure remainder is 0 or 1
        if remainder < 0:
            remainder += 2
            number = number // -2 + 1

        else:
            number = number // -2

        base_negative_two_representation = str(remainder) + base_negative_two_representation

    return base_negative_two_representation

Big(O) Analysis

Time Complexity

O(log n) – The algorithm repeatedly divides the input number n by -2 until the quotient becomes 0. The number of divisions required is proportional to the number of digits in the base -2 representation of n. Since the base is 2 (in absolute value), the number of digits and hence the number of divisions is logarithmic with respect to n. Therefore, the time complexity is O(log n).

Space Complexity

O(log N) – The algorithm repeatedly divides the input number N by -2 until the quotient becomes 0. The adjusted remainders are stored to form the base -2 representation. The number of divisions, and thus the number of remainders stored, is proportional to the number of digits in the base -2 representation, which is logarithmic with respect to the absolute value of N. Therefore, auxiliary space is used to store digits, resulting in a space complexity of O(log N).

Edge Cases

Case	How to Handle
Input n is 0	The base -2 representation of 0 is '0', so return '0'.
Small positive integer (e.g., 1, 2, 3)	The algorithm should correctly convert small positive integers to their base -2 representation, like 1 is '1'.
Small negative integer (e.g., -1, -2, -3)	The algorithm should correctly convert small negative integers to their base -2 representation, like -1 is '11'.
Large positive integer (e.g., 100, 1000)	The algorithm should be able to handle large positive integer inputs without integer overflow or performance issues.
Large negative integer (e.g., -100, -1000)	The algorithm should be able to handle large negative integer inputs correctly without integer overflow or infinite loop.
Integer near the maximum possible value (2^31 - 1)	Ensure the algorithm handles integers near the maximum value without integer overflow issues during computations.
Integer near the minimum possible value (-2^31)	Ensure the algorithm handles integers near the minimum value without integer overflow issues during computations.
Alternating positive and negative results of n % -2	The algorithm must correctly handle the alternating nature of positive and negative remainders when calculating modulo -2, using n % -2 and n = n // -2 (or equivalent) correctly.

Input n is 0

How to Handle:

The base -2 representation of 0 is '0', so return '0'.

Small positive integer (e.g., 1, 2, 3)

How to Handle:

The algorithm should correctly convert small positive integers to their base -2 representation, like 1 is '1'.

Small negative integer (e.g., -1, -2, -3)

How to Handle:

The algorithm should correctly convert small negative integers to their base -2 representation, like -1 is '11'.

Large positive integer (e.g., 100, 1000)

How to Handle:

The algorithm should be able to handle large positive integer inputs without integer overflow or performance issues.

Large negative integer (e.g., -100, -1000)

How to Handle:

The algorithm should be able to handle large negative integer inputs correctly without integer overflow or infinite loop.

Integer near the maximum possible value (2^31 - 1)

How to Handle:

Ensure the algorithm handles integers near the maximum value without integer overflow issues during computations.

Integer near the minimum possible value (-2^31)

How to Handle:

Ensure the algorithm handles integers near the minimum value without integer overflow issues during computations.

Alternating positive and negative results of n % -2

How to Handle:

The algorithm must correctly handle the alternating nature of positive and negative remainders when calculating modulo -2, using n % -2 and n = n // -2 (or equivalent) correctly.

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 range of the input integer `n`? What is the minimum and maximum possible value for n?
What should the output be if the input is zero?
Is the input guaranteed to be an integer, or do I need to handle other data types like floats or strings?
Are there any leading zeros allowed in the output string? If not, how should the most significant digit be determined?
Are there any specific performance constraints I should be aware of, considering the potential size of the input number?

Brute Force Solution

Approach

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

Start by testing the simplest possibilities, like 0, 1, 10, 11, 100, 101, and so on. Each of these represents a number in base -2.
For each of these possible numbers, calculate its value in regular base 10 (our everyday number system), remembering to use the rules for base -2.
Compare the base 10 value you calculated to the original number you are trying to convert. If they are the same, you have found your answer! Stop.
If the base 10 value is not equal to the original number, move on to the next possibility and repeat the calculation and comparison steps.
Continue generating and testing these combinations until you find one that matches the original number. Because we can always add a leading '1100' as many times as we want, we will eventually find an answer (or the original number was 0).

