How do you solve Convert to Base -2 on LeetCode?

Convert to Base -2 (LeetCode #1017) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(n) space complexity. Key insight: Understanding how negative bases work is crucial.

What is the brute force solution for Convert to Base -2?

The brute force approach for Convert to Base -2 is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves repeatedly dividing the number by -2 and keeping track of the remainders. This is similar to how we convert numbers to other bases, but we need to adjust for the negative base. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Convert to Base -2?

Convert to Base -2 uses the following concepts: Math. The recommended patterns to study are: Mathematical Operations, String Manipulation.

What are the interview tips for Convert to Base -2?

Explain your thought process clearly as you write code. Test edge cases like 0 and small numbers to ensure correctness. Be prepared to discuss the mathematical properties of negative bases.

What are common mistakes when solving Convert to Base -2?

Not adjusting the quotient when the remainder is negative. Forgetting to handle the case when n is 0.

#1017

Convert to Base -2

Medium

MathMathematical OperationsString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(log n)Space O(n)

The optimal solution leverages the properties of negative base conversion, efficiently calculating the binary representation by adjusting the quotient and remainder in a single loop without unnecessary operations.

⚙️

Algorithm

6 steps

1Step 1: Initialize an empty string to store the result.
2Step 2: While n is not zero, calculate the remainder and quotient using divmod with -2.
3Step 3: If the remainder is negative, adjust it by adding 2 and increment the quotient.
4Step 4: Prepend the adjusted remainder to the result.
5Step 5: Update n to be the quotient and repeat until n is zero.
6Step 6: Return the result, ensuring no leading zeros unless the result is '0'.

solution.py11 lines

1def baseNeg2(n):
2    if n == 0:
3        return '0'
4    result = ''
5    while n != 0:
6        n, remainder = divmod(n, -2)
7        if remainder < 0:
8            remainder += 2
9            n += 1
10        result = str(remainder) + result
11    return result

ℹ

Complexity note: The time complexity is O(log n) because the number of digits in the base -2 representation grows logarithmically with the value of n. The space complexity is O(n) due to the storage of the result string.

1Understanding how negative bases work is crucial.
2Adjusting remainders correctly is key to getting the right representation.

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