How do you solve Smallest Good Base on LeetCode?

Smallest Good Base (LeetCode #483) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log(n)) time complexity and O(1) space complexity. Key insight: Understanding the geometric series helps in deriving the good base efficiently.

What is the brute force solution for Smallest Good Base?

The brute force approach for Smallest Good Base is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible base starting from 2 up to n-1 to see if the representation of n in that base consists entirely of 1's. This is straightforward but inefficient for large n. The optimal Optimal Solution approach improves this to O(log(n)).

What algorithm or data structure is used to solve Smallest Good Base?

Smallest Good Base uses the following concepts: Math, Binary Search. The recommended patterns to study are: Mathematical properties, Geometric series.

What are the interview tips for Smallest Good Base?

Always start with a brute force solution to understand the problem better. Think about mathematical properties that could simplify your solution. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Smallest Good Base?

Not considering the mathematical properties of numbers, leading to unnecessary brute force checks. Failing to handle large numbers correctly, especially in languages with strict type limits.

#483

Smallest Good Base

Hard

MathBinary SearchMathematical propertiesGeometric series

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(log(n))Space O(1)

The optimal solution leverages mathematical properties of numbers. Instead of checking every base, we can derive the maximum possible base by considering the logarithmic relationship between n and k, significantly reducing the search space.

⚙️

Algorithm

5 steps

1Step 1: Convert n to an integer.
2Step 2: Calculate the maximum possible value for m (the number of digits in base k) using log2(n).
3Step 3: For each m from max_m down to 1, calculate the potential base k using the formula k = floor(n^(1/m)).
4Step 4: Check if the sum of the geometric series (1 + k + k^2 + ... + k^(m-1)) equals n.
5Step 5: If it does, return k as the smallest good base.

solution.py12 lines

1# Full working Python code
2import math
3
4def smallestGoodBase(n: str) -> str:
5    num = int(n)
6    max_m = int(math.log(num, 2))
7    for m in range(max_m, 1, -1):
8        k = int(num ** (1 / m))
9        total = (k ** (m) - 1) // (k - 1)
10        if total == num:
11            return str(k)
12    return str(num - 1)

ℹ

Complexity note: The complexity is O(log(n)) because we reduce the number of checks significantly by leveraging the properties of logarithms and geometric series.

1Understanding the geometric series helps in deriving the good base efficiently.
2Logarithmic properties allow us to limit the number of bases we need to check.

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