Code Implementation

def convert_to_base_negative_two_brute_force(number):
    maximum_sequence_length = 1
    while True:
        for i in range(2**maximum_sequence_length):
            binary_representation = bin(i)[2:].zfill(maximum_sequence_length)
            
            base_negative_two_value = 0
            for digit_index, digit in enumerate(reversed(binary_representation)):
                if digit == '1':
                    # Calculate the value of the digit in base -2
                    base_negative_two_value += int(digit) * (-2)**digit_index
            
            if base_negative_two_value == number:
                # Return if we find a match.
                return binary_representation

        # Increase the sequence length
        maximum_sequence_length += 1

        if maximum_sequence_length > 20:
            # Avoid infinite loops for very large or impossible to convert numbers
            return "Cannot represent the number in base -2"

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Begin by repeatedly dividing the number by -2.
Observe the remainder after each division. If the remainder is negative, adjust it by adding 2 to the remainder and adding 1 to the quotient.
Record the adjusted remainders (which will always be either 0 or 1). These become the digits of the base -2 representation.
Repeat the division and remainder adjustment until the quotient becomes 0.
The digits written down are read in reverse order to obtain the base -2 representation.
In the special case that the original number is 0, the result is simply 0.

Code Implementation

def convert_to_base_negative_two(number):
    if number == 0:
        return "0"

    base_negative_two_representation = ""

    while number != 0:
        remainder = number % -2

        # Adjust remainder and quotient to ensure remainder is 0 or 1
        if remainder < 0:
            remainder += 2
            number = number // -2 + 1

        else:
            number = number // -2

        base_negative_two_representation = str(remainder) + base_negative_two_representation

    return base_negative_two_representation

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input n is 0	The base -2 representation of 0 is '0', so return '0'.
Small positive integer (e.g., 1, 2, 3)	The algorithm should correctly convert small positive integers to their base -2 representation, like 1 is '1'.
Small negative integer (e.g., -1, -2, -3)	The algorithm should correctly convert small negative integers to their base -2 representation, like -1 is '11'.
Large positive integer (e.g., 100, 1000)	The algorithm should be able to handle large positive integer inputs without integer overflow or performance issues.
Large negative integer (e.g., -100, -1000)	The algorithm should be able to handle large negative integer inputs correctly without integer overflow or infinite loop.
Integer near the maximum possible value (2^31 - 1)	Ensure the algorithm handles integers near the maximum value without integer overflow issues during computations.
Integer near the minimum possible value (-2^31)	Ensure the algorithm handles integers near the minimum value without integer overflow issues during computations.
Alternating positive and negative results of n % -2	The algorithm must correctly handle the alternating nature of positive and negative remainders when calculating modulo -2, using n % -2 and n = n // -2 (or equivalent) correctly.

Input n is 0

How to Handle:

The base -2 representation of 0 is '0', so return '0'.

Small positive integer (e.g., 1, 2, 3)

How to Handle:

The algorithm should correctly convert small positive integers to their base -2 representation, like 1 is '1'.

Small negative integer (e.g., -1, -2, -3)

How to Handle:

The algorithm should correctly convert small negative integers to their base -2 representation, like -1 is '11'.

Large positive integer (e.g., 100, 1000)

How to Handle:

The algorithm should be able to handle large positive integer inputs without integer overflow or performance issues.

Large negative integer (e.g., -100, -1000)

How to Handle:

The algorithm should be able to handle large negative integer inputs correctly without integer overflow or infinite loop.

Integer near the maximum possible value (2^31 - 1)

How to Handle:

Ensure the algorithm handles integers near the maximum value without integer overflow issues during computations.

Integer near the minimum possible value (-2^31)

How to Handle:

Ensure the algorithm handles integers near the minimum value without integer overflow issues during computations.

Alternating positive and negative results of n % -2

How to Handle:

The algorithm must correctly handle the alternating nature of positive and negative remainders when calculating modulo -2, using n % -2 and n = n // -2 (or equivalent) correctly.